Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
2. Section 3.2
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Exercise 113 Page 157

Practice makes perfect
a The fractions have the same denominator so we can add them by adding the numerators and keeping the denominator.

2x/3x^2+16x+5+10/3x^2+16x+5
2x+10/3x^2+16x+5
2(x+5)/3x^2+16x+5

Maybe we can factor out a Giant One here. Let's factor the denominator and find out.

2(x+5)/3x^2+16x+5
â–¼
Factor
2(x+5)/3x^2+x+15x+5
2(x+5)/x(3x+1)+15x+5
2(x+5)/x(3x+1)+5(3x+1)
2(x+5)/(x+5)(3x+1)
2(x+5)/(3x+1)(x+5)
2/3x+1*x+5/x+5
2/3x+1*1
2/3x+1

b Before multiplying the fractions, let's factor every polynomial. We can start with the numerator in the first fraction.

x^2-x-12/3x^2-11x-4*3x^2-20x-7/x^2-9 Let's factor the first expression.

x^2-x-12
â–¼
Factor
x^2-4x+3x-12
x(x-4)+3x-12
x(x-4)+3(x-4)
(x+3)(x-4)

Now we move on to the denominator in the first fraction.

(x+3)(x-4)/3x^2-11x-4*3x^2-20x-7/x^2-9 Let's factor the second expression.

3x^2-11x-4
â–¼
Factor
3x^2-12x+x-4
3x(x-4)+x-4
3x(x-4)+(x-4)
(3x+1)(x-4)

Next, the numerator in the second fraction. (x+3)(x-4)/(3x+1)(x-4)*3x^2-20x-7/x^2-9 Let's factor the third expression.

3x^2-20x-7
â–¼
Factor
3x^2-21x+x-7
3x(x-7)+x-7
3x(x-7)+(x-7)
(3x+1)(x-7)

The denominator in the second fraction is a difference of squares. We can use that to factor it. (x+3)(x-4)/(3x+1)(x-4)*(3x+1)(x-7)/x^2-9 Let's factor the fourth expression.

x^2-9
x^2-3^2
(x+3)(x-3)

We have arrived at the fully factored fractions. (x+3)(x-4)/(3x+1)(x-4)*(3x+1)(x-7)/(x+3)(x-3) Let's simplify this.

(x+3)(x-4)/(3x+1)(x-4)*(3x+1)(x-7)/(x+3)(x-3)
(x+3)(x-4)(3x+1)(x-7)/(3x+1)(x-4)(x+3)(x-3)
(x+3)(x-4)(3x+1)(x-7)/(x+3)(x-4)(3x+1)(x-3)
(x+3)(x-4)(3x+1)/(x+3)(x-4)(3x+1)*x-7/x-3
1*x-7/x-3
x-7/x-3

c

Similar to the previous exercise, we want to factor each polynomial before carrying out the division. First, let's do the numerator in the first fraction.

2x^2+8x-10/2x^2+15x+25÷4x^2+10x-24/2x^2+x-10 Let's factor the first expression.

2x^2+8x-10
â–¼
Factor
2(x^2+4x-5)
2(x^2-x+5x-5)
2(x(x-1)+5x-5)
2(x(x-1)+5(x-1))
2(x+5)(x-1)
Next, the denominator in the first fraction. 2(x+5)(x-1)/2x^2+15x+25÷4x^2+10x-24/2x^2+x-10 Let's factor the second expression.

2x^2+15x+25
â–¼
Factor
2x^2+10x+5x+25
2x(x+5)+5x+25
2x(x+5)+5(x+5)
(2x+5)(x+5)

Moving on to the numerator in the second fraction. 2(x+5)(x-1)/(2x+5)(x+5)÷4x^2+10x-24/2x^2+x-10 Let's factor the third expression.

4x^2+10x-24
â–¼
Factor
4(x^2+5x-6)
4(x^2-x+6x-6)
4(x(x-1)+6x-6)
4(x(x-1)+6(x-1))
4(x+6)(x-1)

Lastly, we factor the denominator in the second fraction. 2(x+5)(x-1)/(2x+5)(x+5)÷4(x+6)(x-1)/2x^2+x-10 Let's factor the fourth expression.

2x^2+x-10
â–¼
Factor
2x^2-4x+5x-10
2x(x-2)+5x-10
2x(x-2)+5(x-2)
(2x+5)(x-2)

Now we substitute the factored forms into the original expression and divide. 2(x+5)(x-1)/(2x+5)(x+5)÷4(x+6)(x-1)/(2x+5)(x-2) Remember that when dividing two fractions, the fraction in the denominator is inverted and multiplied by the numerator.

2(x+5)(x-1)/(2x+5)(x+5)÷4(x+6)(x-1)/(2x+5)(x-2)
2(x+5)(x-1)/(2x+5)(x+5)*(2x+5)(x-2)/4(x+6)(x-1)
2(x+5)(x-1)(2x+5)(x-2)/(2x+5)(x+5)* 4(x+6)(x-1)
2(x-2)(2x+5)(x+5)(x-1)/4(x+6)(2x+5)(x+5)(x-1)
2(x-2)/4(x+6)*(2x+5)(x+5)(x-1)/(2x+5)(x+5)(x-1)
2(x-2)/4(x+6)*1
2(x-2)/4(x+6)
x-2/2(x+6)

d To subtract two fractions they need to have the same numerator. 7/x+5-4-6x/x^2+10x+25 These don't. Let's factor the second denominator and see if we can find a common denominator.

x^2+10x+25
â–¼
Factor
x^2+5x+5x+25
x(x+5)+5x+25
x(x+5)+5(x+5)
(x+5)(x+5)

Now we notice that by multiplying the denominator in the first fraction by (x+5) the fractions get the same denominator. But we have to multiply the numerator with the same factor to make sure that we do not change the original expression.

7/x+5-4-6x/x^2+10x+25
7/x+5-4-6x/(x+5)(x+5)
7(x+5)/(x+5)(x+5)-4-6x/(x+5)(x+5)
7(x+5)-(4-6x)/(x+5)(x+5)
7x+35-(4-6x)/(x+5)(x+5)
7x+35-4+6x/(x+5)(x+5)
13x+31/(x+5)(x+5)
13x+31/(x+5)^2