Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 4.2
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Exercise 69 Page 188

Practice makes perfect
a We want to add the given rational expressions.
x-4/2x^2+9x-5 + x+3/x^2+5x To add the fractions they need to have the same denominator. Therefore, let's start by factoring the denominator.
2x^2+9x-5
â–Ľ
Factor
2x^2+10x-x-5
2x(x+5)-x-5
2x(x+5)-1(x+5)
(2x-1)(x+5)
Next, we will rewrite the denominator in the second fraction. Notice that we will substitute the denominator in the first fraction with its factored form.
x-4/(2x-1)(x+5)+x+3/x^2+5x
x-4/(2x-1)(x+5)+x+3/x(x+5)
(x-4)x/(2x-1)(x+5)x+x+3/x(x+5)
(x-4)x/(2x-1)(x+5)x+(x+3)(2x-1)/x(x+5)(2x-1)
x(x-4)/x(2x-1)(x+5)+(x+3)(2x-1)/x(2x-1)(x+5)
x(x-4)+(x+3)(2x-1)/x(2x-1)(x+5)
The fractions have now been added. Since it is not possible to factor out a common factor in the numerator, we cannot simplify the fraction. However, we can simplify the numerator if we distribute and multiply the parentheses.
x(x-4)+(x+3)(2x-1)/x(2x-1)(x+5)
â–Ľ
Simplify numerator
x^2-4x+(x+3)(2x-1)/x(2x-1)(x+5)
x^2-4x+2x^2-x+6x-3/x(2x-1)(x+5)
x^2+2x^2-4x-x+6x-3/x(2x-1)(x+5)
3x^2+x-3/x(2x-1)(x+5)
b We are given the difference between two rational expressions.
4x^2-11x+6/2x^2-x-6 - x+2/2x+3 The first step is to rewrite the fractions so that they have the same denominator. Since the first denominator is a quadratic trinomial, we can try to factor it.
2x^2-x-6
â–Ľ
Factor
2x^2+3x-4x-6
x(2x+3)-4x-6
x(2x+3)-2(2x+3)
(x-2)(2x+3)
To get the same denominator, we need to expand the second fraction with (x-2). Notice that we will substitute the denominator in the first fraction with its factored form.
4x^2-11x+6/(x-2)(2x+3)-x+2/2x+3
4x^2-11x+6/(x-2)(2x+3)-(x+2)(x-2)/(2x+3)(x-2)
4x^2-11x+6/(x-2)(2x+3)-x^2-4/(2x+3)(x-2)
4x^2-11x+6/(x-2)(2x+3)-x^2-4/(x-2)(2x+3)
4x^2-11x+6-(x^2-4)/(x-2)(2x+3)
4x^2-11x+6-x^2+4/(x-2)(2x+3)
4x^2-x^2-11x+6+4/(x-2)(2x+3)
3x^2-11x+10/(x-2)(2x+3)
To simplify the fraction further we want to factor the numerator.
3x^2-11x+10/(x-2)(2x+3)
â–Ľ
Factor
3x^2-6x-5x+10/(x-2)(2x+3)
3x(x-2)-5x+10/(x-2)(2x+3)
3x(x-2)-5(x-2)/(x-2)(2x+3)
(3x-5)(x-2)/(x-2)(2x+3)
(3x-5)(x-2)/(2x+3)(x-2)
3x-5/2x+3*x-2/x-2
3x-5/2x+3*1
3x-5/2x+3
c Since the numerator and denominator are in factored form already, we can carry out the division straight away. Dividing two fractions is the same thing as multiplying the first one with the second fraction's reciprocal.
(x+4)(2x-1)(x-7)/(x+8)(2x-1)(3x-4) Ă· (4x-3)(x-7)/(x+8)(3x-4)
(x+4)(2x-1)(x-7)/(x+8)(2x-1)(3x-4) * (x+8)(3x-4)/(4x-3)(x-7)
(x+4)(2x-1)(x-7)(x+8)(3x-4)/(x+8)(2x-1)(3x-4)(4x-3)(x-7)
(x+4)(x+8)(2x-1)(3x-4)(x-7)/(4x-3)(x+8)(2x-1)(3x-4)(x-7)
x+4/4x-3*(x+8)(2x-1)(3x-4)(x-7)/(x+8)(2x-1)(3x-4)(x-7)
x+4/4x-3*1
x+4/4x-3
d We want to multiply the given rational expressions.
2m^2+7m-15/m^2-16*m^2-6m+8/2m^2-7m+6 Before we multiply the fractions, we want to factor each expression to make the multiplication easier. Let's start with the numerator in the first fraction.
2m^2+7m-15
â–Ľ
Factor
2m^2+10m-3m-15
2m(m+5)-3m-15
2m(m+5)-3(m+5)
(2m-3)(m+5)
We will continue with the denominator, which we can factor to a product of two binomials.
m^2-16
m^2-4^2
(m+4)(m-4)
We now have the second fraction left, and once again we will start with the numerator.
m^2-6m+8
â–Ľ
Factor
m^2-4m-2m+8
m(m-4)-2m+8
m(m-4)-2(m-4)
(m-2)(m-4)
Finally, we want to factor the denominator in the second fraction.
2m^2-7m+6
â–Ľ
Factor
2m^2-4m-3m+6
2m(m-2)-3m+6
2m(m-2)-3(m-2)
(2m-3)(m-2)
We can now substitute the polynomials with their factored form and then multiply the fractions and try to find Giant Ones.
(2m-3)(m+5)/(m+4)(m-4)*(m-2)(m-4)/(2m-3)(m-2)
(2m-3)(m+5)(m-2)(m-4)/(m+4)(m-4)(2m-3)(m-2)
(m+5)(2m-3)(m-2)(m-4)/(m+4)(2m-3)(m-2)(m-4)
m+5/m+4*(2m-3)(m-2)(m-4)/(2m-3)(m-2)(m-4)
m+5/m+4*1
m+5/m+4