Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
2. Section 4.2
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Exercise 71 Page 189

Practice makes perfect
a To factor an expression, we want to find the greatest common factor that's shared by both terms. In the given expression, we can see that both terms contains x. Therefore, we can factor out x.

bx+ax=x(b+a)

b The greatest common factor of the terms in the expression is x. Therefore, we can factor out x.

x+ax=x(1+a)

c To be able to simplify the fraction we must factor both the numerator and denominator. The numerator, we can factor by pulling out an a. The denominator, we can factor since it is a perfect square trinomial.
ax+a/x^2 + 2 x + 1
a(x+1)/x^2 + 2 x + 1
a(x+1)/x^2+2(x)(1) + 1
a(x+1)/x^2+2(x)(1) + 1^2
a(x+1)/(x + 1)^2
Now we can simplify the fraction by pulling out a Giant 1.
ax+a/(x+1)^2
a(x+1)/(x+1)^2
a(x+1)/(x+1)(x+1)
a/x+1*x+1/x+1
a/x+1*1
a/x+1
d Like in previous parts, we simply the fraction by factoring the numerator and denominator. The numerator contains a difference of squares which means it can be factored. In the denominator, we can factor out the a.
x^2-b^2/ax+ab
(x+b)(x-b)/ax+ab
(x+b)(x-b)/a(x+b)
Now we can simplify the fraction by pulling out a Giant 1.
(x+b)(x-b)/a(x+b)
(x-b)(x+b)/a(x+b)
x-b/a*x+b/x+b
x-b/a*1
x-b/a