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McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
8. Differences of Squares
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Exercise 22 Page 61

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

(h^2+16)(h+4)(h-4)

Practice makes perfect

Look closely at the expression h^4-256. It can be expressed as the difference of two perfect squares.

h^4-256
WritePow

Write as a power

(h^2)^2-16^2

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. ( h^2)^2- 16^2 ⇔ ( h^2+ 16)( h^2- 16) Now, let's take a look at the expression h^2-16. Once again, we can express it as a difference of two perfect squares. Let's do it!

h^2-16
WritePow

Write as a power

h^2-4^2

Now, we can apply the above formula to factor our expression completely. (h^2+16)( h^2- 4^2) ⇕ (h^2+16)( h+ 4)( h- 4)

Checking Our Answer

 Check your answer ✓
We can apply the Distributive Property and compare the result with the given expression.

(h^2 + 16) (h + 4)(h-4)
Distr

Distribute (h+4)

(h^2 + 16) (h(h+4) - 4(h+4))
â–¼
Distribute h & -4
Distr

Distribute h

(h^2 + 16) (h^2+4h - 4(h+4))
Distr

Distribute - 4

(h^2 + 16) (h^2+4h - 4h-16)
SubTerm

Subtract term

(h^2 + 16) (h^2-16)
Distr

Distribute (h^2+16)

h^2(h^2+16)-16(h^2+16)
â–¼
Distribute h^2 & - 16
Distr

Distribute h^2

h^4+16h^2-16(h^2+16)
Distr

Distribute - 16

h^4+16h^2-16h^2-256
SubTerm

Subtract term

h^4-256

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!

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arrow_right Exercises p.60-63
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Exercises 2 (Page 60)
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