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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

Mid-Chapter Quiz

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Exercise 14 Page 224

The formula to factor the difference of two squares is a^2-b^2=(a+b)(a-b).

(2y+3)(2y-3)

Practice makes perfect

Look closely at the expression 4y^2-9. It can be expressed as the difference of two perfect squares.

4y^2-9

Write as a power

2^2 y^2-3^2

a^m b^m = (a b)^m

(2y)^2-3^2

Recall the formula to factor a difference of two squares. a^2- b^2 ⇔ ( a+ b)( a- b) We can apply this formula to our expression. ( ( 2y)^2- 3^2 ) ⇔ ( 2y+ 3)( 2y- 3)

Checking Our Answer

Check your answer ✓

We can apply the Distributive Property and compare the result with the given expression.

(2y+3)(2y-3)

Distribute 2y-3

2y(2y-3)+3(2y-3)

Distribute 2y

4y^2 -6y+3(2y-3)

Distribute 3

4y^2 -6y+6y-9

Add terms

4y^2-9

After applying the Distributive Property and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!