Question

How do you factor completely ${{x}^{3}}-4{{x}^{2}}-2x+8$?

Answer 1

In order to solve this question, firstly find a root $x=a$ which fits the equation ${{x}^{3}}-4{{x}^{2}}-2x+8=0$ perfectly. Then, just divide the factor $\left( x-a \right)$ with the equation that we need to factorize and we get a quadratic equation as the remainder. The quadratic equation is factored then.

Complete step-by-step solution:
Let us start solving the question by taking the equation,
${{x}^{3}}-4{{x}^{2}}-2x+8=0$
Now, let us think of a root that completely satisfies this equation,
If

& x=4 \\\ & \Rightarrow {{4}^{3}}-4\centerdot {{4}^{2}}-2\left( 4 \right)+8=64-64-8+8=0 \\\ \end{aligned}$$ So, we can say that the factor $$\left( x-4 \right)$$ satisfies the equation, now dividing the equation$${{x}^{3}}-4{{x}^{2}}-2x+8$$ with $$\left( x-4 \right)$$, we get $$\left( x-4 \right){{\left| \\!{\overline {\, \begin{aligned} & {{x}^{3}}-4{{x}^{2}}-2x+8 \\\ & _{-}{{x}^{3}}{{-}_{+}}4{{x}^{2}}\downarrow +\downarrow \\\ & 0+0 \\\ & -2x+8 \\\ & _{+}-2x{{+}_{-}}8 \\\ & =0 \\\ \end{aligned} \,}} \right. }^{{{x}^{2}}-2}}$$ Therefore, we can say that $$\left( x-4 \right)\left( {{x}^{2}}-2 \right)={{x}^{3}}-4{{x}^{2}}-2x+8$$ So, now, we just need to make the further factors for the factor $$\left( {{x}^{2}}-2 \right)$$ $$\begin{aligned} & \left( {{x}^{2}}-2 \right) \\\ & \Rightarrow \left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right) \\\ \end{aligned}$$ Thus, we can write that the factors for the equation$${{x}^{3}}-4{{x}^{2}}-2x+8$$ are $$\left( x-4 \right)\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)$$ Hence the equation has been factored. **Note:** The given equation is a cubic equation, that is why we get three factors. However, as the factor $$\left( x-4 \right)$$ is repeated twice, its square has been taken but the degree of the equation still remains the same i.e. $$3$$. The factors can be verified by further multiplying them to get the same equation as before.