# How do you simplify (x^3 - 9x) / (x^2 - 7x + 12)?

May 21, 2018

$\frac{x \left(x + 3\right)}{x - 4}$

#### Explanation:

$\text{factorise numerator and denominator}$

$\textcolor{m a \ge n t a}{\text{factor numerator}}$

$\text{take out a "color(blue)"common factor } x$

$= x \left({x}^{2} - 9\right)$

${x}^{2} - 9 \text{ is a "color(blue)"difference of squares}$

$\text{which factors in general as}$

•color(white)(x)a^2-b^2=(a-b)(a+b)

$\text{here "a=x" and } b = 3$

$\Rightarrow {x}^{2} - 9 = \left(x - 3\right) \left(x + 3\right)$

$\Rightarrow {x}^{3} - 9 x = x \left(x - 3\right) \left(x + 3\right) \leftarrow \textcolor{red}{\text{factorised form}}$

$\textcolor{m a \ge n t a}{\text{factor denominator}}$

$\text{the factors of + 12 which sum to - 7 are - 3 and - 4}$

$\Rightarrow {x}^{2} - 7 x + 12 = \left(x - 3\right) \left(x - 4\right) \leftarrow \textcolor{red}{\text{factored form}}$

$\Rightarrow \frac{{x}^{3} - 9 x}{{x}^{2} - 7 x + 12}$

$= \frac{x \left(x - 3\right) \left(x + 3\right)}{\left(x - 3\right) \left(x - 4\right)}$

$\text{cancel the "color(blue)"common factor } \left(x - 3\right)$

$= \frac{x \cancel{\left(x - 3\right)} \left(x + 3\right)}{\cancel{\left(x - 3\right)} \left(x - 4\right)} = \frac{x \left(x + 3\right)}{x - 4}$

$\text{with restriction } x \ne 4$

May 21, 2018

(x(x+3))/(x-4

#### Explanation:

$\frac{{x}^{3} - 9 x}{{x}^{2} - 7 x + 12}$
color(teal)(=(x(x^2-9))/(x^2-3x-4x+12)

color(blue)(=(x(x+3)(x-3))/((x-3)(x-4))

color(magenta)(=(x(x+3)cancel((x-3)))/(cancel((x-3))(x-4))

color(green)(=(x(x+3))/(x-4

P.S.: Isn't the solution $\textcolor{b l u e}{c}$olorful?