Question

Factor Trinomials of the Form $a{x}^2+bx+c$using the ‘ac’ Method.

$6y^2+y-15$

Short answer

The solution is $\begin{array}{rcl}& & (3y+5)(2y-3)\end{array}$.

Step-by-step solution

Step 1. Given information

The given trinomial is $6y^2+y-15$.

Step 2. Find greatest common factor.

There is no greatest common factor.

Step 3. Compare the given trinomial to ax2+bx+c and then find the product of ac.

The values are as follows:

$\begin{array}{rcl}a& =& 6\\ b& =& 1\\ c& =& -15\end{array}$

The product is as follows:

role="math" localid="1644753373463" $\begin{array}{rcl}ac& =& 6(-15)\\ & =& -90\end{array}$

Step 4. Determine two numbers that multiply to ac and add to b.

Two numbers that multiply to $ac$and add to $b$are as follows:

$\begin{array}{rcl}10(-9)& =& -90\\ & =& ac\\ 10-9& =& 1\\ & =& b\end{array}$

So, two numbers are $10$and localid="1644753833782" $-9$.

Now, split the middle term of the given trinomial $6y^2+y-15$as follows:

$6y^2+y-15=6y^2+10y-9y-15$

Step  5. Factor by grouping.

$\begin{array}{rcl}6y^2+y-15& =& 2y(3y+5)-3(3y+5)\\ & =& (3y+5)(2y-3)\end{array}$

Step 6. Check by multiplying (3y+5)(2y-3).

$\begin{array}{rcl}(3y+5)(2y-3)& =& 6y^2+10y-9y-15\\ & =& 6y^2+y-15\end{array}$

Hence, the factor is $\begin{array}{rcl}& & (3y+5)(2y-3)\end{array}$.