Q. 134E - Intermediate Algebra

Question

In the following exercises, factor using substitution ${\left(5y-1\right)}^2-3\left(5y-1\right)-18$

Step-by-Step Solution

Answer

The factors are $\left(3y-7\right)\left(3y+2\right)$

Step 1. Given Information

The trinomial is ${\left(3y-1\right)}^2-3\left(3y-1\right)-18$

Find factors using substitution method. 

Step 2. Find the term for substitution

Rewrite the trinomial to prepare for the substitution  

${\left(3y-1\right)}^2-3\left(3y-1\right)-18$

Substitute $3y-1=u$

So, trinomial becomes 

$u^2-3u-18$

Step 3. Factor the trinomial

Split the middle term of trinomial to find factors

$$\begin{gathered} u^2-3u-18 \\ {u}^2-6u+3u-18 \end{gathered}$$

Factor it

$$\begin{gathered} u\left(u-6\right)+3\left(u-6\right) \\ \left(u-6\right)\left(u+3\right) \end{gathered}$$

Step 4. Substitute back u

Substitute back $u=3y-1$

$$\begin{gathered} \left(3y-1-6\right)\left(3y-1+3\right) \\ \left(3y-7\right)\left(3y+2\right) \end{gathered}$$