Chapter 5.7, Problem 54WE

To determine

To Factor:The quadratic expression completely.

Answer to Problem 54WE

Simplest form is $6a(2a+1)$

Explanation of Solution

Given information: ${(a-4)}^2+5(a-4)(a+2)+6{(a+2)}^2$

Calculation:

Let’s factor using AC method of factoring.

So, rewrite the expression as

$(a-4)(a-4)+5(a-4)(a+2)+6(a+2)(a+2)$

Coefficient of ${(a-4)}^2$ is 1.

Coeffiicient of middle term is 5.

Coefficient of last term is 6.

Now, multiply the 1 and 6 which is 6.

Find two integers whose multiplication is 6 but adds to 5.

To find that, list all two integers whose multiplication is 6.

Here 2,3 works.

Because 2*3=6

2+3=5

Now, rewrite expression as

$(a-4)(a-4)+2(a-4)(a+2)+3(a-4)(a+2)+6(a+2)(a+2)$

Group the terms and factor using GCF method.

  $((a-4)(a-4)+2(a-4)(a+2))(+3(a-4)(a+2)+6(a+2)(a+2))$

Factor each group using GCF method.

  $(a-4)((a-4)+2(a+2))+(3(a+2)((a-4)+2(a+2))$

Simplify each parentheses,

$(a-4)(a-4+2a+4)+3(a+2)(a-4+2a+4)$

Simplify further by combining like terms,

$(a-4)(3a)+3(a+2)(3a)$

Both terms has $(3a)$ common factor.

It gives

  $3a(a-4+3(a+2))$

Simplify the expression further

$3a(a-4+3a+6)$

Combine like terms,

$3a(4a+2)$

Here $4a$ and 2 has common factor as 2.

So, let’s take it out.

  $3a*2(2a+1)$

Simplify it,

$6a(2a+1)$