Q 125. - Intermediate Algebra

Question

In the following exercises, factor using the ‘ac’ method. $48y^2+12y-36$

Step-by-Step Solution

Answer

The factors are $12\left(y+1\right)\left(4y-3\right)$

Step 1. Given information

The trinomial is $48y^2+12y-36$

Find factors of trinomial.

Step 2 . Find the GCF

The common factor is 12.

Take factor from $48y^2+12y-36$

Trinomial becomes $12\left(4y^2+y-3\right)$

Step 3. Compare the trinomial with a x 2 + b x + c

Compare $a{x}^2+bx+c$with $4y^2+y-3$

So $a=4$, $b=1$and $c=-3$

Find product of a and c

$ac=-12$

Step 4. Factor of ac

Find two numbers such as product is $-12$ and sum is 1.

So, 4 and -3 

Step 5. Find factors

Rewrite trinomial $12\left(4y^2+y-3\right)$

$12\left(4y^2+4y-3y-3\right)$

Factor the trinomial

$$\begin{gathered} 12\left(4y\left(y+1\right)-3\left(y+1\right)\right) \\ 12\left(y+1\right)\left(4y-3\right) \end{gathered}$$

Step 6. Check

Multiply $12\left(y+1\right)\left(4y-3\right)$

$$\begin{gathered} 12\left(4y^2+y-3\right) \\ 48y^2+12y-36 \end{gathered}$$