Q. 371 - Intermediate Algebra

Question

Factor Trinomials of the Form $a{x}^2+bx+c$ Using Trial and Error
In the following exercises, factor completely using trial and error.

$-9a^2+153a+18$

Step-by-Step Solution

Answer

By factoring the polynomial $-81a^2+153a+18$ completely by trial and error method, the factors are $-9(9a+1)(a-2)$

Step 1. Given

The polynomial is $-81a^2+153a+18$

To factor the polynomial by trial and error method.

Step 2. Factor any GCF

Factor the GCF
$-81a^2+153a+18=-9(9a^2-19a-2)$

Step 3. Find all the pairs of factors of first term

The factors of the first term are $1,9$ and $3,3$

$9a^2-19a-2=(x )(9x )$

$9a^2-19a-2=(3x )(3x )$

Step 4. Find all the factor pairs of third term

Consider the sign.

Since the last term is $-2$, the factors have opposite sign.

Since the sign of the second term is negative, the greater factor have negative sign. 

The factors of $-2$ are

• $1,-2$
  

Step 5. Test all combination of factors

Test all combination of factors until the correct product is found. 

Possible factors	Product
$(9a-2)(a+1)$	$9a^2+7a-2$
$(9a+1)(a-2)$	$9a^2-17a-2$
$(3a-2)(3a+1)$	$9a^2-3a-2$
$(3a+1)(3a-2)$	$9a^2+3a-2$

The factors are $(9a+1)(a-2)$

Step 6. Check the answer

Check by multiplying the factor. 

$-9(9a+1)(a-2)=-9(9a^2-18a+a-2)$

                            $=-9a(9a^2-17a-2)$

                            $=-81a^2+153a+18$