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 41 Page 61

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

2(a+4)(a-4)(6a+1)

Practice makes perfect

To factor the given expression we will start by identifying and factoring out the greatest common factor (GCF). Then we will use factoring by grouping and the formula for factoring the difference of perfect squares.

Factor Out the GCF

The greatest common factor (GCF) of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. The GCF of the given expression is 2.
12a^3+2a^2-192a-32
2* 6a^3+ 2* a^2 - 2 * 96a - 2 * 16
2(6a^3+a^2-96a-16)

Factoring by Grouping

Now, 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.
2( 6a^3 + a^2-96a-16)
2(a^2( 6a+1 ) - 96a - 16)
2(a^2( 6a+1 ) -16( 6a+1 ))
Notice that (6a+1) is a factor of both terms, so we can factor it out.
2(a^2( 6a+1 ) -16( 6a+1 ))
2( a^2-16 ) ( 6a+1 )

Difference of Squares

Now, look closely at the expression a^2-16. It can be expressed as the difference of two perfect squares.
a^2-16
a^2-4^2
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. 2( a^2- 4^2 )(6a+1) ⇕ 2( a+ 4)( a- 4)(6a+1)

Checking Our Answer

Check your answer âś“
We can apply the Distributive Property and compare the result with the given expression.
2(a + 4)(a - 4) (6a + 1)
2(a(a+4) - 4(a+4)) (6a + 1)
â–Ľ
Distribute a & - 4
2(a^2 + 4a - 4(a+4)) (6a + 1)
2(a^2 + 4a - 4a - 16) (6a + 1)
2(a^2 - 16) (6a + 1)
2(a^2(6a + 1) - 16(6a + 1))
â–Ľ
Distribute a^2 & -16
2(6a^3 + a^2 - 16(6a + 1))
2(6a^3 + a^2 - 96a - 16)
12a^3+2a^2-192a-32
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!