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

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

Consider the trinomial.

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

The trinomial $6p^2-19pq+10q^2$isgiven in descending order.

There is no greatest common factor.

The first term of the given trinomial is $6p^2$.

So, the factors of $6p^2$are as follows:

$\begin{array}{rcl}6p^2& =& p·6p\\ 6p^2& =& 2p·3p\end{array}$

The last term of the given trinomial is $10q^2$.

Since the middle term of the given polynomial is negative, the factors of the last term will both be negative.

The factors of $10q^2$are as follows:

$\begin{array}{rcl}10q^2& =& (-q)·(-10q)\\ 10q^2& =& (-2q)(-5q)\end{array}$

If the trinomial has no common factors, none of the factors can contain the common factors.

This follows that the combination of the factors is not an option.

The table is shown below:

From the table, conclude that the combination $\left(2p-5q\right)\left(3p-2q\right)$is correct.

$\begin{array}{rcl}\left(2p-5q\right)\left(3p-2q\right)& =& 6p^2-4pq-15pq+10q^2\\ & =& 6p^2-19pq+10q^2\end{array}$

Hence, the factor is $\left(2p-5q\right)\left(3p-2q\right)$.