Q 122. - Intermediate Algebra

Question

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

$90n^3+42n^2-216n$

Step-by-Step Solution

Answer

The solution is $\begin{array}{rcl}& & 6n(5n+9)(3n-4)\end{array}$.

Step 1. Given information

The given trinomial is $90n^3+42n^2-216n$.

Step 2. Find greatest common factor.

Factor the greatest common factor.

$90n^3+42n^2-216n=6n(15n^2+7n-36)$

Step 3. Compare the trinomial 15 n 2 + 7 n - 36 to a x 2 + b x + c and then find the product of a c .

The values are as follows:

$\begin{array}{rcl}a& =& 15\\ b& =& 7\\ c& =& -36\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 15(-36)\\ & =& -540\end{array}$

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

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

$\begin{array}{rcl}(-20)27& =& -540\\ & =& ac\\ -20+27& =& 7\\ & =& b\end{array}$

So, two numbers are $-20$ and $27$.

Now, split the middle term of the trinomial $15n^2+7n-36$ as follows:

$15n^2+7n-36=15n^2+27n-20n-36$

Step 5. Factor by grouping.

$\begin{array}{rcl}15n^2+7n-36& =& 3n(5n+9)-4(5n+9)\\ & =& (5n+9)(3n-4)\end{array}$

Step 6. Check by multiplying all the factors 6 n ( 5 n + 9 ) ( 3 n - 4 ) .

$\begin{array}{rcl}6n(5n+9)(3n-4)& =& 6n\left(15n^2+27n-20n-36\right)\\ & =& 6n\left(15n^2+7n-36\right)\\ & =& 90n^3\hspace{0.17em}+\hspace{0.17em}42n^2\hspace{0.17em}-\hspace{0.17em}216n\end{array}$

Hence, the factor is $\begin{array}{rcl}& & 6n(5n+9)(3n-4)\end{array}$.