Q 109. - Intermediate Algebra

Question

Factor Trinomials of the Form $a{x}^2+bx+c$ Using Trial and Error.

$-30q^3-140q^2-80q$

Step-by-Step Solution

Answer

The solution is $-10q(q+4)(3q+2)$.

Step 1. Given information

Consider the trinomial.

$-30q^3-140q^2-80q$

Step 2. Write the trinomial in descending order and then find the greatest common factor.

The trinomial $-30q^3-140q^2-80q$ is given in descending order.

There is a greatest common factor in the given trinomial.

Factor the greatest common factor.

$-30q^3-140q^2-80q=-10q(3q^2+14q+8)$

Step 3. Find the factors of the first term and the last term of the trinomial 3 q 2 + 14 q + 8 .

The first term of the trinomial $3q^2+14q+8$ is $3q^2$.

So, the factors of $3q^2$ are as follows:

$3q^2=q·3q$

The last term of the trinomial $q^2+14q+8$ is $8$.

Since all the terms in the polynomial $q^2+14q+8$ is positive, all the factors of the last term will be positive.

The factors of $8$ are as follows:

$\begin{array}{rcl}8& =& 1·8\\ 8& =& 2·4\end{array}$

Step 4. Make a table for all the combination of factors of 3 q 2 + 14 q + 8 .

Possible factors	Product
$(q+1)(3q+8)$	$3q^2+11q+8$
$(q+8)(3q+1)$	$3q^2+25q+8$
$(q+2)(3q+4)$	$3q^2+10q+8$
$(q+4)(3q+2)$	$3q^2+14q+8$

From the table, conclude that the combination  $(q+4)(3q+2)$ is correct.

Step 5. Check by multiplying all the factors - 10 q ( q + 4 ) ( 3 q + 2 ) .

$\begin{array}{rcl}-10q(q+4)(3q+2)& =& -10q(q^2+2q+12q+8)\\ & =& -10q(q^2+14q+8)\\ & =& -10q^3-140q^2-80q\end{array}$

Hence, the factor is $-10q(q+4)(3q+2)$.