Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 2.2
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Exercise 98 Page 90

Practice makes perfect
a Examining the algebraic expression, we notice that both of terms contain a squared variable.
4 x^2-9 y^2Since both coefficients are also perfect squares, each term can be rewritten as a single power.
4x^2-9y^2
2^2x^2-3^2y^2
(2x)^2-(3y)^2
We now have a difference of squares. Therefore this can be factored.
(2x)^2-(3y)^2
(2x+3y)(2x-3y)
b Examining the expression, we notice that 8x^3 and 2x^7 contain common factors. Therefore, we should start by breaking them down into factors until we identify these common factors.
2* 4 * x* x * x_(8x^3) - 2* x* x* x * x^4_(2x^7)The terms share one 2 and three x. Therefore, the greatest possible factor we can pull out from both of them is 2x^3 8x^3-2x^7 ⇔ 2x^3( 4-x^4 ) Looking at the parentheses, we see that 4 is a perfect square and so is x^4 if we rewrite it as a power of a power. Therefore, like in Part A, we can further factor this by rewriting the expression in the parentheses as a difference of squares.
2x^3( 4-x^4 )
â–Ľ
Simplify
2x^3(2^2-x^4 )
2x^3(2^2-x^(2* 2))
2x^3(2^2-(x^2)^2)
2x^3(2-x^2)(2+x^2)
c Looking at the expression, we notice that both terms can be rewritten as perfect squares. We can then factor this as a difference of squares.
x^4-81y^4
x^(2*2)-81y^(2* 2)
x^(2*2)-9^2y^(2* 2)
(x^2)^2-9^2(y^2)^2
(x^2)^2-(9y^2)^2
(x^2+9y^2)(x^2-9y^2)
Examining the expression inside the second parentheses, we notice that we can factor this once more by rewriting it as a difference of squares.
(x^2+9y^2)(x^2-9y^2)
(x^2+9y^2)(x^2-3^2y^2)
(x^2+9y^2)(x^2-(3y)^2)
(x^2+9y^2)(x+3y)(x-3y)
d If we compare this to Part B, we see that the only thing that is different is the operation between the terms.

8x^3 - 2x^7 8x^3 + 2x^7 In Part B, we identified the greatest common factor between 8x^3 and 2x^7 to be 2x^3. We can still factor our expression by pulling out this greatest common factor but that is how far we can go. 8x^3+2x^7 ⇔ 2x^3 (4+x^4 )

e Whenever we can write an expression as a difference of squares it can be factored to a product of a conjugate pair of binomials.

a^2-b^2 = (a+b)(a-b)