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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 10 Page 60

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

(f+3)(f-3)(f-4)

Practice makes perfect

To factor the given expression we will use factoring by grouping and then the formula for factoring the difference of perfect squares.

Factoring by Grouping

To factor by grouping, we will take the greatest common factor of the first pair of terms and the greatest common factor of the second pair of terms.

f^3 - 4f^2-9f+36
FactorOut

Factor out f^2

f^2( f-4 ) - 9f + 36
FactorOut

Factor out - 9

f^2( f-4 ) -9( f-4 )
Notice that (f-4) is a factor of both terms, so we can factor it out.

f^2( f-4 ) -9( f-4 )
FactorOut

Factor out ( f-4 )

( f^2-9 ) ( f-4 )

Difference of Squares

Now, look closely at the expression f^2-9. It can be expressed as the difference of two perfect squares.

f^2-9
WritePow

Write as a power

f^2-3^2

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. ( f^2- 3^2 )(f-4) ⇕ ( f+ 3)( f- 3)(f-4)

Checking Our Answer

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

(f + 3)(f - 3) (f - 4)
Distr

Distribute (f+3)

(f(f+3) - 3(f+3)) (f - 4)
â–¼
Distribute f & - 3
Distr

Distribute f

(f^2 + 3f - 3(f+3)) (f - 4)
Distr

Distribute - 3

(f^2 + 3f - 3f - 9) (f - 4)
SubTerm

Subtract term

(f^2 - 9) (f - 4)
Distr

Distribute (f-4)

f^2(f - 4) - 9(f - 4)
â–¼
Distribute f^2 & -9
Distr

Distribute f^2

f^3 - 4f^2 - 9(f - 4)
Distr

Distribute - 9

f^3 - 4f^2 - 9f + 36

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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