McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
2. Adding and Subtracting Rational Expressions
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Exercise 20 Page 541

Start by factoring each expression.

(x-5)(x-4)(x+6)

Practice makes perfect
We want to find the least common multiple (LCM) of the given pair of polynomials. x^2-9x+20 and x^2+x-30 To do so, we first need to find the prime factors of each expression. Since the degree of the first polynomial is 2, we can use the Quadratic Formula to help us factor.
x=- b±sqrt(b^2-4ac)/2a
x=- ( - 9)±sqrt(( - 9)^2-4( 1)( 20))/2( 1)
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Simplify right-hand side
x=9±sqrt((- 9)^2-4(1)(20))/2(1)
x=9±sqrt(81-4(1)(20))/2(1)
x=9±sqrt(81-4(20))/2
x=9±sqrt(81-80)/2
x=9±sqrt(1)/2
x=9 ± 1/2

The solutions for this equation are x= 9 ± 12. Let's separate them into the positive and negative cases.

x=9 ± 1/2
x_1=9+1/2 x_2=9-1/2
x_1=10/2 x_2=8/2
x_1=5 x_2=4
Knowing that x= 5 and x= 4 are the zeros of the trinomial, we can write the factored form of the quadratic expression. (x- 5)(x- 4) Let's factor the second expression. Since the degree of the second polynomial is also 2, we can use the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- 1±sqrt(1^2-4( 1)( - 30))/2( 1)
â–Ľ
Simplify right-hand side
x=- 1±sqrt(1-4(1)(- 30))/2(1)
x=- 1±sqrt(1-4(- 30))/2
x=- 1±sqrt(1+120)/2
x=- 1±sqrt(121)/2
x=- 1 ± 11/2
The solutions for this equation are x= - 1 ± 112. Let's separate them into the positive and negative cases.
x=- 1 ± 11/2
x_1=- 1+11/2 x_2=- 1-11/2
x_1=10/2 x_2=- 12/2
x_1=5 x_2=- 6

Knowing that x= 5 and x= - 6 are the zeros of the trinomial, we can write the factored form of the quadratic expression. (x- 5)(x-( - 6)) ⇔ (x-5)(x+6) We can now write both expressions as the product of their prime factors. x^2-9x+20 &= (x-5) (x-4) x^2+x-30 &= (x-5) (x+6) Finally, we write the product of the prime factors, each raised to the greatest power that occurs in their expression. (x-5) (x-4) (x+6)