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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 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

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

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)

Distribute (h+4)

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

▼

Distribute h & -4

Distribute h

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

Distribute - 4

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

Subtract term

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

Distribute (h^2+16)

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

▼

Distribute h^2 & - 16

Distribute h^2

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

Distribute - 16

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

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!

Subchapter links

Exercises p.60-63