Question

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

$5s^2-9s+4$

Short answer

The solution is $\begin{array}{rcl}& & (s-1)(5s-4)\end{array}$.

Step-by-step solution

Step 1. Given information

The given trinomial is $5s^2-9s+4$.

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:

$\begin{array}{rcl}a& =& 5\\ b& =& -9\\ c& =& 4\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 5·4\\ & =& 20\end{array}$

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

Since the middle term in the given polynomial is negative, both factors will be negative.

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

$\begin{array}{rcl}(-5)(-4)& =& 20\\ & =& ac\\ -5-4& =& -9\\ & =& b\end{array}$

So, two numbers are $-5$and $-4$.

Now, split the middle term of the given trinomial $5s^2-9s+4$as follows:

$5s^2-9s+4=5s^2-5s-4s+4$

Step  5. Factor by grouping.

$\begin{array}{rcl}5s^2-9s+4& =& 5s(s-1)-4(s-1)\\ & =& (s-1)(5s-4)\end{array}$

Step 6. Check by multiplying (s-1)(5s-4).

$\begin{array}{rcl}(s-1)(5s-4)& =& 5s^2-5s-4s+4\\ & =& 5s^2-9s+4\end{array}$

Hence, the factor is $\begin{array}{rcl}& & (s-1)(5s-4)\end{array}$.