Question

In the following exercises, factor completely.

$6x^4-19x^2+15$

Short answer

On factoring, we get $\left(3x^2-5\right)\left(2x^2-3\right)$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$6x^4-19x^2+15$

Rewrite the trinomial to prepare for the substitution,

$=6{\left(x^2\right)}^2-19x^2+15......\left(i\right)$

Assume $u=x^2$ and substitute in equation (i),

$=6u^2-19u+15$

Factor out the trinomial,

role="math" localid="1644848571301" $$\begin{gathered} =6u^2-9u-10u+15 \\ =3u\left(2u-3\right)-5\left(2u-3\right) \\ =\left(3u-5\right)\left(2u-3\right)......\left(ii\right) \end{gathered}$$

Substitute $u=x^2$in equation (ii),

$=\left(3x^2-5\right)\left(2x^2-3\right)$

Step 2. Check the answers.

Check by multiplying,

$$\begin{gathered} =\left(3x^2-5\right)\left(2x^2-3\right) \\ =6x^4-9x^2-10x^2+15 \\ =6x^4-19x^2+15 \end{gathered}$$

This is true.