# Factorize x^3-7x+6?

Dec 29, 2016

Factors of ${x}^{3} - 7 x + 6 = 0$ are $\left(x - 1\right) \left(x + 2\right) \left(x - 3\right) = 0$

#### Explanation:

As the coefficients ($1 - 7 + 6 = 0$) in the given equation add up to zero, it is apparent that $x = 1$ is a solution of ${x}^{3} - 7 x + 6 = 0$ and $\left(x - 1\right)$ is a factor of ${x}^{3} - 7 x + 6$.

Dividing ${x}^{3} - 7 x + 6$ by $\left(x - 1\right)$

${x}^{3} - 7 x + 6 = {x}^{2} \left(x - 1\right) + x \left(x - 1\right) - 6 \left(x - 1\right)$

= $\left(x - 1\right) \left({x}^{2} + x - 6\right)$ - now splitting the middle term

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

= $\left(x - 1\right) \left(x \left(x + 2\right) - 3 \left(x + 2\right)\right)$

= $\left(x - 1\right) \left(x + 2\right) \left(x - 3\right)$

Hence factors of ${x}^{3} - 7 x + 6 = 0$ are $\left(x - 1\right) \left(x + 2\right) \left(x - 3\right) = 0$

Dec 29, 2016

The answer is $= \left(x - 1\right) \left(x + 3\right) \left(x - 2\right)$

#### Explanation:

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

$f \left(1\right) = 1 - 7 + 6 = 0$

Therefore,

$\left(x - 1\right)$ is a factor

To find the other factors, we do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{3}$$\textcolor{w h i t e}{a a a a a}$$- 7 x + 6$$\textcolor{w h i t e}{a a a a a a}$∣$x - 1$

$\textcolor{w h i t e}{a a a a}$${x}^{3} - {x}^{2}$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaaaaaaa)∣${x}^{2} + x - 6$

$\textcolor{w h i t e}{a a a a a}$$0 + {x}^{2} - 7 x$

$\textcolor{w h i t e}{a a a a a a a}$$+ {x}^{2} - x$

$\textcolor{w h i t e}{a a a a a a a a}$$+ 0 - 6 x + 6$

$\textcolor{w h i t e}{a a a a a a a a a a a a}$$- 6 x + 6$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$- 0 + 0$

Therefore,

$\frac{{x}^{3} - 7 x + 6}{x - 1} = {x}^{2} + x - 6 = \left(x + 3\right) \left(x - 2\right)$

So,

$\left({x}^{3} - 7 x + 6\right) = \left(x - 1\right) \left(x + 3\right) \left(x - 2\right)$