# How do you factor 3x^4-x^3-3x^2+1?

May 18, 2015

We can see clearly that all elements that contain the variable $x$ are multiplying ${x}^{2}$, so we can factor it out as follows:

${x}^{2} \left(3 {x}^{2} - x - 3\right) + 1$

${x}^{2} \left[x \left(3 x - 1\right) - 3\right] + 1$

May 18, 2015

Answer: $3 {x}^{4} - {x}^{3} - 3 {x}^{2} + 1$ $=$ $\left(x - 1\right) \left(3 {x}^{3} + 2 {x}^{2} - x - 1\right)$

Explanation:

Let's denote our polynomial $3 {x}^{4} - {x}^{3} - 3 {x}^{2} + 1$ by $P \left(x\right)$.

We use the Remainder Theorem: when dividing a polynomial $P \left(x\right)$ by $\left(x - c\right)$, where $c$ is a constant, the remainder equals $P \left(c\right)$.

Hence, for $x - c$ to be a divisor of the polynomial $P \left(x\right)$, the remainder $P \left(c\right)$ needs to be equal to zero.

Therefore, we need to find a constant $c$ such that $P \left(c\right) = 0$.

$P \left(c\right) = 0$ $\iff$ $3 {c}^{4} - {c}^{3} - 3 {c}^{2} + 1$$=$ $0$ $\iff$ $3 {c}^{2} \left({c}^{2} - 1\right) - \left({c}^{3} - 1\right)$$=$$0$ $\iff$ $3 {c}^{2} \left(c + 1\right) \left(c - 1\right) - \left(c - 1\right) \left({c}^{2} + c + 1\right)$$=$$0$ $\iff$ $\left(c - 1\right) \left(3 {c}^{3} + 3 {c}^{2} - {c}^{2} - c - 1\right)$$=$$0$ $\iff$ $\left(c - 1\right) \left(3 {c}^{3} + 2 {c}^{2} - c - 1\right)$$=$$0$

So, we found that, if $c$$=$$1$, then $P \left(c\right) = 0$. Therefore, $x - 1$ is a divisor of the polynomial $P \left(x\right)$.

Verification : divide $P \left(x\right)$$=$$3 {x}^{4} - {x}^{3} - 3 {x}^{2} + 1$ by $x - 1$ and obtain
$P \left(x\right)$$/$$\left(x - 1\right)$$=$$3 {x}^{3} + 2 {x}^{2} - x - 1$