Question

How do you factor $x^3-7x+6=0$?

Answer 1

Find the factors of the given equation:

Let us consider the given equation as,

$f\left(x\right)=x^3-7x+6$

Put $x=1$

$$\begin{gathered} f\left(1\right)=1^3-7\left(1\right)+6 \\ f\left(1\right)=1-7+6 \\ f\left(1\right)=-6+6 \\ f\left(1\right)=0 \end{gathered}$$

Thus,$\left(x-1\right)$ is the factor of $x^3-7x+6=0$

Rewrite the given Equation as,

$$\begin{gathered} f\left(x\right)=\left(x-1\right)\left(x^2+x-6\right) \\ f\left(x\right)=\left(x-1\right)\left(x^2+3x-2x-6\right) \\ f\left(x\right)=\left(x-1\right)\left(x\left(x+3\right)-2\left(x+3\right)\right) \\ f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x+3\right) \end{gathered}$$

Alter:

Given,

$$\begin{gathered} x^3-7x+6=0 \\ \Rightarrow x^3-x-6x+6=0\left[\because -7x=-6x-x\right] \\ \Rightarrow x\left(x^2-1\right)-6\left(x-1\right)=0 \\ \Rightarrow x\left(x^2-1^2\right)-6\left(x-1\right)=0\left[\because 1^2=1\right] \\ \Rightarrow x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=0\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ \Rightarrow \left(x-1\right)\left(x\left(x+1\right)-6\right)=0\left[\text{take}\left(x-1\right)\text{common}\right] \\ \Rightarrow \left(x-1\right)\left(x^2+x-6\right)=0 \\ \Rightarrow \left(x-1\right)\left(x^2-2x+3x-6\right)=0\left[Replacex=3x-2x\right] \\ \Rightarrow \left(x-1\right)\left(x\left(x-2\right)+3\left(x-2\right)\right)=0 \\ \Rightarrow \left(x-1\right)\left(x-2\right)\left(x+3\right)=0 \end{gathered}$$

Therefore, $\left(x-1\right)\left(x-2\right)\left(x+3\right)$ are the factor of $x^3-7x+6=0$.