Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 3.2
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Exercise 107 Page 154

Practice makes perfect
a When multiplying two fractions the numerator and denominator can be multiplied separately. Then we can factor out a Giant One.
(3x-1)(x+7)/4(2x-5)*10(2x-5)/(4x+1)(x+7)
(3x-1)(x+7)* 10(2x-5)/4(2x-5)(4x+1)(x+7)
10(3x-1)(x+7)(2x-5)/4(4x+1)(x+7)(2x-5)
10(3x-1)/4(4x+1)*(x+7)(2x-5)/(x+7)(2x-5)
10(3x-1)/4(4x+1)*1
10(3x-1)/4(4x+1)
5(3x-1)/2(4x+1)
b Let's start by simplifying the numerator and denominator separately.
(m-3)(m+11)/(2m+5)(m-3)Ă·(4m-3)(m+11)/(4m-3)(2m+5)
(m-3)(m+11)/(m-3)(2m+5)Ă·(4m-3)(m+11)/(4m-3)(2m+5)
(m-3/m-3*m+11/2m+1)Ă·(4m-3/4m-3*m+11/2m+5)
(1*m+11/2m+1)Ă·(1*m+11/2m+5)
m+11/2m+5Ă·m+11/2m+5
To divide fractions, we take the reciprocal of the second fraction and multiply them together. m+11/2m+5÷m+11/2m+5 [0.5em] ⇕ [0.5em] m+11/2m+5*2m+5/m+11 Look at the numerator and denominator of the fractions — they're the same. Therefore, this expression can be simplified to 1.
c We have another expression to multiply. To simplify, let's factor all polynomials before we multiply.
2p^2+5p-12/2p^2-5p+3*p^2+8p-9/3p^2-10p-8 Let's factor the first one.
2p^2+5p-12
â–Ľ
Factor
2p^2+8p-3p-12
2p(p+4)-3p-12
2p(p+4)-3(p+4)
(2p-3)(p+4)
The numerator of the first fraction can be simplified to (2p-3)(p+4). (2p-3)(p+4)/2p^2-5p+3*p^2+8p-9/3p^2-10p-8 Let's factor the second.
2p^2-5p+3
â–Ľ
Factor
2p^2-2p-3p+3
2p(p-1)-3p+3
2p(p-1)-3(p-1)
(2p-3)(p-1)
The factored form of the denominator in the first fraction is (2p-3)(p-1). (2p-3)(p+4)/(2p-3)(p-1)*p^2+8p-9/3p^2-10p-8 Let's factor the third.
p^2+8p-9
â–Ľ
Factor
p^2+9p-p-9
p(p+9)-p-9
p(p+9)-(p+9)
(p-1)(p+9)
In the second fraction, the numerator can be factored to (p-1)(p+9). (2p-3)(p-1)/(2p-3)(p-1)*(p-1)(p+9)/3p^2-10p-8 Let's factor the fourth.
3p^2-10p-8
â–Ľ
Factor
3p^2+12p-2p-8
3p(p+4)-2p-8
3p(p+4)-2(p+4)
(3p-2)(p+4)
Now we have the factored forms. (2p-3)(p-1)/(2p-3)(p-1)*(p-1)(p+9)/(3p-2)(p+4) Let's simplify this.
(2p-3)(p+4)/(2p-3)(p-1)*(p-1)(p+9)/(3p-2)(p+4)
(2p-3)(p+4)(p-1)(p+9)/(2p-3)(p-1)(3p-2)(p+4)
(2p-3)(p+4)(p-1)(p+9)/(2p-3)(p+4)(p-1)(3p-2)
(2p-3)(p+4)(p-1)/(2p-3)(p+4)(p-1)* p+9/3p-2
1* p+9/3p-2
p+9/3p-2
d We have one more fraction to divide.
4x-12/x^2+3x-10Ă·2x^2-13x+21/2x^2+3x-35 Looking at the equation, we can tell that the numerator in the first fraction can be factored as 4x-12 = 4(x-3). Now let's factor the rest of the polynomials. 4(x-3)/x^2+3x-10Ă·2x^2-13x+21/2x^2+3x-35Let's factor the first expression.
x^2+3x-10
â–Ľ
Factor
x^2+5x-2x-10
x(x+5)-2x-10
x(x+5)-2(x+5)
(x-2)(x+5)
The denominator in the first fraction is (x-2)(x+5). 4(x-3)/(x-2)(x+5)Ă·2x^2-13x+21/2x^2+3x-35 Let's factor the second expression.
2x^2-13x+21
â–Ľ
Factor
2x^2-6x-7x+21
2x(x-3)-7x+21
2x(x-3)-7(x-3)
(2x-7)(x-3)
The numerator in the second fraction can be factored to (2x-7)(x-3). 4(x-3)/(x-2)(x+5)Ă·(2x-7)(x-3)/2x^2+3x-35 Let's factor the third expression.
2x^2+3x-35
â–Ľ
Factor
2x^2+10x-7x-35
2x(x+5)-7x-35
2x(x+5)-7(x+5)
(2x-7)(x+5)
Now we have the factored form. 4(x-3)/(x-2)(x+5)Ă·(2x-7)(x-3)/(2x-7)(x+5) Let's simplify this.
4(x-3)/(x-2)(x+5)Ă·(2x-7)(x-3)/(2x-7)(x+5)
4(x-3)/(x-2)(x+5)Ă·(2x-7/2x-7*x-3/x+5)
4(x-3)/(x-2)(x+5)Ă·(1*x-3/x+5)
4(x-3)/(x-2)(x+5)Ă·x-3/x+5
4(x-3)/(x-2)(x+5)*x+5/x-3
4(x-3)(x+5)/(x-2)(x+5)(x-3)
4(x-3)(x+5)/(x-2)(x-3)(x+5)
4/x-2*(x-3)(x+5)/(x-3)(x+5)
4/x-2*1
4/x-2