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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 18 Page 61

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

(w^2+25)(w+5)(w-5)

Practice makes perfect

Look closely at the expression w^4-625. It can be expressed as the difference of two perfect squares.

w^4-625
WritePow

Write as a power

(w^2)^2-25^2

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. ( w^2)^2- 25^2 ⇔ ( w^2+ 25)( w^2- 25) Now, let's take a look at the expression w^2-25. Once again, we can express it as a difference of two perfect squares. Let's do it!

w^2-25
WritePow

Write as a power

w^2-5^2

Now, we can apply the above formula to factor our expression completely. (w^2+25^2)( w^2- 5^2) ⇕ (w^2+25)( w+ 5)( w- 5)

Checking Our Answer

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

(w^2 + 25) (w + 5)(w-5)
Distr

Distribute (w+5)

(w^2 + 25) (w(w+5) - 5(w+5))
â–¼
Distribute w & -5
Distr

Distribute w

(w^2 + 25) (w^2+5w - 5(w+5))
Distr

Distribute - 5

(w^2 + 25) (w^2+5w - 5w-25)
SubTerm

Subtract term

(w^2 + 25) (w^2-25)
Distr

Distribute (w^2+25)

w^2(w^2+25)-25(w^2+25)
â–¼
Distribute w^2 & - 25
Distr

Distribute w^2

w^4+25w^2-25(w^2+25)
Distr

Distribute - 25

w^4+25w^2-25w^2-625
SubTerm

Subtract term

w^4-625

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)
Exercises 3 (Page 60)
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