Q. 399 - Intermediate Algebra

Question

In the following exercises, factor completely using the difference of squares pattern, if possible. 

$a^2+6a+9-9b^2$

Step-by-Step Solution

Answer

The Factored form of given polynomial is, 

$(a-3b+3)(a+3b+3)$

Step 1. Given Information

We are given a polynomial,

$a^2+6a+9-9b^2$

The formula used for factoring using the difference of squares pattern is, 

$a^2-b^2=(a+b)(a-b)$

Step 2. Factorizing the polynomial

The given polynomial can be also written as,

$a^2+6a+9-9b^2=(a{)}^2+2·a·3+(3{)}^2-9b^2$

Using ${(a+b)}^2=a^2+2ab+b^2$, we get

$a^2+6a+9-9b^2=(a+3{)}^2-(3b{)}^2$

Using $a^2-b^2=(a+b)(a-b)$, we get

$$\begin{gathered} a^2+6a+9-9b^2=\left(\right(a+3)-3b)\left(\right(a+3)+3b) \\ {a}^2+6a+9-9b^2=(a-3b+3)(a+3b+3) \end{gathered}$$

Step 3. Checking the factorization by multiplying

Multiplying the factors, we get

$$\begin{gathered} (a-3b+3)(a+3b+3)=a^2+6a+9-9b^2 \\ (a-3b)(a+3b)+3(a-3b)+3(a+3b)+3\times 3=a^2+6a+9-9b^2 \\ {a}^2-9b^2+3a-9b+3a+9b+9=a^2+6a+9-9b^2 \\ {a}^2+6a+9-9b^2=a^2+6a+9-9b^2 \\ LHS=RHS \end{gathered}$$

Hence the factorization is correct.