Q 119. - Intermediate Algebra

Question

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

$60y^2+290y-50$

Step-by-Step Solution

Answer

The solution is $10(y+5)(6y-1)$.

Step 1. Given information

The given trinomial is $60y^2+290y-50$.

Step 2. Find greatest common factor.

Factor the greatest common factor.

$60y^2+290y-50=10(6y^2+29y-5)$

Step 3. Compare the trinomial 6 y 2 + 29 y - 5 to a x 2 + b x + c and then find the product of a c .

The values are as follows:

$\begin{array}{rcl}a& =& 6\\ b& =& 29\\ c& =& -5\end{array}$

The product is as follows:

$\begin{array}{rcl}ac& =& 6(-5)\\ & =& -30\end{array}$

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:

$\begin{array}{rcl}30·(-1)& =& -30\\ & =& ac\\ 30-1& =& 29\\ & =& b\end{array}$

So, two numbers are $30$ and $-1$.

Now, split the middle term of the trinomial $6y^2+29y-5$ as follows:

$6y^2+29y-5=6y^2+30y-y-5$

Step 5. Factor by grouping.

$\begin{array}{rcl}6y^2+29y-5& =& 6y(y+5)-1(y+5)\\ & =& (y+5)(6y-1)\end{array}$

Step 6. Check by multiplying all the factors 10 ( y + 5 ) ( 6 y - 1 ) .

$\begin{array}{rcl}10(y+5)(6y-1)& =& 10(6y^2+30y-y-5)\\ & =& 10(6y^2+29y-5)\\ & =& 60y^2+290y-50\end{array}$

Hence, the factor is $10(y+5)(6y-1)$.