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

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

Consider the trinomial.

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

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.

role="math" localid="1645100105345" $-30q^3-140q^2-80q=-10q(3q^2+14q+8)$

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:

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

From the table, conclude that the combinationlocalid="1645100413930" $(q+4)(3q+2)$is correct.

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

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