Q 271 - Intermediate Algebra

Question

In the following exercises, factor completely.

${(3x+1)}^2-6(3x+1)+9$

Step-by-Step Solution

Answer

The expression is factored as ${(3x+1)}^2-6(3x+1)+9=9x^2-12x+4$

Step 1. Given Information

Consider the expression ${(3x+1)}^2-6(3x+1)+9$.

The objective is to factor the expression completely.

Step 2. Factor using perfect square trinomial

The given expression forms a perfect square trinomial of the form $a^2-2ab+b^2$ which is factored as ${(a-b)}^2$. So the given expression can be factored as

$\begin{array}{rcl}{(3x+1)}^2-6(3x+1)+9& =& {(3x+1)}^2-2·(3x+1)·3+{\left(3\right)}^2 \\ & =& {(3x+1-3)}^2 \\ & =& {(3x-2)}^2 \\ & =& 9x^2-12x+4\end{array}$

Step 3. Check the factors

Multiply the factors to check the solution. 

$\begin{array}{rcl}9x^2-12x+4& =& {\left(3x\right)}^2-2·3x·2+{\left(2\right)}^2 \\ & =& {(3x+2)}^2 \\ & =& {(3x+1-3)}^2 \\ & =& {(3x+1)}^2-2(3x+1)·3+{\left(3\right)}^2 \\ & =& {(3x+1)}^2-6(3x+1)+9\end{array}$

And we get the given expression. So the expression is correctly factored.