Q. 127E - Intermediate Algebra

Question

In the following exercises, factor using substitution. $x^4-x^2-12$

Step-by-Step Solution

Answer

The factors are $\left(x^2+3\right)\left(x-2\right)\left(x+2\right)$

Step 1. Given information

The trinomial to factor is $x^4-x^2-12$.

Use substitution method.

Step 2. Find term for substitution

Rewrite the trinomial to prepare for the substitution 

${\left(x^2\right)}^2-x^2-12$

Substitute $x^2=u$and rewrite the trinomial

$u^2-u-12$

Step 3. Factor the trinomial

Split the middle term of trinomial to find factors

$u^2+3x-4x-12$

$$\begin{gathered} u\left(u+3\right)-4\left(u+3\right) \\ \left(u+3\right)\left(u-4\right) \end{gathered}$$

Step 4. Substitute back u

Substitute back $u=x^2$

$\left(x^2+3\right)\left(x^2-4\right)$

$x^2-4$further factor as $\left(x+2\right)\left(x-2\right)$

So factors are $\left(x^2+3\right)\left(x-2\right)\left(x+2\right)$

 

Step 5. Check

Multiply all the factors $\left(x^2+3\right)\left(x-2\right)\left(x+2\right)$

$$\begin{gathered} \left(x^2+3\right)\left(x-2\right)\left(x+2\right) \\ \left(x^2+3\right)\left(x^2-4\right) \end{gathered}$$

again multiply

$$\begin{gathered} x^4+3x^2-4x^2-12 \\ {x}^4-x^2-12 \end{gathered}$$