McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
8. Differences of Squares
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Exercise 8 Page 60

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

(16n^2+c^2)(4n+c)(4n-c)

Practice makes perfect
Look closely at the expression 256n^4-c^4. It can be expressed as the difference of two perfect squares.
256n^4-c^4
16^2(n^2)^2-(c^2)^2
(16n^2)^2-(c^2)^2
Recall the formula to factor a difference of squares. a^2- b^2 ⇔ ( a+ b)( a- b) We can apply this formula to our expression. ( 16n^2)^2-( c^2)^2 ⇕ ( 16n^2+ c^2)( 16n^2- c^2) Now, let's take a look at the expression 16n^2-c^2. Once again, we can express it as a difference of two perfect squares. Let's do it!
16n^2-c^2
4^2n^2-c^2
(4n)^2-c^2
Now, we can apply the above formula to factor our expression completely. (16n^2+c^2)(( 4n)^2- c^2) ⇕ (16n^2+c^2)( 4n+ c)( 4n- c)

Checking Our Answer

Check your answer âś“
We can apply the Distributive Property and compare the result with the given expression.
(16n^2 + c^2) (4n + c)(4n - c)
(16n^2 + c^2) (4n(4n + c) - c(4n+c))
â–Ľ
Distribute 4n & - c
(16n^2 + c^2) (16n^2 + 4cn - c(4n+c))
(16n^2 + c^2) (16n^2 + 4cn - 4cn - c^2)
(16n^2 + c^2) (16n^2 - c^2)
16n^2 (16n^2 + c^2 ) - c^2 (16n^2 + c^2)
â–Ľ
Distribute 16n^2 & - c^2
256n^4 + 16c^2n^2 - c^2 (16n^2 + c^2)
256n^4 + 16c^2n^2 - 16c^2n^2 - c^4
256n^4 - c^4
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!