Question

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

$12z^2-41z-11$

Short answer

The solution is $\begin{array}{rcl}& & (4z+1)(3z-11)\end{array}$.

Step-by-step solution

Step 1. Given information

The given trinomial is $12z^2-41z-11$.

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:

$\begin{array}{rcl}a& =& 12\\ b& =& -41\\ c& =& -11\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 12(-11)\\ & =& -132\end{array}$

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:

$\begin{array}{rcl}3(-44)& =& -132\\ & =& ac\\ 3-44& =& -41\\ & =& b\end{array}$

So, two numbers are $3$and $-44$.

Now, split the middle term of the given trinomial $12z^2-41z-11$as follows:

$12z^2-41z-11=12z^2+3z-44z-11$

Step  5. Factor by grouping.

$\begin{array}{rcl}12z^2-41z-11& =& 3z(4z+1)-11(4z+1)\\ & =& (4z+1)(3z-11)\end{array}$

Step 6. Check by multiplying (4z+1)(3z-11).

$\begin{array}{rcl}(4z+1)(3z-11)& =& 12z^2+3z-44z-11\\ & =& 12z^2+41z-11\end{array}$

Hence, the factor is $\begin{array}{rcl}& & (4z+1)(3z-11)\end{array}$.