Q. 370 - 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.

$6p^2-19pq+10q^2$

Step-by-Step Solution

Answer

By factoring the polynomial $6p^2-19pq+10q^2$ completely by trial and error method, the factors are $(3p-2q)(2p-5q)$

Step 1. Given

The polynomial is $6p^2-19pq+10q^2$

To factor the polynomial by trial and error method.

Step 2. Factor any GCF

Factor the GCF.

There is no GCF in the polynomial.

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

The factors of the first term are $1,6$ and $2,3$

$6p^2-19pq+10q^2=(6p q)(p q)$

$6p^2-19pq+10q^2=(p q)(6p q)$

$6p^2-19pq+10q^2=(2p q)(3p q)$

$6p^2-19pq+10q^2=(3p q)(2p q)$

Step 4. Find all the factor pairs of third term

Consider the sign.

Since the last term is $10$, the factors have same sign either positive or negative.

Since the sign of the second term is negative, both the factors have negative sign. 

The factors of $10$ are

• $-1,-10$
  
• $-2,-5$
  

Step 5. Test all combination of factors

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

Possible outcomes	Product
$(6p-10q)(p-q)$	$6p^2-16pq+10q^2$
$(6p-q)(p-10q)$	$6p^2-61pq+10q^2$
$(3p-10q)(2p-q)$	$6p^2-23pq+10q^2$
$(3p-q)(2p-10q)$	$6p^2-32pq+10q^2$
$(6p-5q)(p-2q)$	$6p^2-17pq+10q^2$
$(6p-2q)(p-5q)$	$6p^2-32pq+10q^2$
$(3p-2q)(2p-5q)$	$6p^2-19pq+10q^2$
$(3p-q)(2p-10q)$	$6p^2-32pq+10q^2$

The factors of the trinomial is $(3p-2q)(2p-5q)$

Step 6. Check the answer

Check by multiplying the factor. $(3p-2q)(2p-5q)=6p^2-15pq-4pq+10q^2$

                              $=6p^2-19pq+10q^2$