Table of contents These are my solutions to the problems in the book Maths: A Student's Survival Guide .
Intro
Types of mathematical symbols (p. 101 ):
Symbols for numbers, quantities, variables, objects: 2 2 2 a a a x 2 x^2 x 2 \sqrt{} log \log log x 1 x_1 x 1 x 2 x_2 x 2
Symbols of operation: + − ∗ ÷ ∑ + - * \div \sum + − ∗ ÷ ∑
Symbols of relation: = ≠ a b = \neq \frac{a}{b} = = b a
a , b , c , . . . a,b,c,... a , b , c , . . .
x , y , z , . . . x,y,z,... x , y , z , . . .
Identity:
x + 0 = x x − 0 = x x ∗ 1 = x x / 1 = x
x + 0 = x \\
x - 0 = x \\
x * 1 = x \\
x / 1 = x
x + 0 = x x − 0 = x x ∗ 1 = x x / 1 = x
Terminology
Remember: read through these every now and then to remember the terminology.
1.A.3
2 x − ( x − 2 y ) + 5 y = 2 x − x + 2 y + 5 y = x + 7 y
2x-(x-2y)+5y = \\
2x-x+2y+5y = \\
x+7y
2 x − ( x − 2 y ) + 5 y = 2 x − x + 2 y + 5 y = x + 7 y
4 ( 3 a − 2 b ) − 6 ( 2 a − b ) = 12 a − 8 b − 12 a + 6 b = − 2 b
4(3a-2b) - 6(2a-b) = \\
12a-8b-12a+6b = \\
-2b
4 ( 3 a − 2 b ) − 6 ( 2 a − b ) = 1 2 a − 8 b − 1 2 a + 6 b = − 2 b
6 a − 2 ( 3 a − 5 b ) − ( a + 4 b ) = 6 a − 6 a + 10 b − a − 4 b = 6 b − a
6a - 2(3a-5b) - (a+4b) = \\
6a-6a+10b-a-4b = \\
6b-a
6 a − 2 ( 3 a − 5 b ) − ( a + 4 b ) = 6 a − 6 a + 1 0 b − a − 4 b = 6 b − a
2 x y ( 3 x − 4 y ) − 5 x y ( 2 x − y ) = 6 x 2 y − 8 x y 2 − 10 x 2 y + 5 x y 2 = − 4 x 2 y − 3 x y 2
2xy(3x-4y) - 5xy(2x-y) = \\
6x^2y - 8xy^2 - 10x^2y + 5xy^2 = \\
-4x^2y - 3xy^2
2 x y ( 3 x − 4 y ) − 5 x y ( 2 x − y ) = 6 x 2 y − 8 x y 2 − 1 0 x 2 y + 5 x y 2 = − 4 x 2 y − 3 x y 2
2 a 2 ( 3 a − 2 a b ) − 5 a b ( 2 a 2 − 4 a b ) = 6 a 3 − 4 a 3 b − 10 a 3 b + 20 a 2 b 2 = 6 a 3 − 14 a 3 b + 20 a 2 b 2
2a^2(3a-2ab) - 5ab(2a^2 - 4ab) = \\
6a^3 - 4a^3b - 10a^3b + 20a^2b^2 = \\
6a^3 - 14a^3b + 20a^2b^2
2 a 2 ( 3 a − 2 a b ) − 5 a b ( 2 a 2 − 4 a b ) = 6 a 3 − 4 a 3 b − 1 0 a 3 b + 2 0 a 2 b 2 = 6 a 3 − 1 4 a 3 b + 2 0 a 2 b 2
− 3 p − ( p + q ) + 2 q ( p − 3 ) = − 3 p − p − q + 2 p q − 6 q = − 4 p − 7 q + 2 p q
-3p-(p+q)+2q(p-3) = \\
-3p-p-q+2pq-6q = \\
-4p-7q+2pq
− 3 p − ( p + q ) + 2 q ( p − 3 ) = − 3 p − p − q + 2 p q − 6 q = − 4 p − 7 q + 2 p q
Factoring
x y + x z = x ( y + z )
\boxed{xy + xz = x(y + z)}
x y + x z = x ( y + z )
x y + x ≠ x ( y + 0 ) x y + x = x ( y + 1 )
xy + x \neq x(y + 0) \\
xy + x = x(y + 1)
x y + x = x ( y + 0 ) x y + x = x ( y + 1 )
1.A.4
5 a + 10 b = 5 ( a + 2 b ) 3 a 2 + 2 a b = a ( 3 a + 2 b ) 3 a 2 − 6 a b = 3 a ( a − 2 b ) 5 x y + 8 x z = x ( 5 y + 8 z ) 5 x y − 10 x z = 5 x ( y − 2 z ) a 2 b + 3 a b 2 = a b ( a + 3 b ) 4 p q 2 − 6 p 2 q = 2 p q ( 2 q − 3 p ) 3 x 2 y 3 + 5 x 3 y 2 = x 2 y 2 ( 3 y + 5 x ) 4 p 2 q + 2 p q 2 − 6 p 2 q 2 = 2 p q ( 2 p + q − 3 p q ) 2 a 2 b 3 + 3 a 3 b 2 − 6 a 2 b 2 = a 2 b 2 ( 2 b + 3 a − 6 )
5a+10b = 5(a+2b) \\
3a^2 + 2ab = a(3a + 2b) \\
3a^2 - 6ab = 3a(a - 2b) \\
5xy + 8xz = x(5y + 8z) \\
5xy - 10xz = 5x(y - 2z) \\
a^2b + 3ab^2 = ab(a + 3b) \\
4pq^2 - 6p^2q = 2pq(2q - 3p) \\
3x^2y^3 + 5x^3y^2 = x^2y^2(3y + 5x) \\
4p^2q + 2pq^2 - 6p^2q^2 = 2pq(2p + q - 3pq) \\
2a^2b^3 + 3a^3b^2 - 6a^2b^2 = a^2b^2(2b + 3a - 6)
5 a + 1 0 b = 5 ( a + 2 b ) 3 a 2 + 2 a b = a ( 3 a + 2 b ) 3 a 2 − 6 a b = 3 a ( a − 2 b ) 5 x y + 8 x z = x ( 5 y + 8 z ) 5 x y − 1 0 x z = 5 x ( y − 2 z ) a 2 b + 3 a b 2 = a b ( a + 3 b ) 4 p q 2 − 6 p 2 q = 2 p q ( 2 q − 3 p ) 3 x 2 y 3 + 5 x 3 y 2 = x 2 y 2 ( 3 y + 5 x ) 4 p 2 q + 2 p q 2 − 6 p 2 q 2 = 2 p q ( 2 p + q − 3 p q ) 2 a 2 b 3 + 3 a 3 b 2 − 6 a 2 b 2 = a 2 b 2 ( 2 b + 3 a − 6 )
1.B.(a)
( 2 x + 3 y ) ( x + 5 y ) = 2 x 2 + 10 x y + 3 x y + 15 y 2 = 2 x 2 + 13 y + 15 y 2
(2x + 3y)(x + 5y) = \\
2x^2 + 10xy + 3xy + 15y^2 = \\
2x^2 + 13y + 15y^2
( 2 x + 3 y ) ( x + 5 y ) = 2 x 2 + 1 0 x y + 3 x y + 1 5 y 2 = 2 x 2 + 1 3 y + 1 5 y 2
( 3 a − 5 b ) ( 2 a − b ) = 6 a 2 − 3 a b − 10 a b + 5 b 2 = 6 a 2 − 13 a b + 5 b 2
(3a - 5b)(2a - b) = \\
6a^2 - 3ab - 10ab + 5b^2 = \\
6a^2 - 13ab + 5b^2
( 3 a − 5 b ) ( 2 a − b ) = 6 a 2 − 3 a b − 1 0 a b + 5 b 2 = 6 a 2 − 1 3 a b + 5 b 2
( 3 x + 2 ) 2 = ( 3 x + 2 ) ( 3 x + 2 ) = 9 x 2 + 6 x + 6 x + 2 2 = 9 x 2 + 12 x + 4
(3x + 2)^2 = \\
(3x + 2)(3x + 2) = \\
9x^2 + 6x + 6x + 2^2 = \\
9x^2 + 12x + 4
( 3 x + 2 ) 2 = ( 3 x + 2 ) ( 3 x + 2 ) = 9 x 2 + 6 x + 6 x + 2 2 = 9 x 2 + 1 2 x + 4
( 2 y − 5 ) 2 = ( 2 y − 5 ) ( 2 y − 5 ) = 4 y 2 − 10 y − 10 y + 25 = 4 y 2 − 20 y + 25
(2y - 5)^2 = \\
(2y - 5)(2y - 5) = \\
4y^2 - 10y - 10y + 25 = \\
4y^2 - 20y + 25
( 2 y − 5 ) 2 = ( 2 y − 5 ) ( 2 y − 5 ) = 4 y 2 − 1 0 y − 1 0 y + 2 5 = 4 y 2 − 2 0 y + 2 5
( 2 p 2 + 3 p q ) ( q 2 − 2 p q ) = 2 p 2 q 2 − 4 p 3 q + 3 p q 3 − 6 p 2 q 2 = − 4 p 2 q 2 − 4 p 3 q + 3 p q 3
(2p^2 + 3pq)(q^2 - 2pq) = \\
2p^2q^2 - 4p^3q + 3pq^3 - 6p^2q^2 = \\
-4p^2q^2 - 4p^3q + 3pq^3
( 2 p 2 + 3 p q ) ( q 2 − 2 p q ) = 2 p 2 q 2 − 4 p 3 q + 3 p q 3 − 6 p 2 q 2 = − 4 p 2 q 2 − 4 p 3 q + 3 p q 3
Factoring by grouping
x 2 + 9 x + 14 = x 2 + 2 x + 7 x + 14 = ( x 2 + 2 x ) + ( 7 x + 14 ) = x ( x + 2 ) + 7 ( x + 2 )
x^2 + 9x + 14 = \\
x^2 + 2x + 7x + 14 = \\
(x^2 + 2x) + (7x + 14) = \\
x(x + 2) + 7(x + 2)
x 2 + 9 x + 1 4 = x 2 + 2 x + 7 x + 1 4 = ( x 2 + 2 x ) + ( 7 x + 1 4 ) = x ( x + 2 ) + 7 ( x + 2 )
Now ( x + 2 ) (x + 2) ( x + 2 ) summands , and can be factored further:
( x + 2 ) ( x + 7 )
(x + 2)(x + 7)
( x + 2 ) ( x + 7 )
This is easier to see if we say that y = ( x + 2 ) y = (x + 2) y = ( x + 2 ) ( x + 2 ) (x + 2) ( x + 2 ) y y y
x ( x + 2 ) + 7 ( x + 2 ) = x y + 7 y = y ( x + 7 ) = ( x + 2 ) ( x + 7 )
x(x + 2) + 7(x + 2) = \\
xy + 7y = \\
y(x + 7) = \\
(x + 2)(x + 7)
x ( x + 2 ) + 7 ( x + 2 ) = x y + 7 y = y ( x + 7 ) = ( x + 2 ) ( x + 7 )
The process
The function a x 2 + b x + c \boxed{ax^2 + bx + c} a x 2 + b x + c p and q, so that p q = a c \boxed{pq = ac} p q = a c p + q = b \boxed{p+q = b} p + q = b c is negative, find p − q = b \boxed{p-q = b} p − q = b
Above these values were 2 and 7, which is why we split 9 x 9x 9 x 2 x + 7 x 2x + 7x 2 x + 7 x
Remember: when factoring trinomials, rather than memorising shortcuts (e.g. the "ac" method), make sure you can always factor by grouping. Factoring by grouping is a reliable way especially when a > 1 a > 1 a > 1
Find which p q pq p q a c ac a c
Split b b b p p p q q q
Then group. Then factor further.
6 x 2 − 7 x + 2 6 x 2 − ( 3 x + 4 x ) + 2 Split 6 x 2 − 3 x − 4 x + 2 ( 6 x 2 − 3 x ) + ( − 4 x + 2 ) Group 3 x ( 2 x − 1 ) + 2 ( − 2 x + 1 ) Factor 3 x ( 2 x − 1 ) + 2 ( − 1 ) ( 2 x − 1 ) Factor 3 x ( 2 x − 1 ) − 2 ( 2 x − 1 ) ( 2 x − 1 ) ( 3 x − 2 ) Factor
6x^2 - 7x + 2 \\
6x^2 - (3x + 4x) + 2 \qquad \text{Split} \\
6x^2 - 3x - 4x + 2 \\
(6x^2 - 3x) + (-4x + 2) \qquad \text{Group} \\
3x(2x - 1) + 2(-2x + 1) \qquad \text{Factor} \\
3x(2x - 1) + 2(-1)(2x - 1) \qquad \text{Factor} \\
3x(2x - 1) - 2(2x - 1) \\
(2x - 1)(3x - 2) \qquad \text{Factor}
6 x 2 − 7 x + 2 6 x 2 − ( 3 x + 4 x ) + 2 Split 6 x 2 − 3 x − 4 x + 2 ( 6 x 2 − 3 x ) + ( − 4 x + 2 ) Group 3 x ( 2 x − 1 ) + 2 ( − 2 x + 1 ) Factor 3 x ( 2 x − 1 ) + 2 ( − 1 ) ( 2 x − 1 ) Factor 3 x ( 2 x − 1 ) − 2 ( 2 x − 1 ) ( 2 x − 1 ) ( 3 x − 2 ) Factor
y 2 + 8 y + 12 = ( y + 2 ) ( y + 6 ) x 2 + 8 x + 16 = ( x + 4 ) ( x + 4 ) = ( x + 4 ) 2 p 2 + 13 p + 22 = ( p + 2 ) ( p + 11 )
y^2 + 8y + 12 = (y + 2)(y + 6) \\
x^2 + 8x + 16 = (x + 4)(x + 4) = (x + 4)^2 \\
p^2 + 13p + 22 = (p + 2)(p + 11)
y 2 + 8 y + 1 2 = ( y + 2 ) ( y + 6 ) x 2 + 8 x + 1 6 = ( x + 4 ) ( x + 4 ) = ( x + 4 ) 2 p 2 + 1 3 p + 2 2 = ( p + 2 ) ( p + 1 1 )
p q = a c p + q = b
pq = ac \\
p+q = b
p q = a c p + q = b
Therefore:
a x 2 + b x + c = a x 2 + ( p + q ) x + c = Substituted b with (p + q) a x 2 + p x + q x + c = ( a x 2 + p x ) + ( q x + c ) = x ( a x + p ) + ( q x + c ) = Factored out x, i.e. divided by x a x ( x + p / a ) + ( q x + c ) = Factored out a a x ( x + p / a ) + q ( x + c / q ) Factored out q
ax^2 + bx + c = \\
ax^2 + (p + q)x + c = \qquad \text{Substituted b with (p + q)} \\
ax^2 + px + qx + c = \\
(ax^2 + px) + (qx + c) = \\
x(ax + p) + (qx + c) = \qquad \text{Factored out x, i.e. divided by x} \\
ax(x + p/a) + (qx + c) = \qquad \text{Factored out a} \\
ax(x + p/a) + q(x + c/q) \qquad \text{Factored out q}
a x 2 + b x + c = a x 2 + ( p + q ) x + c = Substituted b with (p + q) a x 2 + p x + q x + c = ( a x 2 + p x ) + ( q x + c ) = x ( a x + p ) + ( q x + c ) = Factored out x, i.e. divided by x a x ( x + p / a ) + ( q x + c ) = Factored out a a x ( x + p / a ) + q ( x + c / q ) Factored out q
But this can be factored further.
p q = a c pq = ac p q = a c
p = a c / q Divide both sides by q p / a = c / q Divide both sides by a
p = ac/q \qquad \text{Divide both sides by q} \\
p/a = c/q \qquad \text{Divide both sides by a}
p = a c / q Divide both sides by q p / a = c / q Divide both sides by a
Therefore
( x + p / a ) = ( x + c / q ) (x + p/a) = (x + c/q) ( x + p / a ) = ( x + c / q )
which allows us to factor further:
a x ( x + p / a ) + q ( x + c / q ) = ( a x + q ) ( x + p / a )
ax(x + p/a) + q(x + c/q) = \\
\boxed{(ax + q)(x + p/a)}
a x ( x + p / a ) + q ( x + c / q ) = ( a x + q ) ( x + p / a )
Example:
2 x 2 + 7 x + 3 a = 2 c = 3 b = 7 p + q = 7 p + q = b p = 6 q = 1 a c = 2 ∗ 3 = 6 p q = 6 ∗ 1 = 6 p q = a c
2x^2 + 7x + 3 \\
a = 2 \\
c = 3 \\
b = 7 \qquad p + q = 7 \qquad p+q = b \\
p = 6 \\
q = 1 \\
ac = 2 * 3 = 6 \qquad pq = 6 * 1 = 6 \qquad pq=ac \\
2 x 2 + 7 x + 3 a = 2 c = 3 b = 7 p + q = 7 p + q = b p = 6 q = 1 a c = 2 ∗ 3 = 6 p q = 6 ∗ 1 = 6 p q = a c
2 x 2 + 7 x + 3 = ( a x + q ) ( x + p / a ) = ( 2 x + 1 ) ( x + 6 / 2 ) = ( 2 x + 1 ) ( x + 3 )
2x^2 + 7x + 3 = \\
\boxed{(ax + q)(x + p/a)} = \\
(2x + 1)(x + 6/2) = \\
(2x + 1)(x + 3)
2 x 2 + 7 x + 3 = ( a x + q ) ( x + p / a ) = ( 2 x + 1 ) ( x + 6 / 2 ) = ( 2 x + 1 ) ( x + 3 )
Therefore, to use the "ac" method, find p and q and then divide one of them with a, and place a to the opposite group:
( a x + q ) ⏟ Place a here ( x + p / a ) ⏟ Divide by a here
\boxed{\underbrace{(ax + q)}_{\text{Place a here }}\space\underbrace{(x + p/a)}_{\text{Divide by a here}}}
Place a here ( a x + q ) Divide by a here ( x + p / a )
1.B.(a) cont.
3 a 2 + 16 a + 5 = ( 3 a + 1 ) ( a + 5 )
3a^2 + 16a + 5 = (3a + 1)(a + 5)
3 a 2 + 1 6 a + 5 = ( 3 a + 1 ) ( a + 5 )
3 b 2 + 10 b + 7 = p and q are 3 or 7 here 3 b 2 + 3 b + 7 b + 7 = ( 3 b + q ) ( b + p / a ) = Is p 3 or 7? Which of 3 and 7 can be divided by a? ( 3 b + 7 ) ( b + 1 ) Only 3 can be divided by a. Therefore q is 7.
3b^2 + 10b + 7 = \qquad \text{p and q are 3 or 7 here} \\
3b^2 + 3b + 7b + 7 = \\
(3b + q)(b + p/a) = \qquad \text{Is p 3 or 7? Which of 3 and 7 can be divided by a?} \\
(3b + 7)(b + 1) \qquad \text{Only 3 can be divided by a. Therefore q is 7.}
3 b 2 + 1 0 b + 7 = p and q are 3 or 7 here 3 b 2 + 3 b + 7 b + 7 = ( 3 b + q ) ( b + p / a ) = Is p 3 or 7? Which of 3 and 7 can be divided by a? ( 3 b + 7 ) ( b + 1 ) Only 3 can be divided by a. Therefore q is 7.
5 x 2 + 8 x + 3 = ( 5 x + 3 ) ( x + 1 )
5x^2 + 8x + 3 = (5x + 3)(x + 1)
5 x 2 + 8 x + 3 = ( 5 x + 3 ) ( x + 1 )
Next some functions with the form a x 2 + b x − c \boxed{ax^2 + bx - c} a x 2 + b x − c c is negative, so these will factorise into an function like ( x + y ) ( x − z ) \boxed{(x + y)(x - z)} ( x + y ) ( x − z ) p and q such that p q = a c \boxed{pq = ac} p q = a c p − q = b \boxed{p - q = b} p − q = b c is negative.
x 2 + x − 2 = ( x + 1 ) ( x − 2 ) p and q are 1 or 2
x^2 + x - 2 = (x + 1)(x - 2) \qquad \text{p and q are 1 or 2}
x 2 + x − 2 = ( x + 1 ) ( x − 2 ) p and q are 1 or 2
2 x 2 + 10 x − 3 x − 15 = ( 2 x 2 + 10 x ) + ( − 3 x − 15 )
2x^2 + 10x - 3x - 15 = \\
(2x^2 + 10x) + (-3x - 15)
2 x 2 + 1 0 x − 3 x − 1 5 = ( 2 x 2 + 1 0 x ) + ( − 3 x − 1 5 )
2 a 2 + a − 15 = 2 a 2 + ( 6 − 5 ) a − 15 p and q are 6 and 5 = 2 a 2 + 6 a − 5 a − 15 = ( 2 a 2 + 6 a ) + ( − 5 a − 15 ) = 2 a ( a + 3 ) + 5 ( − a − 3 ) Can we make these positive? = 2 a ( a + 3 ) − 5 ( a + 3 ) = ( 2 a − 5 ) ( a + 3 )
\begin{aligned}
2a^2 + a - 15 &= 2a^2 + (6 - 5)a - 15 \quad \text{p and q are 6 and 5} \\
&= 2a^2 + 6a - 5a - 15 \\
&= (2a^2 + 6a) + (-5a - 15) \\
&= 2a(a + 3) + 5(-a - 3) \quad \text{Can we make these positive?} \\
&= 2a(a + 3) - 5(a + 3) \\
&= (2a - 5)(a + 3)
\end{aligned}
2 a 2 + a − 1 5 = 2 a 2 + ( 6 − 5 ) a − 1 5 p and q are 6 and 5 = 2 a 2 + 6 a − 5 a − 1 5 = ( 2 a 2 + 6 a ) + ( − 5 a − 1 5 ) = 2 a ( a + 3 ) + 5 ( − a − 3 ) Can we make these positive? = 2 a ( a + 3 ) − 5 ( a + 3 ) = ( 2 a − 5 ) ( a + 3 )
I want to see if turning + 5 ( − a − 3 ) +5(-a - 3) + 5 ( − a − 3 ) − 5 ( a + 3 ) -5(a + 3) − 5 ( a + 3 )
a = 1 5 ( − a − 3 ) = − 5 − 15 = − 20 − 5 ( a + 3 ) = − 5 − 15 = − 20
a = 1 \\
5(-a - 3) = -5 - 15 = -20 \\
-5(a + 3) = -5 - 15 = -20
a = 1 5 ( − a − 3 ) = − 5 − 1 5 = − 2 0 − 5 ( a + 3 ) = − 5 − 1 5 = − 2 0
Looks like it. (This is the anticommutative property of subtraction.)
2 x 2 + 5 x − 12 = ( 2 x − 3 ) ( x + 4 ) p 2 − q 2 = ( p + q ) ( p − q ) Difference of two squares
2x^2 + 5x - 12 = (2x - 3)(x + 4) \\
p^2 - q^2 = (p + q)(p - q) \qquad \boxed{\text{Difference of two squares}}
2 x 2 + 5 x − 1 2 = ( 2 x − 3 ) ( x + 4 ) p 2 − q 2 = ( p + q ) ( p − q ) Difference of two squares
Next some functions with the form a x 2 − b x + c \boxed{ax^2 - bx + c} a x 2 − b x + c c is positive, so these will factorise into an function like ( x − y ) ( x − z ) \boxed{(x - y)(x - z)} ( x − y ) ( x − z ) p + q = b \boxed{p+q = b} p + q = b c is positive.
6 y 2 − 19 y + 10 = Are p and q 4 and 15? 4 / 6 ∉ N Cannot divide 4 by a 15 / 6 ∉ N Cannot divide 15 by a. Wat do?
6y^2 - 19y + 10 = \qquad \text{Are p and q 4 and 15?} \\
4/6 \notin \natnums \qquad \text{Cannot divide 4 by a} \\
15/6 \notin \natnums \qquad \text{Cannot divide 15 by a. Wat do?}
6 y 2 − 1 9 y + 1 0 = Are p and q 4 and 15? 4 / 6 ∈ / N Cannot divide 4 by a 1 5 / 6 ∈ / N Cannot divide 15 by a. Wat do?
Well, I can kind of hack my way through by trying different values for a, b, c and d in
( a y − b ) ( c y − d ) (ay - b)(cy - d) ( a y − b ) ( c y − d )
and finally end up with:
6 y 2 − 19 y + 10 = ( 3 y − 2 ) ( 2 y − 5 ) Correct
6y^2 - 19y + 10 = (3y - 2)(2y - 5) \qquad \text{Correct}
6 y 2 − 1 9 y + 1 0 = ( 3 y − 2 ) ( 2 y − 5 ) Correct
How can I find this easier?
a c = p q = 60 b = p + q = 15 + 4
ac = pq = 60 \\
b = p + q = 15 + 4
a c = p q = 6 0 b = p + q = 1 5 + 4
6 y 2 − 19 y + 10 = 6 y 2 − ( 15 + 4 ) y + 10 = 6 y 2 − ( 15 y + 4 y ) + 10 = 6 y 2 − 15 y − 4 y + 10 = ( 6 y 2 − 15 y ) − ( 4 y + 10 ) = Be careful when grouping [note 1] ( 6 y 2 − 15 y ) + ( − 4 y + 10 ) = ( 6 y 2 − 15 y ) − ( 4 y − 10 ) = Can we do this? [note 2] 3 y ( 2 y − 5 ) − 2 ( 2 y − 5 ) = 3 y ( 2 y − 5 ) + − 2 ( 2 y − 5 ) = Found factors 3y and -2 3 y ( 2 y − 5 ) − 2 ( 2 y − 5 ) = ( 3 y − 2 ) ( 2 y − 5 )
6y^2 - 19y + 10 = \\
6y^2 - (15 + 4)y + 10 = \\
6y^2 - (15y + 4y) + 10 = \\
6y^2 - 15y - 4y + 10 = \\
\cancel{(6y^2 - 15y) - (4y + 10)} = \quad \text{Be careful when grouping}^{\text{[note 1]}} \\
(6y^2 - 15y) + (-4y + 10) = \\
\cancel{(6y^2 - 15y) - (4y - 10)} = \quad \text{Can we do this?}^{\text{[note 2]}} \\
\cancel{3y(2y - 5) - 2(2y - 5)} = \\
3y(2y - 5) + -2(2y - 5) = \quad \text{Found factors 3y and -2} \\
3y(2y - 5) - 2(2y - 5) = \\
(3y - 2)(2y - 5)
6 y 2 − 1 9 y + 1 0 = 6 y 2 − ( 1 5 + 4 ) y + 1 0 = 6 y 2 − ( 1 5 y + 4 y ) + 1 0 = 6 y 2 − 1 5 y − 4 y + 1 0 = ( 6 y 2 − 1 5 y ) − ( 4 y + 1 0 ) = Be careful when grouping [note 1] ( 6 y 2 − 1 5 y ) + ( − 4 y + 1 0 ) = ( 6 y 2 − 1 5 y ) − ( 4 y − 1 0 ) = Can we do this? [note 2] 3 y ( 2 y − 5 ) − 2 ( 2 y − 5 ) = 3 y ( 2 y − 5 ) + − 2 ( 2 y − 5 ) = Found factors 3y and -2 3 y ( 2 y − 5 ) − 2 ( 2 y − 5 ) = ( 3 y − 2 ) ( 2 y − 5 )
Remember : once you have found p and q, you can always "go the long way" and factor by grouping , rather than try to use the "ac" method (which would be a shortcut).
[Note 1] Remember : be careful when grouping operands, so that you preserve their signs:
10 − 2 + 3 = 11 10 + − 2 + 3 = 11 10 − ( 2 + 3 ) ≠ 11 Remember: subtraction isn’t associative 10 + ( − 2 + 3 ) = 11 But addition is ( 10 + − 2 ) + 3 = 11
10 - 2 + 3 = 11 \\
10 + -2 + 3 = 11 \\
10 - (2 + 3) \neq 11 \quad \boxed{\text{Remember: subtraction isn't associative}} \\
10 + (-2 + 3) = 11 \quad \text{But addition is} \\
(10 + -2) + 3 = 11
1 0 − 2 + 3 = 1 1 1 0 + − 2 + 3 = 1 1 1 0 − ( 2 + 3 ) = 1 1 Remember: subtraction isn’t associative 1 0 + ( − 2 + 3 ) = 1 1 But addition is ( 1 0 + − 2 ) + 3 = 1 1
A − B A - B A − B A + ( − B ) A + (-B) A + ( − B ) A − B A - B A − B A A A − B -B − B
[Note 2] Let's try:
a + ( − b x + c ) = a − ( b x − c ) ? a = 2 b = 3 c = 7 x = 1 2 + ( − 3 + 7 ) = 6 2 − ( 3 − 7 ) = 6
a + (-bx + c) = a - (bx - c)? \\
a = 2 \\
b = 3 \\
c = 7 \\
x = 1 \\
2 + (-3 + 7) = 6 \\
2 - (3 - 7) = 6
a + ( − b x + c ) = a − ( b x − c ) ? a = 2 b = 3 c = 7 x = 1 2 + ( − 3 + 7 ) = 6 2 − ( 3 − 7 ) = 6
Looks like we can.
a + ( − b + c ) = a − ( b − c )
a + (-b + c) = a - (b - c)
a + ( − b + c ) = a − ( b − c )
Looks like this is the anticommutative property
( a − b ) = − ( − a + b )
\boxed{(a - b) = -(-a + b)}
( a − b ) = − ( − a + b )
(Trivia : division is only right-distributive , not left-distributive:)
a / ( b + c ) ≠ a / b + a / c ( a + b ) / c = a / c + b / c
a / (b+c) \neq a/b + a/c \\
(a+b) / c = a/c + b/c
a / ( b + c ) = a / b + a / c ( a + b ) / c = a / c + b / c
4 x 2 − 81 y 2 = ( 2 x + 9 y ) ( 2 x − 9 y ) Difference of two squares
4x^2 - 81y^2 = (2x + 9y)(2x - 9y) \quad \text{Difference of two squares} \\
4 x 2 − 8 1 y 2 = ( 2 x + 9 y ) ( 2 x − 9 y ) Difference of two squares
6 x 2 − 19 x + 10 = ( 6 x 2 − 4 x ) + ( − 15 x + 10 ) = 2 x ( 3 x − 2 ) + 5 ( − 3 x + 2 ) = 2 x ( 3 x − 2 ) − 5 ( 3 x − 2 ) = ( 2 x − 5 ) ( 3 x − 2 )
\begin{aligned}
6x^2 - 19x + 10 &= (6x^2 - 4x) + (-15x + 10) \\
&= 2x(3x - 2) + 5(-3x + 2) \\
&= 2x(3x - 2) - 5(3x - 2) \\
&= (2x - 5)(3x - 2)
\end{aligned}
6 x 2 − 1 9 x + 1 0 = ( 6 x 2 − 4 x ) + ( − 1 5 x + 1 0 ) = 2 x ( 3 x − 2 ) + 5 ( − 3 x + 2 ) = 2 x ( 3 x − 2 ) − 5 ( 3 x − 2 ) = ( 2 x − 5 ) ( 3 x − 2 )
4 x 2 − 12 x + 9 = ( 4 x 2 − 6 x ) + ( 6 x + 9 ) = 2 x ( 2 x − 3 ) − 3 ( 2 x − 3 ) = ( 2 x − 3 ) ( 2 x − 3 ) = ( 2 x − 3 ) 2
4x^2 - 12x + 9 = \\
(4x^2 - 6x) + (6x + 9) = \\
2x(2x - 3) - 3(2x - 3) = \\
(2x - 3)(2x - 3) = \\
(2x - 3)^2
4 x 2 − 1 2 x + 9 = ( 4 x 2 − 6 x ) + ( 6 x + 9 ) = 2 x ( 2 x − 3 ) − 3 ( 2 x − 3 ) = ( 2 x − 3 ) ( 2 x − 3 ) = ( 2 x − 3 ) 2
1.B.(b)
The distributive property :
( a + b ) ( c + d ) = ( c ( a + b ) + d ( a + b ) ) = a c + b c + a d + b d
\boxed{(a+b)(c+d) = (c(a+b) + d(a+b)) = ac + bc + ad + bd}
( a + b ) ( c + d ) = ( c ( a + b ) + d ( a + b ) ) = a c + b c + a d + b d
The total area of the rectangle ( a + b ) ( c + d )
\boxed{\text{The total area of the rectangle}} \\
(a+b)(c+d)
The total area of the rectangle ( a + b ) ( c + d )
( a + b ) 2 = ( a + b ) ( a + b ) = a 2 + 2 a b + b 2 The diagram shows that ( a + b ) 2 ≠ a 2 + b 2
(a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2 \\
\text{ } \\
\text{The diagram shows that }(a + b)^2 \neq a^2 + b^2
( a + b ) 2 = ( a + b ) ( a + b ) = a 2 + 2 a b + b 2 The diagram shows that ( a + b ) 2 = a 2 + b 2
Difference of two squares. Remember: make sure you’re able to spot these in the wild! ( a + b ) ( a − b ) = a 2 − a b + a b − b 2 = a 2 − b 2
\boxed{\text{Difference of two squares.}} \\
\text{Remember: make sure you're able to spot these in the wild!} \\
(a+b)(a-b) = \\
a^2 - ab + ab - b^2 = \\
a^2 - b^2
Difference of two squares. Remember: make sure you’re able to spot these in the wild! ( a + b ) ( a − b ) = a 2 − a b + a b − b 2 = a 2 − b 2
See also https://en.wikipedia.org/wiki/Difference_of_two_squares#Geometrical_demonstrations .
Remember: you might sometimes see expressions like x 2 − 3 x^2 - 3 x 2 − 3 ( x + 3 ) ( x − 3 ) (x + \sqrt{3})(x - \sqrt{3}) ( x + 3 ) ( x − 3 )
1.B.1
( x + 2 ) ( x + 3 ) = x 2 + 5 x + 6 ( a + 3 ) ( a − 4 ) = a 2 − a − 12 ( x − 2 ) ( x − 3 ) = x 2 − 5 a + 6 ( p + 3 ) ( 2 p + 1 ) = 2 p 2 + 7 p + 3 ( 3 x − 2 ) ( 3 x + 2 ) = 9 x 2 − 2 2 Difference of two squares ( 2 x − 3 y ) ( x + 2 y ) = 2 x 2 + 4 x y − 3 x y + 6 y 2 = 2 x 2 + x y + 6 y 2 ( 3 a − 2 b ) ( 2 a − 5 b ) = 6 a 2 − 19 b + 10 b 2 ( 3 x + 4 y ) 2 = ( 3 x + 4 y ) ( 3 x + 4 y ) = 9 x 2 + 24 x y + 16 y 2 ( 3 x − 4 y ) 2 = ( 3 x − 4 y ) ( 3 x − 4 y ) = 9 x 2 − 24 x y + 16 y 2 ( 3 x + 4 y ) ( 3 x − 4 y ) = 9 x 2 − 16 y 2 Difference of two squares ( 2 p 2 + 3 p q ) ( 5 p + 3 q ) = 10 p 3 + 6 p 2 q + 15 p 2 q + 9 p q 2 = 10 p 3 + 21 p 2 q + 9 p q 2 ( 2 a b − b 2 ) ( a 2 − 3 a b ) = 2 a 3 b − 6 a 2 b 2 − a 2 b 2 + 3 a b 3 = 2 a 3 b − 7 a 2 b 2 + 3 a b 3 ( a + b ) ( a 2 − a b + b 2 ) = a 3 − a 2 b + a b 2 + a 2 b − a b 2 + b 3 = a 3 + b 3 ( a − b ) ( a 2 + a b + b 2 ) = a 3 + a 2 b + a b 2 − a 2 b − a b 2 − b 3 = a 3 − b 3
(x + 2)(x + 3) = x^2 + 5x + 6 \\
(a + 3)(a - 4) = a^2 - a - 12 \\
(x - 2)(x - 3) = x^2 - 5a + 6 \\
(p + 3)(2p + 1) = 2p^2 + 7p + 3 \\
(3x - 2)(3x + 2) = 9x^2 - 2^2 \quad \text{Difference of two squares} \\
(2x - 3y)(x + 2y) = 2x^2 + 4xy - 3xy + 6y^2 = 2x^2 + xy + 6y^2 \\
(3a - 2b)(2a - 5b) = 6a^2 - 19b + 10b^2 \\
(3x + 4y)^2 = \boxed{(3x + 4y)(3x + 4y) = 9x^2 + 24xy + 16y^2} \\
(3x - 4y)^2 = \boxed{(3x - 4y)(3x - 4y) = 9x^2 - 24xy + 16y^2} \\
\boxed{(3x + 4y)(3x - 4y) = 9x^2 - 16y^2} \quad \text{Difference of two squares} \\
(2p^2 + 3pq)(5p + 3q) = 10p^3 + 6p^2q + 15p^2q + 9pq^2 = 10p^3 + 21p^2q + 9pq^2 \\
(2ab - b^2)(a^2 - 3ab) = 2a^3b - 6a^2b^2 - a^2b^2 + 3ab^3 = 2a^3b - 7a^2b^2 + 3ab^3 \\
\begin{aligned}
(a+b)(a^2 - ab + b^2) &= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 \\
&= a^3 + b^3
\end{aligned} \\
\begin{aligned}
(a-b)(a^2 + ab + b^2) &= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 \\
&= a^3 - b^3
\end{aligned}
( x + 2 ) ( x + 3 ) = x 2 + 5 x + 6 ( a + 3 ) ( a − 4 ) = a 2 − a − 1 2 ( x − 2 ) ( x − 3 ) = x 2 − 5 a + 6 ( p + 3 ) ( 2 p + 1 ) = 2 p 2 + 7 p + 3 ( 3 x − 2 ) ( 3 x + 2 ) = 9 x 2 − 2 2 Difference of two squares ( 2 x − 3 y ) ( x + 2 y ) = 2 x 2 + 4 x y − 3 x y + 6 y 2 = 2 x 2 + x y + 6 y 2 ( 3 a − 2 b ) ( 2 a − 5 b ) = 6 a 2 − 1 9 b + 1 0 b 2 ( 3 x + 4 y ) 2 = ( 3 x + 4 y ) ( 3 x + 4 y ) = 9 x 2 + 2 4 x y + 1 6 y 2 ( 3 x − 4 y ) 2 = ( 3 x − 4 y ) ( 3 x − 4 y ) = 9 x 2 − 2 4 x y + 1 6 y 2 ( 3 x + 4 y ) ( 3 x − 4 y ) = 9 x 2 − 1 6 y 2 Difference of two squares ( 2 p 2 + 3 p q ) ( 5 p + 3 q ) = 1 0 p 3 + 6 p 2 q + 1 5 p 2 q + 9 p q 2 = 1 0 p 3 + 2 1 p 2 q + 9 p q 2 ( 2 a b − b 2 ) ( a 2 − 3 a b ) = 2 a 3 b − 6 a 2 b 2 − a 2 b 2 + 3 a b 3 = 2 a 3 b − 7 a 2 b 2 + 3 a b 3 ( a + b ) ( a 2 − a b + b 2 ) = a 3 − a 2 b + a b 2 + a 2 b − a b 2 + b 3 = a 3 + b 3 ( a − b ) ( a 2 + a b + b 2 ) = a 3 + a 2 b + a b 2 − a 2 b − a b 2 − b 3 = a 3 − b 3
a = 2 a 2 = 4 ( a + 1 ) ( a − 1 ) = 3 ( a + 1 ) ( a − 1 ) = a 2 − 1
a = 2 \\
a^2 = 4 \\
(a + 1)(a - 1) = 3 \\
(a + 1)(a - 1) = a^2 - 1
a = 2 a 2 = 4 ( a + 1 ) ( a − 1 ) = 3 ( a + 1 ) ( a − 1 ) = a 2 − 1
1.B.2
x 2 + 8 x + 7 = ( x + 1 ) ( x + 7 ) p 2 + 6 p + 5 = ( p + 1 ) ( p + 5 ) x 2 + 7 x + 6 = ( x + 1 ) ( x + 6 ) x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) y 2 + 6 y + 9 = ( y + 3 ) ( y + 3 ) = ( y + 3 ) 2 x 2 + 6 x + 8 = ( x + 2 ) ( x + 4 ) a 2 + 7 a + 10 = ( a + 2 ) ( a + 5 ) x 2 + 9 x + 20 = ( x + 4 ) ( x + 5 ) x 2 + 13 x + 36 = ( x + 4 ) ( x + 9 )
x^2 + 8x + 7 = (x + 1)(x + 7) \\
p^2 + 6p + 5 = (p + 1)(p + 5) \\
x^2 + 7x + 6 = (x + 1)(x + 6) \\
x^2 + 5x + 6 = (x + 2)(x + 3) \\
y^2 + 6y + 9 = (y + 3)(y + 3) = (y + 3)^2 \\
x^2 + 6x + 8 = (x + 2)(x + 4) \\
a^2 + 7a + 10 = (a + 2)(a + 5) \\
x^2 + 9x + 20 = (x + 4)(x + 5) \\
x^2 + 13x + 36 = (x + 4)(x + 9)
x 2 + 8 x + 7 = ( x + 1 ) ( x + 7 ) p 2 + 6 p + 5 = ( p + 1 ) ( p + 5 ) x 2 + 7 x + 6 = ( x + 1 ) ( x + 6 ) x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) y 2 + 6 y + 9 = ( y + 3 ) ( y + 3 ) = ( y + 3 ) 2 x 2 + 6 x + 8 = ( x + 2 ) ( x + 4 ) a 2 + 7 a + 1 0 = ( a + 2 ) ( a + 5 ) x 2 + 9 x + 2 0 = ( x + 4 ) ( x + 5 ) x 2 + 1 3 x + 3 6 = ( x + 4 ) ( x + 9 )
1.B.3
3 x 2 + 8 x + 5 = ( 3 x + 5 ) ( x + 1 ) 2 y 2 + 15 y + 7 = ( 2 y + 1 ) ( y + 7 ) 3 a 2 + 11 a + 6 = ( 3 a 2 + 9 a ) + ( 2 a + 6 ) = 3 a ( a + 3 ) + 2 ( a + 3 ) = ( 3 a + 2 ) ( a + 3 ) 3 x 2 + 19 x + 6 = ( 3 x + 1 ) ( x + 6 ) 5 p 2 + 23 p + 12 = ( 5 p + 3 ) ( p + 4 ) 5 x 2 + 16 x + 12 = ( 5 x + 6 ) ( x + 2 )
3x^2 + 8x + 5 = (3x + 5)(x + 1) \\
2y^2 + 15y + 7 = (2y + 1)(y + 7) \\
3a^2 + 11a + 6 = (3a^2 + 9a) + (2a + 6) = 3a(a + 3) + 2(a + 3) = (3a + 2)(a + 3) \\
3x^2 + 19x + 6 = (3x + 1)(x + 6) \\
5p^2 + 23p + 12 = (5p + 3)(p + 4) \\
5x^2 + 16x + 12 = (5x + 6)(x + 2)
3 x 2 + 8 x + 5 = ( 3 x + 5 ) ( x + 1 ) 2 y 2 + 1 5 y + 7 = ( 2 y + 1 ) ( y + 7 ) 3 a 2 + 1 1 a + 6 = ( 3 a 2 + 9 a ) + ( 2 a + 6 ) = 3 a ( a + 3 ) + 2 ( a + 3 ) = ( 3 a + 2 ) ( a + 3 ) 3 x 2 + 1 9 x + 6 = ( 3 x + 1 ) ( x + 6 ) 5 p 2 + 2 3 p + 1 2 = ( 5 p + 3 ) ( p + 4 ) 5 x 2 + 1 6 x + 1 2 = ( 5 x + 6 ) ( x + 2 )
1.B.4
a x 2 − b x + c b = p + q p q = a c x 2 − 11 x + 24 = ( x − 3 ) ( x − 8 ) y 2 − 9 y + 18 = ( y − 3 ) ( y − 6 ) x 2 − 11 x + 18 = ( x − 2 ) ( x − 9 )
\boxed{ax^2 - bx + c} \qquad \boxed{b = p + q}\boxed{pq = ac} \\
x^2 - 11x + 24 = (x - 3)(x - 8) \\
y^2 - 9y + 18 = (y - 3)(y - 6) \\
x^2 - 11x + 18 = (x - 2)(x - 9)
a x 2 − b x + c b = p + q p q = a c x 2 − 1 1 x + 2 4 = ( x − 3 ) ( x − 8 ) y 2 − 9 y + 1 8 = ( y − 3 ) ( y − 6 ) x 2 − 1 1 x + 1 8 = ( x − 2 ) ( x − 9 )
a x 2 + b x − c b = p − q p q = a c p 2 + 5 p − 24 = ( p + 8 ) ( p − 3 ) x 2 + 4 x − 12 = ( x + 6 ) ( x − 2 )
\boxed{ax^2 + bx - c} \qquad \boxed{b = p - q}\boxed{pq = ac} \\
p^2 + 5p - 24 = (p + 8)(p - 3) \\
x^2 + 4x - 12 = (x + 6)(x - 2)
a x 2 + b x − c b = p − q p q = a c p 2 + 5 p − 2 4 = ( p + 8 ) ( p − 3 ) x 2 + 4 x − 1 2 = ( x + 6 ) ( x − 2 )
a x 2 − b x − c b = p − q p q = a c 2 q 2 − 5 q − 3 = ( 2 q + 1 ) ( q − 3 ) 3 x 2 − 10 x − 8 = ( 3 x + 2 ) ( x − 4 ) 2 a 2 − 3 a − 5 = ( 2 x − 5 ) ( x + 1 ) 2 x 2 − 5 x − 12 = ( 2 x + 3 ) ( x − 4 )
\boxed{ax^2 - bx - c} \qquad \boxed{b = p - q}\boxed{pq = ac} \\
2q^2 - 5q - 3 = (2q + 1)(q - 3) \\
3x^2 - 10x - 8 = (3x + 2)(x - 4) \\
2a^2 - 3a - 5 = (2x - 5)(x + 1) \\
2x^2 - 5x - 12 = (2x + 3)(x - 4)
a x 2 − b x − c b = p − q p q = a c 2 q 2 − 5 q − 3 = ( 2 q + 1 ) ( q − 3 ) 3 x 2 − 1 0 x − 8 = ( 3 x + 2 ) ( x − 4 ) 2 a 2 − 3 a − 5 = ( 2 x − 5 ) ( x + 1 ) 2 x 2 − 5 x − 1 2 = ( 2 x + 3 ) ( x − 4 )
3 b 2 − 20 b + 12 = ( 3 b − 2 ) ( b − 6 ) 9 x 2 − 25 y 2 = ( 3 x + 5 y ) ( 3 x + 5 y ) Diff. of two squares 16 x 4 − 81 y 4 = ( 4 x 2 + 9 y 2 ) ( 4 x 2 − 9 y 2 ) = ( 4 x 2 + 9 y 2 ) ( 2 x + 3 y ) ( 2 x − 3 y )
3b^2 - 20b + 12 = (3b - 2)(b - 6) \\
9x^2 - 25y^2 = (3x + 5y)(3x + 5y) \quad \text{Diff. of two squares} \\
16x^4 - 81y^4 = (4x^2 + 9y^2)(4x^2 - 9y^2) = (4x^2 + 9y^2)(2x + 3y)(2x - 3y)
3 b 2 − 2 0 b + 1 2 = ( 3 b − 2 ) ( b − 6 ) 9 x 2 − 2 5 y 2 = ( 3 x + 5 y ) ( 3 x + 5 y ) Diff. of two squares 1 6 x 4 − 8 1 y 4 = ( 4 x 2 + 9 y 2 ) ( 4 x 2 − 9 y 2 ) = ( 4 x 2 + 9 y 2 ) ( 2 x + 3 y ) ( 2 x − 3 y )
Notice how the ( 4 x 2 − 9 y 2 ) (4x^2 - 9y^2) ( 4 x 2 − 9 y 2 ) ( 4 x 2 + 9 y 2 ) (4x^2 + 9y^2) ( 4 x 2 + 9 y 2 )
a x 2 + b x + c \boxed{ax^2 + bx + c} a x 2 + b x + c
There seem to be at least two ways to factorise quadratic functions quickly:
Find p and q such that p q = a c pq = ac p q = a c p + q = b p+q = b p + q = b p and q you can divide with a, and place it to the opposite group of a, and divide it by a. Example:
3 x 2 + 11 a + 6 a c = 18 p + q = 9 + 2 p q = 9 ∗ 2 (p (9) can be divided by a (3)) ( 3 x + 2 ) ( x + 9 / 3 ) (Place a and q left, and p/a right) 3x^2 + 11a + 6 \\
ac = 18 \\
p+q = 9+2 \\
pq = 9*2 \qquad \text{(p (9) can be divided by a (3))} \\
(3x + 2)(x + 9/3) \qquad \text{(Place a and q left, and p/a right)} 3 x 2 + 1 1 a + 6 a c = 1 8 p + q = 9 + 2 p q = 9 ∗ 2 (p (9) can be divided by a (3)) ( 3 x + 2 ) ( x + 9 / 3 ) (Place a and q left, and p/a right)
This doesn't work with some quadratic functions, though. For example:
6 x 2 − 19 x + 10 a c = 60 p + q = 15 + 4 p q = 15 ∗ 4 (p nor q can be divided by a) 6x^2 - 19x + 10 \\
ac = 60 \\
p+q = 15 + 4 \\
pq = 15 * 4 \qquad \text{(p nor q can be divided by a)} 6 x 2 − 1 9 x + 1 0 a c = 6 0 p + q = 1 5 + 4 p q = 1 5 ∗ 4 (p nor q can be divided by a)
But both p and q can be divided by factors of a . Here the factors
are 3 and 2, so the result looks something like ( 3 x − . . . ) ( 2 x − . . . ) (3x - ...)(2x - ...) ( 3 x − . . . ) ( 2 x − . . . ) ( 6 x − . . . ) ( x − . . . ) (6x - ...)(x - ...) ( 6 x − . . . ) ( x − . . . )
So, more generally, it looks like whatever you leave as the multiplier on
one side, you have to divide by on the opposite side.
6 x 2 − 19 x + 10 = ( 3 ∗ 2 ) x 2 − ( 15 + 4 ) x + 19 = ( 3 x − Divide by 2 here ) ( 2 x − Divide by 3 here ) = ( 3 x − 4 / 2 ) ( 2 x − 15 / 3 ) = ( 3 x − 2 ) ( 2 x − 5 ) 6x^2 - 19x + 10 = \\
(3*2)x^2 - (15 + 4)x + 19 = \\
(3x - \boxed{\text{Divide by 2 here}})(2x - \boxed{\text{Divide by 3 here}}) = \\
(3x - 4/2)(2x - 15/3) = \\
(3x - 2)(2x - 5) 6 x 2 − 1 9 x + 1 0 = ( 3 ∗ 2 ) x 2 − ( 1 5 + 4 ) x + 1 9 = ( 3 x − Divide by 2 here ) ( 2 x − Divide by 3 here ) = ( 3 x − 4 / 2 ) ( 2 x − 1 5 / 3 ) = ( 3 x − 2 ) ( 2 x − 5 )
( 6 x 2 − 15 x ) + ( − 4 x + 10 ) = 3 x ( 2 x − 5 ) − 2 ( 2 x − 5 ) = ( 3 x − 2 ) ( 2 x − 5 ) (6x^2 - 15x) + (-4x + 10) = \\
3x(2x - 5) - 2(2x - 5) = \\
(3x - 2)(2x - 5) ( 6 x 2 − 1 5 x ) + ( − 4 x + 1 0 ) = 3 x ( 2 x − 5 ) − 2 ( 2 x − 5 ) = ( 3 x − 2 ) ( 2 x − 5 )
Find p and q such that p q = c pq = c p q = c a, and which one you have to add, to get b — in other words, b = a p + q b = ap + q b = a p + q
3 x 2 + 11 a + 6 p q = c 3 ∗ 2 = 6 Now we know the result looks like ( 3 x + 3 or 2 ) ( x + 3 or 2 ) a p + q = c 3 ∗ 3 + 2 = 11 a q + p = c 3 ∗ 2 + 3 = 9 Now we know a must by multiplied by p, not q ( 3 x + 2 ) ( x + 3 ) 3x^2 + 11a + 6 \\
pq = c \qquad 3*2 = 6 \\
\text{Now we know the result looks like }(3x + \boxed{3\text{ or }2})(x + \boxed{3\text{ or }2}) \\
ap + q = c \qquad 3*3 + 2 = 11 \\
aq + p = c \qquad 3*2 + 3 = 9 \\
\text{Now we know a must by multiplied by p, not q} \\
(3x + 2)(x + 3) 3 x 2 + 1 1 a + 6 p q = c 3 ∗ 2 = 6 Now we know the result looks like ( 3 x + 3 or 2 ) ( x + 3 or 2 ) a p + q = c 3 ∗ 3 + 2 = 1 1 a q + p = c 3 ∗ 2 + 3 = 9 Now we know a must by multiplied by p, not q ( 3 x + 2 ) ( x + 3 )
Let's try a harder one:
6 x 2 − 19 x + 10 = p q = c 2 ∗ 5 = 10 a p + q = b 6 ∗ 2 + 5 = 17 a q + p = b 6 ∗ 5 + 2 = 32 6x^2 - 19x + 10 = \\
pq = c \qquad 2*5 = 10 \\
ap + q = b \qquad 6*2 + 5 = 17 \\
aq + p = b \qquad 6*5 + 2 = 32 6 x 2 − 1 9 x + 1 0 = p q = c 2 ∗ 5 = 1 0 a p + q = b 6 ∗ 2 + 5 = 1 7 a q + p = b 6 ∗ 5 + 2 = 3 2
Doesn't work so well here, so we're back to having to factorise a into
two values, like in option 1 above.
Another quick way might be to find factors of a \boxed{a} a c \boxed{c} c b b b 6 x 2 − 19 x + 10 6x^2 - 19x + 10 6 x 2 − 1 9 x + 1 0
Factors of a:
Factors of c:
( 6 x − 2 ) ( x − 5 ) b = 6 ∗ 5 + 2 ∗ 1 = 32 ( 6 x − 5 ) ( x − 2 ) b = 6 ∗ 2 + 5 ∗ 1 = 17 ( 3 x − 2 ) ( 2 x − 5 ) b = 3 ∗ 5 + 2 ∗ 2 = 19 ( 3 x − 5 ) ( 2 x − 2 ) b = 3 ∗ 2 + 5 ∗ 2 = 16
(6x - 2)(x - 5) \qquad b = 6*5 + 2*1 = 32 \\
(6x - 5)(x - 2) \qquad b = 6*2 + 5*1 = 17 \\
(3x - 2)(2x - 5) \qquad b = 3*5 + 2*2 = \boxed{19} \\
(3x - 5)(2x - 2) \qquad b = 3*2 + 5*2 = 16
( 6 x − 2 ) ( x − 5 ) b = 6 ∗ 5 + 2 ∗ 1 = 3 2 ( 6 x − 5 ) ( x − 2 ) b = 6 ∗ 2 + 5 ∗ 1 = 1 7 ( 3 x − 2 ) ( 2 x − 5 ) b = 3 ∗ 5 + 2 ∗ 2 = 1 9 ( 3 x − 5 ) ( 2 x − 2 ) b = 3 ∗ 2 + 5 ∗ 2 = 1 6
1.C Fractions
n u m e r a t o r d e n o m i n a t o r
\dfrac{numerator}{denominator}
d e n o m i n a t o r n u m e r a t o r
Division is right-distributive :
a + b c = a c + b c that is, ( a + b ) / c = a / c + b / c
\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c} \qquad \text{that is,} \qquad (a+b)/c = a/c + b/c
c a + b = c a + c b that is, ( a + b ) / c = a / c + b / c
But not left-distributive:
a b + c ≠ a b + a c that is, a / ( b + c ) ≠ a / b + a / c
\dfrac{a}{b + c} \neq \dfrac{a}{b} + \dfrac{a}{c} \qquad \text{that is,} \qquad a/(b+c) \neq a/b + a/c
b + c a = b a + c a that is, a / ( b + c ) = a / b + a / c
The line in the fraction is effectively working as a bracket.
Dividing by c is the same as multiplying by 1/c:
( a + b ) c = ( a + b ) 1 c
\dfrac{(a+b)}{c} = (a+b)\dfrac{1}{c}
c ( a + b ) = ( a + b ) c 1
A fraction keeps the same value if its top and bottom are multiplied or divided by the same value:
a b = a c b c a b = a / c b / c
\dfrac{a}{b} = \dfrac{ac}{bc} \qquad \dfrac{a}{b} = \dfrac{a/c}{b/c}
b a = b c a c b a = b / c a / c
Doing this is called cancelling
a c b c = a b Divide both sides by c
\dfrac{a\cancel{c}}{b\cancel{c}} = \dfrac{a}{b} \quad \text{Divide both sides by c}\\
b c a c = b a Divide both sides by c
a ÷ c b ÷ c = a b Multiply both sides by c
\dfrac{a \div \cancel{c}}{b \div \cancel{c}} = \dfrac{a}{b} \quad \text{Multiply both sides by c}
b ÷ c a ÷ c = b a Multiply both sides by c
It doesn't matter where you put the sign:
− a b = − a b = a − b
-\dfrac{a}{b} = \dfrac{-a}{b} = \dfrac{a}{-b}
− b a = b − a = − b a
1.C.1
9 12 = 9 / 3 12 / 3 = 3 4 \dfrac{9}{12} = \dfrac{9/3}{12/3} = \dfrac{3}{4} 1 2 9 = 1 2 / 3 9 / 3 = 4 3
6 30 = 6 / 6 30 / 6 = 1 5 \dfrac{6}{30} = \dfrac{6/6}{30/6} = \dfrac{1}{5} 3 0 6 = 3 0 / 6 6 / 6 = 5 1
25 95 = 5 19 \dfrac{25}{95} = \dfrac{5}{19} 9 5 2 5 = 1 9 5
24 64 = 3 8 \dfrac{24}{64} = \dfrac{3}{8} 6 4 2 4 = 8 3
5 x 8 x = 5 8 \dfrac{5x}{8x} = \dfrac{5}{8} 8 x 5 x = 8 5
a b a c = b c \dfrac{ab}{ac} = \dfrac{b}{c} a c a b = c b
3 y 2 2 y = 3 y y 2 y = 3 y 2 \dfrac{3y^2}{2y} = \dfrac{3y\bcancel{y}}{2\bcancel{y}} = \dfrac{3y}{2} 2 y 3 y 2 = 2 y 3 y y = 2 3 y
8 p q 2 q = 4 p 1 = 4 p \dfrac{8pq}{2q} = \dfrac{4p}{1} = 4p 2 q 8 p q = 1 4 p = 4 p
4 a 2 2 a b = 2 a b \dfrac{4a^2}{2ab} = \dfrac{2a}{b} 2 a b 4 a 2 = b 2 a
3 x 2 y 3 2 x y 4 = 3 x 2 y \dfrac{3x^2y^3}{2xy^4} = \dfrac{3x}{2y} 2 x y 4 3 x 2 y 3 = 2 y 3 x
6 p 2 q 5 p q 2 = 6 p 5 q \dfrac{6p^2q}{5pq^2} = \dfrac{6p}{5q} 5 p q 2 6 p 2 q = 5 q 6 p
5 a b b 3 = 5 a b 2 \dfrac{5ab}{b^3} = \dfrac{5a}{b^2} b 3 5 a b = b 2 5 a
Sometimes, the process of factoring will be very important in simplifying fractions.
x y + x z x w = x ( y + x ) x w = y + z w \dfrac{xy + xz}{xw} = \dfrac{\cancel{x}(y + x)}{\cancel{x}w} = \dfrac{y + z}{w} x w x y + x z = x w x ( y + x ) = w y + z
a b + a c b + c = a ( b + c ) b + c = a \dfrac{ab + ac}{b + c} = \dfrac{a\cancel{(b + c)}}{\cancel{b + c}} = a b + c a b + a c = b + c a ( b + c ) = a
1.C.2
a b + a c a d = a ( b + c ) a d = b + c d Factorise and cancel or = a b a d + a c a d = b + c d Right-distribute ad and cancel
\begin{aligned}
\dfrac{ab + ac}{ad} &= \dfrac{a(b + c)}{ad} = \dfrac{b + c}{d} \quad \text{Factorise and cancel} \\
& \text{or} \\
&= \dfrac{ab}{ad} + \dfrac{ac}{ad} = \dfrac{b + c}{d} \quad \text{Right-distribute ad and cancel}
\end{aligned}
a d a b + a c = a d a ( b + c ) = d b + c Factorise and cancel or = a d a b + a d a c = d b + c Right-distribute ad and cancel
Remember: the line of a fraction basically splits the fraction into two brackets.
a b + c a ≠ b + c
\dfrac{ab + c}{a} \neq b + c
a a b + c = b + c
a b + c a = ( a b + c ) / a = a b / a + c / a Right-distribute = b + c a
\begin{aligned}
\dfrac{ab + c}{a} &= (ab + c)/a \\
&= ab/a + c/a \qquad \text{Right-distribute} \\
&= b + \dfrac{c}{a}
\end{aligned}
a a b + c = ( a b + c ) / a = a b / a + c / a Right-distribute = b + a c
2 x + 6 y 6 x − 8 y = 2 ( x + 3 y ) 2 ( 3 x − 4 y ) = x + 3 y 3 x − 4 y
\dfrac{2x + 6y}{6x - 8y} = \dfrac{2(x + 3y)}{2(3x - 4y)} = \dfrac{x + 3y}{3x - 4y}
6 x − 8 y 2 x + 6 y = 2 ( 3 x − 4 y ) 2 ( x + 3 y ) = 3 x − 4 y x + 3 y
6 a − 9 b 4 a − 6 b = 3 ( 2 a − 3 b ) 2 ( 2 a − 3 b ) = 3 2
\dfrac{6a - 9b}{4a - 6b} = \dfrac{3(2a - 3b)}{2(2a - 3b)} = \dfrac{3}{2}
4 a − 6 b 6 a − 9 b = 2 ( 2 a − 3 b ) 3 ( 2 a − 3 b ) = 2 3
p x − p q p 2 − p x = x − q p − x
\dfrac{px - pq}{p^2 - px} = \dfrac{x - q}{p - x}
p 2 − p x p x − p q = p − x x − q
3 x + 2 y 6 x = 3 x 6 x + 2 y 6 x = 1 2 + 2 ∗ 1 y 2 ∗ 3 x = 1 2 + y 3 x Can’t simplify
\dfrac{3x + 2y}{6x} = \dfrac{3x}{6x} + \dfrac{2y}{6x} = \dfrac{1}{2} + \dfrac{2*1y}{2*3x} = \dfrac{1}{2} + \dfrac{y}{3x} \quad \text{Can't simplify}
6 x 3 x + 2 y = 6 x 3 x + 6 x 2 y = 2 1 + 2 ∗ 3 x 2 ∗ 1 y = 2 1 + 3 x y Can’t simplify
2 x y + 5 x z 6 x = 2 y + 5 z 6
\dfrac{2xy + 5xz}{6x} = \dfrac{2y + 5z}{6}
6 x 2 x y + 5 x z = 6 2 y + 5 z
4 x z + 6 y z 2 x + 3 y = 2 z ( 2 x + 3 y ) 2 x + 3 y = 2 z
\dfrac{4xz + 6yz}{2x + 3y} = \dfrac{2z(2x + 3y)}{2x + 3y} = 2z
2 x + 3 y 4 x z + 6 y z = 2 x + 3 y 2 z ( 2 x + 3 y ) = 2 z
2 p − 3 q 2 p + 3 q Can’t simplify
\dfrac{2p - 3q}{2p + 3q} \quad \text{Can't simplify}
2 p + 3 q 2 p − 3 q Can’t simplify
x 2 − y 2 ( x + y ) 2 = ( x + y ) ( x − y ) ( x + y ) ( x + y ) = x − y x + y Diff. of squares
\dfrac{x^2 - y^2}{(x + y)^2} = \dfrac{(x + y)(x - y)}{(x + y)(x + y)} = \dfrac{x-y}{x+y} \quad \text{Diff. of squares}
( x + y ) 2 x 2 − y 2 = ( x + y ) ( x + y ) ( x + y ) ( x − y ) = x + y x − y Diff. of squares
x 2 + 5 x + 6 x 2 + x − 2 = ( x + 3 ) ( x + 2 ) ( x + 2 ) ( x − 1 ) = x + 3 x − 1
\dfrac{x^2 + 5x + 6}{x^2 + x - 2} = \dfrac{(x+3)(x+2)}{(x+2)(x-1)} = \dfrac{x+3}{x-1}
x 2 + x − 2 x 2 + 5 x + 6 = ( x + 2 ) ( x − 1 ) ( x + 3 ) ( x + 2 ) = x − 1 x + 3
Adding and subtracting fractions
A B + C D = A D B D + B C B D = A D + B C B D
\dfrac{A}{B} + \dfrac{C}{D} = \dfrac{AD}{BD} + \dfrac{BC}{BD} = \dfrac{AD + BC}{BD}
B A + D C = B D A D + B D B C = B D A D + B C
The same applies to subtraction.
A B − C D = A D B D − B C B D = A D − B C B D
\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{AD}{BD} - \dfrac{BC}{BD} = \dfrac{AD - BC}{BD}
B A − D C = B D A D − B D B C = B D A D − B C
Remember: always treat the numerator and denominator as bracketed expressions, so that you properly distribute signs.
a b − c − d e ≠ a e − c b − d b b e
\dfrac{a}{b} - \dfrac{c - d}{e} \neq \dfrac{ae - cb - db}{be}
b a − e c − d = b e a e − c b − d b
a b − c − d e = e ( a ) − b ( c − d ) ( b ) ( e ) = a e − c b + d b b e
\dfrac{a}{b} - \dfrac{c - d}{e} = \dfrac{e(a) - b(c - d)}{(b)(e)} = \dfrac{ae - cb + db}{be}
b a − e c − d = ( b ) ( e ) e ( a ) − b ( c − d ) = b e a e − c b + d b
Remember: if both the numerator and the denominator are negative, the result is a positive number, since they both have − 1 -1 − 1
− 5 − 3 = − 1 ( 5 ) − 1 ( 3 ) = 5 3
\dfrac{-5}{-3} = \dfrac{\cancel{-1}(5)}{\cancel{-1}(3)} = \dfrac{5}{3}
− 3 − 5 = − 1 ( 3 ) − 1 ( 5 ) = 3 5
1.C.3
3 x − 2 x + 3 + 2 x − 3 x + 1 = ( 3 x − 2 ) ( x + 1 ) + ( 2 x − 3 ) ( x + 3 ) ( x + 3 ) ( x + 1 )
\dfrac{3x - 2}{x + 3} + \dfrac{2x - 3}{x + 1} = \dfrac{(3x - 2)(x + 1) + (2x - 3)(x + 3)}{(x + 3)(x + 1)}
x + 3 3 x − 2 + x + 1 2 x − 3 = ( x + 3 ) ( x + 1 ) ( 3 x − 2 ) ( x + 1 ) + ( 2 x − 3 ) ( x + 3 )
= 3 x 2 + x − 2 + 2 x 2 + 3 x − 9 ( x + 1 ) ( x + 3 ) = \dfrac{3x^2 + x - 2 + 2x^2 + 3x - 9}{(x + 1)(x + 3)} = ( x + 1 ) ( x + 3 ) 3 x 2 + x − 2 + 2 x 2 + 3 x − 9
= 5 x 2 + 4 x − 11 x 2 + 4 x + 3 = \dfrac{5x^2 + 4x - 11}{x^2 + 4x + 3} = x 2 + 4 x + 3 5 x 2 + 4 x − 1 1
Not multiplying out the demoninator makes cancelling easier, so we usually leave the denominator factored:
= 5 x 2 + 4 x − 11 ( x + 1 ) ( x + 3 ) = \dfrac{5x^2 + 4x - 11}{(x + 1)(x + 3)} = ( x + 1 ) ( x + 3 ) 5 x 2 + 4 x − 1 1
Repeated factors in adding fractions
When adding fractions, you need to find a common denominator. One way is to multiply diagonally (cross-multiply):
3 4 + 5 6 = 3 ∗ 6 4 ∗ 6 + 5 ∗ 4 4 ∗ 6 = 18 + 20 24 = 19 12
\dfrac{3}{4} + \dfrac{5}{6} = \dfrac{3*6}{4*6} + \dfrac{5*4}{4*6} = \dfrac{18 + 20}{24} = \dfrac{19}{12}
4 3 + 6 5 = 4 ∗ 6 3 ∗ 6 + 4 ∗ 6 5 ∗ 4 = 2 4 1 8 + 2 0 = 1 2 1 9
But if the denominators share a repeated factor, then you can use a shortcut:
A B C + D B E = A E B C E + C D B C E
\dfrac{A}{BC} + \dfrac{D}{BE} = \dfrac{AE}{BCE} + \dfrac{CD}{BCE}
B C A + B E D = B C E A E + B C E C D
3 4 + 5 6 = 3 2 ( 2 ) + 5 2 ( 3 ) = 3 ( 3 ) + 5 ( 2 ) 2 ( 2 ) ( 3 ) = 19 12
\dfrac{3}{4} + \dfrac{5}{6} = \dfrac{3}{2(2)} + \dfrac{5}{2(3)} = \dfrac{3(3) + 5(2)}{2(2)(3)} = \dfrac{19}{12}
4 3 + 6 5 = 2 ( 2 ) 3 + 2 ( 3 ) 5 = 2 ( 2 ) ( 3 ) 3 ( 3 ) + 5 ( 2 ) = 1 2 1 9
Also, when there is a repeated factor in the denominators, you can ignore the repeated factor in one of the denominators when multiplying the denominators:
3 4 + 5 6 = 3 2 ( 2 ) + 5 2 ( 3 ) = 3 ( 3 ) + 5 ( 2 ) 4 ( 3 ) = 19 12
\dfrac{3}{4} + \dfrac{5}{6} = \dfrac{3}{2(2)} + \dfrac{5}{\cancel{2}(3)} = \dfrac{3(3) + 5(2)}{4(3)} = \dfrac{19}{12}
4 3 + 6 5 = 2 ( 2 ) 3 + 2 ( 3 ) 5 = 4 ( 3 ) 3 ( 3 ) + 5 ( 2 ) = 1 2 1 9
(Remember: ) Or more simply, ask yourself: what do I need to multiply the left summand's top and bottom, and the right summand's top and bottom, so that the denominators are the same?
3 4 + 5 6 = 3 4 ( 3 3 ) + 5 6 ( 2 2 ) = 19 12
\dfrac{3}{4} + \dfrac{5}{6} = \dfrac{3}{4} \bigg(\dfrac{3}{3}\bigg) + \dfrac{5}{6} \bigg(\dfrac{2}{2}\bigg) = \dfrac{19}{12}
4 3 + 6 5 = 4 3 ( 3 3 ) + 6 5 ( 2 2 ) = 1 2 1 9
All of this applies to more complicated expressions as well — here x is a repeated factor:
2 x ( x + 3 ) + 3 x ( 2 x − 1 ) = 2 ( 2 x − 1 ) + 3 ( x + 3 ) x ( x + 3 ) ( 2 x − 1 ) = 7 x + 7 x ( x + 3 ) ( 2 x − 1 )
\dfrac{2}{x(x+3)} + \dfrac{3}{x(2x - 1)} = \dfrac{2(2x - 1) + 3(x + 3)}{x(x+3)(2x-1)} = \dfrac{7x + 7}{x(x+3)(2x-1)}
x ( x + 3 ) 2 + x ( 2 x − 1 ) 3 = x ( x + 3 ) ( 2 x − 1 ) 2 ( 2 x − 1 ) + 3 ( x + 3 ) = x ( x + 3 ) ( 2 x − 1 ) 7 x + 7
1.C.4
2 9 + 7 15 = 2 ∗ 5 3 ∗ 3 + 7 ∗ 3 3 ∗ 5 = 31 45 \dfrac{2}{9} + \dfrac{7}{15} = \dfrac{2*5}{3*3} + \dfrac{7*3}{3*5} = \dfrac{31}{45} 9 2 + 1 5 7 = 3 ∗ 3 2 ∗ 5 + 3 ∗ 5 7 ∗ 3 = 4 5 3 1
5 6 + 3 8 = 20 48 + 9 48 = 29 24 \dfrac{5}{6} + \dfrac{3}{8} = \dfrac{20}{48} + \dfrac{9}{48} = \dfrac{29}{24} 6 5 + 8 3 = 4 8 2 0 + 4 8 9 = 2 4 2 9
1 3 + 3 4 + 5 6 = 7 6 + 3 4 = 23 12 \dfrac{1}{3} + \dfrac{3}{4} + \dfrac{5}{6} = \dfrac{7}{6} + \dfrac{3}{4} = \dfrac{23}{12} 3 1 + 4 3 + 6 5 = 6 7 + 4 3 = 1 2 2 3
3 x y ( 2 x − y ) + 5 y x ( 2 x − y ) = 3 x 2 + 5 y 2 x y ( 2 x − y ) \dfrac{3x}{y(2x - y)} + \dfrac{5y}{x(2x - y)} = \dfrac{3x^2 + 5y^2}{xy(2x - y)} y ( 2 x − y ) 3 x + x ( 2 x − y ) 5 y = x y ( 2 x − y ) 3 x 2 + 5 y 2
2 x ( 3 x + 1 ) + 5 x ( 2 x − 1 ) = 2 ( 2 x − 1 ) + 5 ( 3 x + 1 ) x ( 3 x + 1 ) ( 2 x − 1 ) = 19 x + 3 x ( 3 x + 1 ) ( 2 x − 1 ) \dfrac{2}{x(3x + 1)} + \dfrac{5}{x(2x - 1)} = \dfrac{2(2x - 1) + 5(3x + 1)}{x(3x + 1)(2x - 1)} = \dfrac{19x + 3}{x(3x + 1)(2x - 1)} x ( 3 x + 1 ) 2 + x ( 2 x − 1 ) 5 = x ( 3 x + 1 ) ( 2 x − 1 ) 2 ( 2 x − 1 ) + 5 ( 3 x + 1 ) = x ( 3 x + 1 ) ( 2 x − 1 ) 1 9 x + 3
4 x 2 − y 2 + 3 ( x + y ) 2 = 4 ( x + y ) ( x − y ) + 3 ( x + y ) 2 = 4 ( x + y ) + 3 ( x − y ) ( x + y ) 2 ( x − y ) = 7 x + y ( x + y ) 2 ( x − y ) \dfrac{4}{x^2 - y^2} + \dfrac{3}{(x+y)^2} = \dfrac{4}{(x+y)(x-y)} + \dfrac{3}{(x+y)^2} = \dfrac{4(x+y) + 3(x-y)}{(x+y)^2(x-y)} = \dfrac{7x + y}{(x+y)^2(x-y)} x 2 − y 2 4 + ( x + y ) 2 3 = ( x + y ) ( x − y ) 4 + ( x + y ) 2 3 = ( x + y ) 2 ( x − y ) 4 ( x + y ) + 3 ( x − y ) = ( x + y ) 2 ( x − y ) 7 x + y
1.C.5
3 x − 5 10 + 2 x − 3 15 = 3 ( 3 x − 5 ) + 2 ( 2 x − 3 ) 10 ∗ 3 ⏟ Or 15 ∗ 2 = 13 x − 21 30 \dfrac{3x - 5}{10} + \dfrac{2x - 3}{15} = \dfrac{3(3x - 5) + 2(2x - 3)}{\underbrace{10*3}_{\text{Or }15*2}} = \dfrac{13x - 21}{30} 1 0 3 x − 5 + 1 5 2 x − 3 = Or 1 5 ∗ 2 1 0 ∗ 3 3 ( 3 x − 5 ) + 2 ( 2 x − 3 ) = 3 0 1 3 x − 2 1
3 a + 5 b 4 − a − 3 b 2 = 1 ( 3 a + 5 b ) − 2 ( a − 3 b ) 4 ∗ 1 = a + 11 b 4 \dfrac{3a + 5b}{4} - \dfrac{a - 3b}{2} = \dfrac{1(3a + 5b) - 2(a -3b)}{4*1} = \dfrac{a + 11b}{4} 4 3 a + 5 b − 2 a − 3 b = 4 ∗ 1 1 ( 3 a + 5 b ) − 2 ( a − 3 b ) = 4 a + 1 1 b
3 m − 5 n 6 − 3 m − 7 n 2 = ( 3 m − 5 n ) − 3 ( 3 m − 7 n ) 6 = − 6 m + 16 n 6 = 2 ( 8 n − 3 m ) 2 ∗ 3 = 8 n − 3 m 3 \dfrac{3m - 5n}{6} - \dfrac{3m - 7n}{2} = \dfrac{(3m - 5n) - 3(3m - 7n)}{6} = \dfrac{-6m + 16n}{6} = \dfrac{2(8n - 3m)}{2*3} = \dfrac{8n - 3m}{3} 6 3 m − 5 n − 2 3 m − 7 n = 6 ( 3 m − 5 n ) − 3 ( 3 m − 7 n ) = 6 − 6 m + 1 6 n = 2 ∗ 3 2 ( 8 n − 3 m ) = 3 8 n − 3 m
2 b a ( 2 a + b ) + 3 a b ( 2 a + b ) = ( 2 b a ( 2 a + b ) ∗ b b ) + ( 3 a b ( 2 a + b ) ∗ a a ) = 2 b 2 + 3 a 2 a b ( 2 a + b ) \dfrac{2b}{a(2a + b)} + \dfrac{3a}{b(2a + b)} = \bigg(\dfrac{2b}{a(2a + b)} * \dfrac{b}{b}\bigg) + \bigg(\dfrac{3a}{b(2a + b)} * \dfrac{a}{a}\bigg) = \dfrac{2b^2 + 3a^2}{ab(2a+b)} a ( 2 a + b ) 2 b + b ( 2 a + b ) 3 a = ( a ( 2 a + b ) 2 b ∗ b b ) + ( b ( 2 a + b ) 3 a ∗ a a ) = a b ( 2 a + b ) 2 b 2 + 3 a 2
2 a ( a + b ) ( 3 a + b ) + 3 b ( a − b ) ( 3 a + b ) = 2 a 2 + a b + 3 b 2 ( a 2 − b 2 ) ( 3 a + b ) \dfrac{2a}{(a+b)(3a+b)} + \dfrac{3b}{(a-b)(3a+b)} = \dfrac{2a^2 + ab + 3b^2}{(a^2 - b^2)(3a + b)} ( a + b ) ( 3 a + b ) 2 a + ( a − b ) ( 3 a + b ) 3 b = ( a 2 − b 2 ) ( 3 a + b ) 2 a 2 + a b + 3 b 2
5 x 2 − y 2 − 2 x ( x + y ) = 5 ( x + y ) ( x − y ) − 2 x ( x + y ) = 5 x − 2 x + 2 y x ( x + y ) ( x − y ) = 3 x + 2 y x ( x 2 − y 2 ) \dfrac{5}{x^2 - y^2} - \dfrac{2}{x(x + y)} = \dfrac{5}{(x+y)(x-y)} - \dfrac{2}{x(x + y)} = \dfrac{5x - 2x + 2y}{x(x+y)(x-y)} = \dfrac{3x + 2y}{x(x^2 - y^2)} x 2 − y 2 5 − x ( x + y ) 2 = ( x + y ) ( x − y ) 5 − x ( x + y ) 2 = x ( x + y ) ( x − y ) 5 x − 2 x + 2 y = x ( x 2 − y 2 ) 3 x + 2 y
Multiplying and dividing fractions
A B ∗ C D = A C B D A B C = A 1 ∗ B C = A B C
\dfrac{A}{B} * \dfrac{C}{D} = \dfrac{AC}{BD} \qquad\qquad A\dfrac{B}{C} = \dfrac{A}{1} * \dfrac{B}{C} = \dfrac{AB}{C}
B A ∗ D C = B D A C A C B = 1 A ∗ C B = C A B
A B ÷ C D = A B ∗ D C Flip top and bottom
\dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} * \dfrac{D}{C} \quad \text{Flip top and bottom}
B A ÷ D C = B A ∗ C D Flip top and bottom
Remember: dividing by A 1 \frac{A}{1} 1 A 1 A \frac{1}{A} A 1
1.C.6
2 x ( 2 x − 3 y ) − 3 2 x ( x + 4 y ) = 4 x ( x + 4 y ) − 3 x ( 2 x − 3 y ) 2 x 2 ( 2 x − 3 y ) ( x + 4 y ) = \dfrac{2}{x(2x - 3y)} - \dfrac{3}{2x(x + 4y)} = \dfrac{4x(x + 4y) - 3x(2x - 3y)}{2x^2(2x - 3y)(x + 4y)} = x ( 2 x − 3 y ) 2 − 2 x ( x + 4 y ) 3 = 2 x 2 ( 2 x − 3 y ) ( x + 4 y ) 4 x ( x + 4 y ) − 3 x ( 2 x − 3 y ) =
− 2 x 2 + 25 x y 2 x 2 ( 2 x − 3 y ) ( x + 4 y ) = − 2 x + 25 y 2 x ( 2 x − 3 y ) ( x + 4 y ) \dfrac{-2x^2 + 25xy}{2x^2(2x - 3y)(x + 4y)} = \dfrac{-2x + 25y}{2x(2x - 3y)(x + 4y)} 2 x 2 ( 2 x − 3 y ) ( x + 4 y ) − 2 x 2 + 2 5 x y = 2 x ( 2 x − 3 y ) ( x + 4 y ) − 2 x + 2 5 y
You didn't notice that x in the denominators was a repeated factor. We can do as follows:
2 x ( 2 x − 3 y ) − 3 2 x ( x + 4 y ) = 4 ( x + 4 y ) − 3 ( 2 x − 3 y ) 2 x ( 2 x − 3 y ) ( x + 4 y ) = − 2 x + 25 y 2 x ( 2 x − 3 y ) ( x + 4 y ) \dfrac{2}{x(2x - 3y)} - \dfrac{3}{2x(x + 4y)} = \dfrac{4(x + 4y) - 3(2x - 3y)}{2x(2x - 3y)(x + 4y)} = \dfrac{-2x + 25y}{2x(2x - 3y)(x + 4y)} x ( 2 x − 3 y ) 2 − 2 x ( x + 4 y ) 3 = 2 x ( 2 x − 3 y ) ( x + 4 y ) 4 ( x + 4 y ) − 3 ( 2 x − 3 y ) = 2 x ( 2 x − 3 y ) ( x + 4 y ) − 2 x + 2 5 y
2 x − 1 3 − x − 7 5 = ( 10 x − 5 ) − ( 3 x − 21 ) 15 = 7 x + 16 15 \dfrac{2x - 1}{3} - \dfrac{x - 7}{5} = \dfrac{(10x - 5) - (3x - 21)}{15} = \dfrac{7x + 16}{15} 3 2 x − 1 − 5 x − 7 = 1 5 ( 1 0 x − 5 ) − ( 3 x − 2 1 ) = 1 5 7 x + 1 6
3 a 2 2 b ∗ a b 6 c = 3 a 3 b 12 b c = a 3 4 c \dfrac{3a^2}{2b} * \dfrac{ab}{6c} = \dfrac{3a^3b}{12bc} = \dfrac{a^3}{4c} 2 b 3 a 2 ∗ 6 c a b = 1 2 b c 3 a 3 b = 4 c a 3
2 a 3 b ÷ b 2 9 a 2 = 18 a 3 3 b 3 = 6 a 3 b 3 \dfrac{2a}{3b} \div \dfrac{b^2}{9a^2} = \dfrac{18a^3}{3b^3} = \dfrac{6a^3}{b^3} 3 b 2 a ÷ 9 a 2 b 2 = 3 b 3 1 8 a 3 = b 3 6 a 3
3 x y 2 z ÷ 2 x 2 5 y z 2 = 15 x y z 2 2 x 2 y 2 z = 15 z 2 x y \dfrac{3x}{y^2z} \div \dfrac{2x^2}{5yz^2} = \dfrac{15xyz^2}{2x^2y^2z} = \dfrac{15z}{2xy} y 2 z 3 x ÷ 5 y z 2 2 x 2 = 2 x 2 y 2 z 1 5 x y z 2 = 2 x y 1 5 z
3 x 2 ( 2 x + 3 y ) 2 y ( x − y ) ∗ y 2 ( x − y ) x ( x + 3 y ) = 3 x y ( 2 x + 3 y ) 2 ( x + 3 y ) = 6 x 2 y + 9 x y 2 2 ( x + 3 y ) \dfrac{3x\cancel{^2}(2x + 3y)}{2\cancel{y}\cancel{(x - y)}} * \dfrac{y\cancel{^2}\cancel{(x - y)}}{\cancel{x}(x + 3y)} = \dfrac{3xy(2x + 3y)}{2(x + 3y)} = \dfrac{6x^2y + 9xy^2}{2(x + 3y)} 2 y ( x − y ) 3 x 2 ( 2 x + 3 y ) ∗ x ( x + 3 y ) y 2 ( x − y ) = 2 ( x + 3 y ) 3 x y ( 2 x + 3 y ) = 2 ( x + 3 y ) 6 x 2 y + 9 x y 2
5 p q ( p + q ) 3 p + 2 q ∗ 3 p + 2 q q 2 ( 5 p − q ) = 5 p ( p + q ) q ( 5 p − q ) \dfrac{5p\cancel{q}(p + q)}{\cancel{3p + 2q}} * \dfrac{\cancel{3p + 2q}}{q\cancel{^2}(5p - q)} = \dfrac{5p(p + q)}{q(5p - q)} 3 p + 2 q 5 p q ( p + q ) ∗ q 2 ( 5 p − q ) 3 p + 2 q = q ( 5 p − q ) 5 p ( p + q )
( a 2 − b 2 ) 4 a 2 + b 2 ∗ a 4 − b 4 ( a + b ) 4 = ( a + b ) 4 ( a − b ) 4 a 2 + b 2 ∗ ( a 2 + b 2 ) ( a 2 − b 2 ) ( a + b ) 4 \dfrac{(a^2 - b^2)^4}{a^2 + b^2} * \dfrac{a^4 - b^4}{(a + b)^4} = \dfrac{\cancel{(a+b)^4}(a-b)^4}{\cancel{a^2 + b^2}} * \dfrac{\cancel{(a^2 + b^2)}(a^2 - b^2)}{\cancel{(a + b)^4}} a 2 + b 2 ( a 2 − b 2 ) 4 ∗ ( a + b ) 4 a 4 − b 4 = a 2 + b 2 ( a + b ) 4 ( a − b ) 4 ∗ ( a + b ) 4 ( a 2 + b 2 ) ( a 2 − b 2 )
= ( a − b ) 4 ( a 2 − b 2 ) = ( a + b ) ( a − b ) 5 = (a-b)^4(a^2 - b^2) = (a+b)(a-b)^5 = ( a − b ) 4 ( a 2 − b 2 ) = ( a + b ) ( a − b ) 5
The rules for working with powers
a m ∗ a n = a m + n 1 0 2 ∗ 1 0 3 = 1 0 5
\boxed{a^m * a^n = a^{m+n}} \qquad 10^2 * 10^3 = 10^5
a m ∗ a n = a m + n 1 0 2 ∗ 1 0 3 = 1 0 5
a m ÷ a n = a m − n 1 0 3 ÷ 1 0 2 = 1 0 1
\boxed{a^m \div a^n = a^{m-n}} \qquad 10^3 \div 10^2 = 10^1
a m ÷ a n = a m − n 1 0 3 ÷ 1 0 2 = 1 0 1
( a m ) n = a m n ( 1 0 2 ) 3 = 1 0 6
\boxed{(a^m)^n = a^{mn}} \qquad (10^2)^3 = 10^6
( a m ) n = a m n ( 1 0 2 ) 3 = 1 0 6
a m + a m ≠ ( 2 a ) m 1 2 + 1 2 ≠ ( 2 ∗ 1 ) 2 a m + a m = 2 ( a m ) 1 2 + 1 2 = 2 ( 1 2 )
a^m + a^m \neq (2a)^m \qquad 1^2 + 1^2 \neq (2*1)^2 \\
a^m + a^m = 2(a^m) \qquad 1^2 + 1^2 = 2(1^2)
a m + a m = ( 2 a ) m 1 2 + 1 2 = ( 2 ∗ 1 ) 2 a m + a m = 2 ( a m ) 1 2 + 1 2 = 2 ( 1 2 )
Special cases:
a 3 ÷ a 2 = a ∗ a ∗ a a ∗ a = a 1 = a
a^3 \div a^2 = \frac{a*a*a}{a*a} = \boxed{a^1 = a}
a 3 ÷ a 2 = a ∗ a a ∗ a ∗ a = a 1 = a
a 3 ÷ a 3 = a ∗ a ∗ a a ∗ a ∗ a = a 0 = 1
a^3 \div a^3 = \frac{a*a*a}{a*a*a} = \boxed{a^0 = 1}
a 3 ÷ a 3 = a ∗ a ∗ a a ∗ a ∗ a = a 0 = 1
a 2 ÷ a 3 = a ∗ a a ∗ a ∗ a = a − 1 = 1 a and a − n = 1 a n
a^2 \div a^3 = \frac{a*a}{a*a*a} = \boxed{a^{-1} = \frac{1}{a}}\quad\text{ and }\quad\boxed{a^{-n} = \frac{1}{a^n}}
a 2 ÷ a 3 = a ∗ a ∗ a a ∗ a = a − 1 = a 1 and a − n = a n 1
A fraction exponent is the same as taking a root:
a 1 / 2 ∗ a 1 / 2 = a 1 = a Therefore a 1 / 2 = ± a Square root
a^{1/2} * a^{1/2} = a^1 = a \quad \text{Therefore }\boxed{a^{1/2} = \pm \sqrt{a}} \quad \text{Square root}
a 1 / 2 ∗ a 1 / 2 = a 1 = a Therefore a 1 / 2 = ± a Square root
a 1 / 3 ∗ a 1 / 3 ∗ a 1 / 3 = a 1 = a Therefore a 1 / n = a n Nth root
a^{1/3} * a^{1/3} * a^{1/3} = a^1 = a \quad \text{Therefore }\boxed{a^{1/n} = \sqrt[n]{a}} \quad \text{Nth root}
a 1 / 3 ∗ a 1 / 3 ∗ a 1 / 3 = a 1 = a Therefore a 1 / n = n a Nth root
( a b ) 3 = a b ∗ a b ∗ a b = a 3 b 3 Therefore ( a b ) n = a n b n and a b = a b
\Big(\dfrac{a}{b}\Big)^3 = \dfrac{a}{b} * \dfrac{a}{b} * \dfrac{a}{b} = \dfrac{a^3}{b^3} \quad \text{Therefore } \boxed{\Big(\dfrac{a}{b}\Big)^n = \dfrac{a^n}{b^n}} \text{ and } \boxed{\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}}
( b a ) 3 = b a ∗ b a ∗ b a = b 3 a 3 Therefore ( b a ) n = b n a n and b a = b a
An even root has two solutions. An odd root has one solution.
By convention, a \sqrt{a} a only the positive value, a.k.a. the principal root , but a 1 / 2 = ± a a^{1/2} = \pm \sqrt{a} a 1 / 2 = ± a \sqrt{} not implicitly contain the negative value as well, and you have to explicitly say ± \pm ±
a 1 / 2 = ± a = { a , − a } a 1 / 3 = a 3 ( a ) 2 = a ( a 1 / 2 ) 2 = ± a
a^{1/2} = \pm \sqrt{a} = \{\sqrt{a}, -\sqrt{a}\} \\
a^{1/3} = \sqrt[3]{a} \\
(\sqrt{a})^2 = a \\
(a^{1/2})^2 = \pm a
a 1 / 2 = ± a = { a , − a } a 1 / 3 = 3 a ( a ) 2 = a ( a 1 / 2 ) 2 = ± a
4 1 / 2 = ± 4 = { 2 − 2 = ± 2
4^{1/2} = \pm \sqrt{4} = \begin{cases}2 \\ -2\end{cases} = \pm 2
4 1 / 2 = ± 4 = { 2 − 2 = ± 2
8 1 / 3 = 8 3 = 2
8^{1/3} = \sqrt[3]{8} = 2
8 1 / 3 = 3 8 = 2
2 7 2 / 3 = 2 7 1 / 3 ∗ 2 7 1 / 3 = ( 27 3 ) 2 = 3 2 = 9
27^{2/3} = 27^{1/3} * 27^{1/3} = (\sqrt[3]{27})^2 = 3^2 = 9
2 7 2 / 3 = 2 7 1 / 3 ∗ 2 7 1 / 3 = ( 3 2 7 ) 2 = 3 2 = 9
1 6 1 / 4 = ± 16 4 = ± 2
16^{1/4} = \pm \sqrt[4]{16} = \pm 2
1 6 1 / 4 = ± 4 1 6 = ± 2
Remember: in general, each even root (e.g. ± x 2 \pm \sqrt[2]{x} ± 2 x ± x 4 \pm \sqrt[4]{x} ± 4 x positive number has two possible solutions. An odd root (e.g. x 3 \sqrt[3]{x} 3 x
Remember to always include the ± \pm ± even roots : a 2 = b a^2 = b a 2 = b a = ± b a = \pm\sqrt{b} a = ± b
At present, we cannot find any even roots of negative numbers , although in Chapter 10 [Complex numbers] we will find out how it is possible to extend the number system so that we can have roots for these numbers too. Have a guess at how many fourth roots of 16 we shall then have .
Based on the chapter name it has something to do with complex numbers. I have no clear idea what to do yet, but I'll try something out here (I think − 1 \sqrt{-1} − 1
− 1 = ( − 1 ) 1 / 2 = − ( 1 1 / 4 ∗ 1 1 / 4 ) = − 1 1 / 4 ∗ − 1 1 / 4 = ( − 1 ) 1 / 4 ( − 1 ) 1 / 4 ± 1 = 1 1 / 2 = 1 ± 2 = 2 1 / 2 ± − 2 = ( − 2 ) 1 / 2 = ± i ? ± 8 = 8 1 / 2 = 8 1 / 2 1 = 8 1 / 2 ∗ 8 1 / 2 1 ∗ 8 1 / 2 = 8 2 8 1 / 2 = 8 2 ÷ 8 2 8 1 / 2 ÷ 8 2 = 1 8 − 3 / 2 ± 16 = ± 4 ± 16 4 = ± 2 ± − 16 = ± i ?
\sqrt{-1} = (-1)^{1/2} = \cancel{-(1^{1/4} * 1^{1/4}) = -1^{1/4} * -1^{1/4}} = (-1)^{1/4}(-1)^{1/4} \\
\pm \sqrt{1} = 1^{1/2} = 1 \\
\pm \sqrt{2} = 2^{1/2} \\
\pm \sqrt{-2} = (-2)^{1/2} = \pm i? \\
\pm \sqrt{8} = 8^{1/2} = \frac{8^{1/2}}{1} = \frac{8^{1/2} * 8^{1/2}}{1 * 8^{1/2}} = \frac{8^2}{8^{1/2}} = \frac{8^2 \div 8^2}{8^{1/2} \div 8^2} = \frac{1}{8^{-3/2}} \\
\pm \sqrt{16} = \pm 4 \\
\pm \sqrt[4]{16} = \pm 2 \\
\pm \sqrt{-16} = \pm i?
− 1 = ( − 1 ) 1 / 2 = − ( 1 1 / 4 ∗ 1 1 / 4 ) = − 1 1 / 4 ∗ − 1 1 / 4 = ( − 1 ) 1 / 4 ( − 1 ) 1 / 4 ± 1 = 1 1 / 2 = 1 ± 2 = 2 1 / 2 ± − 2 = ( − 2 ) 1 / 2 = ± i ? ± 8 = 8 1 / 2 = 1 8 1 / 2 = 1 ∗ 8 1 / 2 8 1 / 2 ∗ 8 1 / 2 = 8 1 / 2 8 2 = 8 1 / 2 ÷ 8 2 8 2 ÷ 8 2 = 8 − 3 / 2 1 ± 1 6 = ± 4 ± 4 1 6 = ± 2 ± − 1 6 = ± i ?
My guess is we'll still have two roots for 16 4 \sqrt[4]{16} 4 1 6 16 16 1 6 − 16 -16 − 1 6
Yes, it is most satisfyingly four.
(╯°□°)╯︵ ┻━┻
I guess it'll be something like:
± 16 4 = { 2 − 2 i − i
\pm \sqrt[4]{16} = \begin{cases}2 \\ -2 \\ i \\ -i\end{cases}
± 4 1 6 = ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ 2 − 2 i − i
1.D.1
3 − 1 = 1 3 1 1 6 1 / 2 = ± 16 = 4 = ± 4 9 3 / 2 = 9 2 / 2 ∗ 9 1 / 2 = 9 ∗ ± 9 = ± 27 2 7 − 1 / 3 = 1 27 3 = 1 3 4 0 = 4 n ÷ 4 n = 1 7 1 = 7 7 − 2 = 1 7 2 = 1 49 4 − 1 / 2 = 1 ± 4 = ± 1 2 3 2 1 / 5 = 32 5 = 2
3^{-1} = \frac{1}{3^1} \\
16^{1/2} = \pm \sqrt{16} = \cancel{4} = \pm 4 \\
9^{3/2} = 9^{2/2} * 9^{1/2} = 9 * \pm \sqrt{9} = \pm 27 \\
27^{-1/3} = \frac{1}{\sqrt[3]{27}} = \frac{1}{3} \\
4^{0} = 4^{n} \div 4^{n} = 1 \\
7^{1} = 7 \\
7^{-2} = \frac{1}{7^2} = \frac{1}{49} \\
4^{-1/2} = \frac{1}{\pm \sqrt{4}} = \pm \frac{1}{2} \\
32^{1/5} = \sqrt[5]{32} = 2
3 − 1 = 3 1 1 1 6 1 / 2 = ± 1 6 = 4 = ± 4 9 3 / 2 = 9 2 / 2 ∗ 9 1 / 2 = 9 ∗ ± 9 = ± 2 7 2 7 − 1 / 3 = 3 2 7 1 = 3 1 4 0 = 4 n ÷ 4 n = 1 7 1 = 7 7 − 2 = 7 2 1 = 4 9 1 4 − 1 / 2 = ± 4 1 = ± 2 1 3 2 1 / 5 = 5 3 2 = 2
1 6 − 3 / 4 = 1 6 − 1 / 4 ∗ 1 6 − 1 / 4 ∗ 1 6 − 1 / 4 = ( 1 ± 16 4 ) 3 = ( ± 1 2 ) 3 = ± 1 8
16^{-3/4} = 16^{-1/4} * 16^{-1/4} * 16^{-1/4} = (\frac{1}{\pm \sqrt[4]{16}})^3 = (\pm \frac{1}{2})^3 = \pm \frac{1}{8}
1 6 − 3 / 4 = 1 6 − 1 / 4 ∗ 1 6 − 1 / 4 ∗ 1 6 − 1 / 4 = ( ± 4 1 6 1 ) 3 = ( ± 2 1 ) 3 = ± 8 1
or
1 6 − 3 / 4 = 1 1 6 3 / 4 = 1 ( 1 6 1 / 4 ) 3 = 1 ( ± 16 4 ) 3 = ( 1 ± 2 ) 3 = ± 1 8
16^{-3/4} = \frac{1}{16^{3/4}} = \frac{1}{(16^{1/4})^3} = \frac{1}{(\pm \sqrt[4]{16})^3} = (\frac{1}{\pm 2})^3 = \pm \frac{1}{8}
1 6 − 3 / 4 = 1 6 3 / 4 1 = ( 1 6 1 / 4 ) 3 1 = ( ± 4 1 6 ) 3 1 = ( ± 2 1 ) 3 = ± 8 1
2 5 3 / 2 = 25 ∗ 25 = ± 125 4 9 − 1 / 2 = 1 ± 49 = ± 1 7
25^{3/2} = 25 * \sqrt{25} = \pm 125 \\
49^{-1/2} = \frac{1}{\pm \sqrt{49}} = \pm \frac{1}{7}
2 5 3 / 2 = 2 5 ∗ 2 5 = ± 1 2 5 4 9 − 1 / 2 = ± 4 9 1 = ± 7 1
We don’t very often have to say that we have none of something. So why is having a symbol for zero so important?
So that we can say we have none of something when we need to?
So that we can create generalizations (axioms) like the additive identity ?
To act as a step towards the concept of negative numbers (which are useful to describe losses etc.)?
It makes it possible to put in all the necessary place values in our system for writing numbers, for example 301. Having a place value system means that once the symbols for 1 to 9 are learnt, a number of any size can be written.
1001 in base-10 is ( 1 ∗ 1000 ) + ( 0 ∗ 100 ) + ( 0 ∗ 10 ) + ( 1 ∗ 1 ) {(1*1000) + (0*100) + (0*10) + (1*1)} ( 1 ∗ 1 0 0 0 ) + ( 0 ∗ 1 0 0 ) + ( 0 ∗ 1 0 ) + ( 1 ∗ 1 ) ( 1 ∗ 1 0 3 ) + ( 0 ∗ 1 0 2 ) + ( 0 ∗ 1 0 1 ) + ( 1 ∗ 1 0 0 ) {(1 * 10^3) + (0 * 10^2) + (0 * 10^1) + (1 * 10^0)} ( 1 ∗ 1 0 3 ) + ( 0 ∗ 1 0 2 ) + ( 0 ∗ 1 0 1 ) + ( 1 ∗ 1 0 0 )
1001 in base-2 is ( 1 ∗ 8 ) + ( 0 ∗ 4 ) + ( 0 ∗ 2 ) + ( 1 ∗ 1 ) {(1*8) + (0*4) + (0*2) + (1*1)} ( 1 ∗ 8 ) + ( 0 ∗ 4 ) + ( 0 ∗ 2 ) + ( 1 ∗ 1 ) ( 1 ∗ 2 3 ) + ( 0 ∗ 2 2 ) + ( 0 ∗ 2 1 ) + ( 1 ∗ 2 0 ) {(1 * 2^3) + (0 * 2^2) + (0 * 2^1) + (1 * 2^0)} ( 1 ∗ 2 3 ) + ( 0 ∗ 2 2 ) + ( 0 ∗ 2 1 ) + ( 1 ∗ 2 0 )
910 _{10} 1 0 10012 _2 2
If we put little tiny points for the value of each possible fraction on a number line how close will these points be together? Will there be any gaps?
They will be infinitesimally close. There won't be any gaps, because the gap between any two points can always be made smaller by adding a new point between them, ad infinitum.
Think of a fraction which lies between F 1 = 1 100 F_1 = \frac{1}{100} F 1 = 1 0 0 1 F 2 = 1 101 F_2 = \frac{1}{101} F 2 = 1 0 1 1
|
F1
|
F2
| | | |
F3
Midway between F 1 F_1 F 1 F 2 F_2 F 2
F 3 = ( 1 100 + 1 101 ) / 2 = 201 100 ∗ 101 / 2 = 201 ∗ 1 10100 ∗ 2 = 201 20200
F_3 = (\frac{1}{100} + \frac{1}{101}) / 2 = \frac{201}{100 * 101} / 2 = \frac{201 * 1}{10100 * 2} = \frac{201}{20200}
F 3 = ( 1 0 0 1 + 1 0 1 1 ) / 2 = 1 0 0 ∗ 1 0 1 2 0 1 / 2 = 1 0 1 0 0 ∗ 2 2 0 1 ∗ 1 = 2 0 2 0 0 2 0 1
Or easier — this is close to, but not exactly the midway point:
F 3 = 1 100 + 1 / 2 = 1 ∗ 2 ( 100 + 1 / 2 ) ∗ 2 = 2 201
F_3 = \frac{1}{100 + 1/2} = \frac{1 * 2}{(100 + 1/2) * 2} = \frac{2}{201}
F 3 = 1 0 0 + 1 / 2 1 = ( 1 0 0 + 1 / 2 ) ∗ 2 1 ∗ 2 = 2 0 1 2
Why isn't this the midway point? Isn't 100.5 halfway between 100 and 101?
Example: 1 3 \frac{1}{3} 3 1 1 2 \frac{1}{2} 2 1 1 4 \frac{1}{4} 4 1
1 4 = 0.25 + 0.08333 … 1 3 = 0.333 … + 0.1666 … 1 2 = 0.5
\frac{1}{4} = 0.25 \qquad +0.08333\dots \\
\frac{1}{3} = 0.333\dots \qquad +0.1666\dots \\
\frac{1}{2} = 0.5
4 1 = 0 . 2 5 + 0 . 0 8 3 3 3 … 3 1 = 0 . 3 3 3 … + 0 . 1 6 6 6 … 2 1 = 0 . 5
1/1: |---------++++++++++|
1/2: |------+++| |
1/3: |----++| | |
1/4: |---+| | | |
1/5: | | | | | |
The midpoint is exactly at:
( 1 2 + 1 4 ) / 2 = 3 4 / 2 = 3 8
(\frac{1}{2} + \frac{1}{4}) / 2 = \frac{3}{4} / 2 = \frac{3}{8}
( 2 1 + 4 1 ) / 2 = 4 3 / 2 = 8 3
1/2: |---------------| |
1/4: |-------| | | |
3/8: |---|---|---| | | | | |
Does there exist a fraction that equals 2 \sqrt{2} 2
a b = 2 ( a b ) 2 = a 2 b 2 = 2 a 2 = 2 b 2
\frac{a}{b} = \sqrt{2} \qquad (\frac{a}{b})^2 = \frac{a^2}{b^2} = 2 \qquad a^2 = 2b^2
b a = 2 ( b a ) 2 = b 2 a 2 = 2 a 2 = 2 b 2
What kind of number must 2 b 2 2b^2 2 b 2
Couldn't say.
It must be even, so a 2 a^2 a 2
What happens if you square (a) even numbers (b) odd numbers?
(a) The result stays even
(b) The result stays odd
( 2 n ) 2 = 4 n 2 ( 2 n + 1 ) 2 = 4 n 2 + 4 n + 1
(2n)^2 = 4n^2 \\
(2n + 1)^2 = 4n^2 + 4n + 1
( 2 n ) 2 = 4 n 2 ( 2 n + 1 ) 2 = 4 n 2 + 4 n + 1
Remember: even numbers are 2 n 2n 2 n 2 n + 1 2n + 1 2 n + 1
2 b 2 2b^2 2 b 2 a 2 a^2 a 2 a a a
We see that the number a a a 2 a 1 2a_1 2 a 1
a 2 = ( 2 a 1 ) ( 2 a 1 ) = 4 a 1 2 = 2 b 2
a^2 = (2a_1)(2a_1) = 4a_1^2 = 2b^2
a 2 = ( 2 a 1 ) ( 2 a 1 ) = 4 a 1 2 = 2 b 2
Note: here a 1 a_1 a 1 a a a a = 2 a 1 a = 2a_1 a = 2 a 1 a 1 a_1 a 1 a a a x x x a 1 a_1 a 1
a 2 = ( 2 x ) ( 2 x ) = 4 x 2 = 2 b 2
a^2 = (2x)(2x) = 4x^2 = 2b^2
a 2 = ( 2 x ) ( 2 x ) = 4 x 2 = 2 b 2
Remember: often people seem to say x x x y y y a set or a range of possible values , and x 1 x_1 x 1 y 0 y_0 y 0 specific value in that set. See the line equations for examples.
Continuing:
4 x 2 = 2 b 2 2 x 2 = b 2
4x^2 = 2b^2 \\
2x^2 = b^2
4 x 2 = 2 b 2 2 x 2 = b 2
Now, by the same argument as before, b b b a a a b b b
a b = 2
\frac{a}{b} = \sqrt{2}
b a = 2
If both a a a b b b a 2 = 2 b 2 a^2 = 2b^2 a 2 = 2 b 2 b 2 = 2 x 2 b^2 = 2x^2 b 2 = 2 x 2 a a a b b b
2 / 2 = 1 / 1 6 / 4 = 3 / 2 8 / 6 = 4 / 3 14 / 10 = 7 / 5 etc.
2/2 = 1/1 \\
6/4 = 3/2 \\
8/6 = 4/3 \\
14/10 = 7/5 \\
\text{etc.}
2 / 2 = 1 / 1 6 / 4 = 3 / 2 8 / 6 = 4 / 3 1 4 / 1 0 = 7 / 5 etc.
But if we cancel them, we can use exactly the same argument to show that they would cancel down again, and so on for ever. So there is no fraction which is exactly equal to 2 \sqrt{2} 2
Remember:
Rational numbers: a b \frac{a}{b} b a a a a b b b Irrational numbers: π \pi π 2 \sqrt{2} 2 Real numbers: a set of { Rational numbers , Irrational numbers } \{\text{Rational numbers}, \text{Irrational numbers}\} { Rational numbers , Irrational numbers } Imaginary numbers: − 4 \sqrt{-4} − 4 Complex numbers: a set of { Real numbers , Imaginary numbers } \{\text{Real numbers}, \text{Imaginary numbers}\} { Real numbers , Imaginary numbers }
1.F.1
1011 1 2 = 1 + 2 + 4 + 16 = 2 3 10 111 1 2 = 1 5 10 11101 1 2 = 64 − 1 − 4 = 5 9 10 ( 11111 1 2 = 64 − 1 )
10111_2 = 1 + 2 + 4 + 16 = 23_{10} \\
1111_2 = 15_{10} \\
111011_2 = 64 - 1 - 4 = 59_{10} \qquad (111111_2 = 64 - 1)
1 0 1 1 1 2 = 1 + 2 + 4 + 1 6 = 2 3 1 0 1 1 1 1 2 = 1 5 1 0 1 1 1 0 1 1 2 = 6 4 − 1 − 4 = 5 9 1 0 ( 1 1 1 1 1 1 2 = 6 4 − 1 )
A way to convert from base-10 to other bases: continuously divide a number by the new base and keep track of the remainder. When the quotient is 0, the result is in the remainders (read bottom-up).
10 9 10 109_{10} 1 0 9 1 0 109 109 1 0 9
Quotient
Base
Remainder
109
10
10
10
9
1
10
0
0
10
1
10 9 10 109_{10} 1 0 9 1 0 1101101 1101101 1 1 0 1 1 0 1
Quotient
Base
Remainder
109
2
54
2
1
27
2
0
13
2
1
6
2
1
3
2
0
1
2
1
0
2
1
10 9 10 109_{10} 1 0 9 1 0 155 155 1 5 5
Quotient
Base
Remainder
109
8
13
8
5
1
8
5
0
8
1
Why does this work? It's a bit clearer when you add a column for the powers.
Quotient
Base
Power
Remainder
13
2
6
2
0 ^0 0 1
3
2
1 ^1 1 0
1
2
2 ^2 2 1
0
2
3 ^3 3 1
It's even easier to understand when you realize that the table method above is basically chopping off the least power of base-n in each step, and leaving you the next power.
1234510 _{10} 1 0 —> 10 _{10} 1 0 ^
1234510 _{10} 1 0 —> 8 _{8} 8 ^
1234510 _{10} 1 0 —> 2 _{2} 2 ^
Many numbers can be written as products of smaller numbers.
57 = 3 ∗ 19 54 = 2 ∗ 27 = 6 ∗ 9 = 3 ∗ 18 = 3 ∗ 3 ∗ 6 = 2 ∗ 3 ∗ 3 ∗ 3
57 = 3 * 19 \\
54 = 2 * 27 = 6 * 9 = 3 * 18 = 3 * 3 * 6 = 2 * 3 * 3 * 3
5 7 = 3 ∗ 1 9 5 4 = 2 ∗ 2 7 = 6 ∗ 9 = 3 ∗ 1 8 = 3 ∗ 3 ∗ 6 = 2 ∗ 3 ∗ 3 ∗ 3
Remember: numbers which have no factors other than themselves and 1 1 1 prime numbers .
There is only one even prime number: 2 2 2
Every number can be written as a product of its prime factors.
Every number can only be broken down into a product of prime factors in one way.
These are the only prime factorisations for these two numbers — there are no other prime factorisations possible:
57 = 3 ∗ 19 54 = 2 ∗ 3 ∗ 3 ∗ 3 = 2 ∗ 3 3
57 = 3 * 19 \\
54 = 2 * 3 * 3 * 3 = 2 * 3^3
5 7 = 3 ∗ 1 9 5 4 = 2 ∗ 3 ∗ 3 ∗ 3 = 2 ∗ 3 3
Square roots can be simplified with prime numbers. Factoring the number inside the square root into primes allows you to calculate the square roots of the factors.
8 = 2 ∗ 2 ∗ 2 = 4 ∗ 2 = 2 2 54 = 2 ∗ 3 ∗ 3 ∗ 3 = 9 ∗ 2 ∗ 3 = 3 6 27 = 3 2 ∗ 3 = 3 2 ∗ 3 = 3 3
\sqrt{8} = \sqrt{2 * 2 * 2} = \sqrt{4} * \sqrt{2} = 2\sqrt{2} \\
\sqrt{54} = \sqrt{2 * 3 * 3 * 3} = \sqrt{9} * \sqrt{2 * 3} = 3\sqrt{6} \\
\sqrt{27} = \sqrt{3^2 * 3} = \sqrt{3^2} * \sqrt{3} = 3\sqrt{3}
8 = 2 ∗ 2 ∗ 2 = 4 ∗ 2 = 2 2 5 4 = 2 ∗ 3 ∗ 3 ∗ 3 = 9 ∗ 2 ∗ 3 = 3 6 2 7 = 3 2 ∗ 3 = 3 2 ∗ 3 = 3 3
Btw, can a ∗ b \sqrt{a * b} a ∗ b a ∗ b \sqrt{a} * \sqrt{b} a ∗ b
16 = 4 2 = 4 = 4 4 = ( 4 ) 2 = ( 2 ∗ 2 ) = 4 16 = 1 6 1 / 2 = 4 1 / 2 ∗ 4 1 / 2 = ( 4 1 ) = 4 = 2 1 / 2 ∗ 2 1 / 2 ∗ 2 1 / 2 ∗ 2 1 / 2 = ( 2 2 ) = 4
\sqrt{16} = \sqrt{4^2} = 4 = \sqrt{4}\sqrt{4} = (\sqrt{4})^2 = (2*2) = 4 \\
\sqrt{16} = 16^{1/2} = 4^{1/2} * 4^{1/2} = (4^1) = 4 = 2^{1/2} * 2^{1/2} * 2^{1/2} * 2^{1/2} = (2^2) = 4
1 6 = 4 2 = 4 = 4 4 = ( 4 ) 2 = ( 2 ∗ 2 ) = 4 1 6 = 1 6 1 / 2 = 4 1 / 2 ∗ 4 1 / 2 = ( 4 1 ) = 4 = 2 1 / 2 ∗ 2 1 / 2 ∗ 2 1 / 2 ∗ 2 1 / 2 = ( 2 2 ) = 4
Looks like it :
For all nonnegative real numbers x and y, x y = x y \sqrt{xy} = \sqrt{x}\sqrt{y} x y = x y
1.F.3
28 = 2 ∗ 2 ∗ 7 = 2 7 45 = 3 ∗ 3 ∗ 5 = 3 5 50 = 2 ∗ 5 ∗ 5 = 5 2 44 = 2 ∗ 2 ∗ 11 = 2 11 63 = 3 ∗ 3 ∗ 7 = 3 7 40 = 2 ∗ 2 ∗ 2 ∗ 5 = 2 2 5 = 2 10
\sqrt{28} = \sqrt{2 * 2 * 7} = 2\sqrt{7} \\
\sqrt{45} = \sqrt{3 * 3 * 5} = 3\sqrt{5} \\
\sqrt{50} = \sqrt{2 * 5 * 5} = 5\sqrt{2} \\
\sqrt{44} = \sqrt{2 * 2 * 11} = 2\sqrt{11} \\
\sqrt{63} = \sqrt{3 * 3 * 7} = 3\sqrt{7} \\
\sqrt{40} = \sqrt{2 * 2 * 2 * 5} = 2\sqrt{2}\sqrt{5} = 2\sqrt{10}
2 8 = 2 ∗ 2 ∗ 7 = 2 7 4 5 = 3 ∗ 3 ∗ 5 = 3 5 5 0 = 2 ∗ 5 ∗ 5 = 5 2 4 4 = 2 ∗ 2 ∗ 1 1 = 2 1 1 6 3 = 3 ∗ 3 ∗ 7 = 3 7 4 0 = 2 ∗ 2 ∗ 2 ∗ 5 = 2 2 5 = 2 1 0
Seems easiest to first try to divide by 2, then by 3, then by 5, etc. — i.e. by incrementally larger primes.
Most square roots are irrational, the exception being perfect squares , like 2 = 4 2 = \sqrt{4} 2 = 4 5 = 25 5 = \sqrt{25} 5 = 2 5 surds
1.F.(d)
( 3 + 5 ) + ( 3 − 5 ) = 6 ( 3 + 5 ) ( 3 − 5 ) = 3 2 − 5 2 = 9 − 5 = 4 ( 1 + 2 ) ( 1 − 2 ) = 1 − 2 = − 1
(3 + \sqrt{5}) + (3 - \sqrt{5}) = 6 \\
(3 + \sqrt{5})(3 - \sqrt{5}) = 3^2 - \sqrt{5}^2 = 9 - 5 = 4 \\
(1 + \sqrt{2})(1 - \sqrt{2}) = 1 - 2 = -1
( 3 + 5 ) + ( 3 − 5 ) = 6 ( 3 + 5 ) ( 3 − 5 ) = 3 2 − 5 2 = 9 − 5 = 4 ( 1 + 2 ) ( 1 − 2 ) = 1 − 2 = − 1
We can use the difference of two squares to simplify square roots in fractions.
3 − 2 5 − 2 = ( 3 − 2 ) ( 5 + 2 ) ( 5 − 2 ) ( 5 + 2 ) = 3 5 + 3 2 − 10 − 2 3
\dfrac{3 - \sqrt{2}}{\sqrt{5} - \sqrt{2}} = \dfrac{(3 - \sqrt{2})(\sqrt{5} + \sqrt{2})}{(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})} = \dfrac{3\sqrt{5} + 3\sqrt{2} - \sqrt{10} - 2}{3}
5 − 2 3 − 2 = ( 5 − 2 ) ( 5 + 2 ) ( 3 − 2 ) ( 5 + 2 ) = 3 3 5 + 3 2 − 1 0 − 2
This is called rationalising the denominator .
1.F.4
5 3 + 2 = 5 ( 3 − 2 ) ( 3 + 2 ) ( 3 − 2 ) = 15 − 5 2 7
\dfrac{5}{3 + \sqrt{2}} = \dfrac{5(3 - \sqrt{2})}{(3 + \sqrt{2})(3 - \sqrt{2})} = \dfrac{15 - 5\sqrt{2}}{7}
3 + 2 5 = ( 3 + 2 ) ( 3 − 2 ) 5 ( 3 − 2 ) = 7 1 5 − 5 2
3 − 5 3 + 5 = ( 3 − 5 ) ( 3 − 5 ) ( 3 + 5 ) ( 3 − 5 ) = 9 − 3 5 − 3 5 + 5 9 − 5 = 14 − 6 5 4 = 7 − 3 5 2
\dfrac{3 - \sqrt{5}}{3 + \sqrt{5}} = \dfrac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \dfrac{9 - 3\sqrt{5} - 3\sqrt{5} + 5}{9 - 5} = \dfrac{14 - 6\sqrt{5}}{4} = \dfrac{7 - 3\sqrt{5}}{2}
3 + 5 3 − 5 = ( 3 + 5 ) ( 3 − 5 ) ( 3 − 5 ) ( 3 − 5 ) = 9 − 5 9 − 3 5 − 3 5 + 5 = 4 1 4 − 6 5 = 2 7 − 3 5
3 − 2 3 5 + 3 2 = … ?
\dfrac{3 - 2\sqrt{3}}{5 + 3\sqrt{2}} = \dots?
5 + 3 2 3 − 2 3 = … ?
( 3 2 ) 2 = ( 3 1 ∗ 2 1 / 2 ) 2 3 ( 3 − 2 3 ) = 9 − 6 3 3 ( 5 + 3 2 ) = 15 + 9 2 2 ( 3 − 2 3 ) = 6 − 4 3 2 ( 5 + 3 2 ) = 10 + 6 2 3 ( 3 − 2 3 ) = 3 3 − 2 ∗ 3 = 3 3 − 6 2 ( 5 + 3 2 ) = 5 2 + 6 3 ( 5 + 3 2 ) = 5 3 + 3 2 3 = 5 3 + 3 5 ( 5 + 3 2 ) ( 5 − 3 2 ) = 25 − ( 3 2 ) ( 3 2 ) = 25 − 9 ∗ 2
(3\sqrt{2})^2 = (3^1 * 2^{1/2})^2 \\
3(3 - 2\sqrt{3}) = 9 - 6\sqrt{3} \\
3(5 + 3\sqrt{2}) = 15 + 9\sqrt{2} \\
2(3 - 2\sqrt{3}) = 6 - 4\sqrt{3} \\
2(5 + 3\sqrt{2}) = 10 + 6\sqrt{2} \\
\sqrt{3}(3 - 2\sqrt{3}) = 3\sqrt{3} - 2*3 = 3\sqrt{3} - 6 \\
\sqrt{2}(5 + 3\sqrt{2}) = 5\sqrt{2} + 6 \\
\sqrt{3}(5 + 3\sqrt{2}) = 5\sqrt{3} + 3\sqrt{2}\sqrt{3} = 5\sqrt{3} + 3\sqrt{5} \\
(5 + 3\sqrt{2})(5 - 3\sqrt{2}) = 25 - (3\sqrt{2})(3\sqrt{2}) = 25 - 9*2
( 3 2 ) 2 = ( 3 1 ∗ 2 1 / 2 ) 2 3 ( 3 − 2 3 ) = 9 − 6 3 3 ( 5 + 3 2 ) = 1 5 + 9 2 2 ( 3 − 2 3 ) = 6 − 4 3 2 ( 5 + 3 2 ) = 1 0 + 6 2 3 ( 3 − 2 3 ) = 3 3 − 2 ∗ 3 = 3 3 − 6 2 ( 5 + 3 2 ) = 5 2 + 6 3 ( 5 + 3 2 ) = 5 3 + 3 2 3 = 5 3 + 3 5 ( 5 + 3 2 ) ( 5 − 3 2 ) = 2 5 − ( 3 2 ) ( 3 2 ) = 2 5 − 9 ∗ 2
I was unsure there if ( a b ) 2 = a 2 b (a\sqrt{b})^2 = a^2b ( a b ) 2 = a 2 b ( a b ) ( a b ) = a b a b = a a b b = a 2 b 2 (ab)(ab) = abab = aabb = a^2b^2 ( a b ) ( a b ) = a b a b = a a b b = a 2 b 2
3 − 2 3 5 + 3 2 = ( 3 − 2 3 ) ( 5 − 3 2 ) ( 5 + 3 2 ) ( 5 − 3 2 ) = 15 − 9 2 − 10 3 + 6 2 3 25 − 9 ∗ 2 =
\dfrac{3 - 2\sqrt{3}}{5 + 3\sqrt{2}} = \dfrac{(3 - 2\sqrt{3})(5 - 3\sqrt{2})}{(5 + 3\sqrt{2})(5 - 3\sqrt{2})} = \dfrac{15 - 9\sqrt{2} - 10\sqrt{3} + 6\sqrt{2}\sqrt{3}}{25 - 9*2} =
5 + 3 2 3 − 2 3 = ( 5 + 3 2 ) ( 5 − 3 2 ) ( 3 − 2 3 ) ( 5 − 3 2 ) = 2 5 − 9 ∗ 2 1 5 − 9 2 − 1 0 3 + 6 2 3 =
15 − 9 2 − 10 3 + 6 6 7
\dfrac{15 - 9\sqrt{2} - 10\sqrt{3} + 6\sqrt{6}}{7}
7 1 5 − 9 2 − 1 0 3 + 6 6
Remember: x y = x y \boxed{\sqrt{x}\sqrt{y} = \sqrt{xy}} x y = x y x + y ≠ x + y \sqrt{x} + \sqrt{y} \neq \sqrt{x + y} x + y = x + y
Remember : ( a b ) 2 = ( a b ) ( a b ) = a b a b = a a b b = a 2 b 2 (ab)^2 = (ab)(ab) = abab = aabb = a^2b^2 ( a b ) 2 = ( a b ) ( a b ) = a b a b = a a b b = a 2 b 2
