Q 117. - Intermediate Algebra

Question

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

$2n^2-27n-45$

Step-by-Step Solution

Answer

The solution is $\begin{array}{rcl}& & (n-15)(2n+3)\end{array}$.

Step 1. Given information

The given trinomial is $2n^2-27n-45$.

Step 2. Find greatest common factor.

There is no greatest common factor.

Step 3. Compare the given trinomial to a x 2 + b x + c and then find the product of a c .

The values are as follows:

$\begin{array}{rcl}a& =& 2\\ b& =& -27\\ c& =& -45\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 2(-45)\\ & =& -90\end{array}$

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:

$\begin{array}{rcl}3(-30)& =& -90\\ & =& ac\\ 3-30& =& -27\\ & =& b\end{array}$

So, two numbers are $3$ and $-30$.

Now, split the middle term of the given trinomial $2n^2-27n-45$ as follows:

$2n^2-27n-45=2n^2-30n+3n-45$

Step 5. Factor by grouping.

$\begin{array}{rcl}2n^2-27n-45& =& 2n(n-15)+3(n-15)\\ & =& (n-15)(2n+3)\end{array}$

Step 6. Check by multiplying ( n - 15 ) ( 2 n + 3 ) .

$\begin{array}{rcl}(n-15)(2n+3)& =& 2n^2-30n+3n-45\\ & =& 2n^2-27n-45\end{array}$

Hence, the factor is $\begin{array}{rcl}& & (n-15)(2n+3)\end{array}$.