# How to factor x^3-8x^2+15x?

Apr 9, 2015

We can see that $x$ is common to all the terms. We can write it as

$x \cdot \left({x}^{2} - 8 x + 15\right)$

Now we need to factorise color(blue)(x^2-8x+15

We can use Splitting the Middle Term technique to factorise this.

It is in the form $a {x}^{2} + b x + c$ where $a = 1 , b = - 8 , c = 15$

To split the middle term, we need to think of two numbers ${N}_{1} \mathmr{and} {N}_{2}$ such that:
${N}_{1} \cdot {N}_{2} = a \cdot c \mathmr{and} {N}_{1} + {N}_{2} = b$
${N}_{1} \cdot {N}_{2} = \left(1\right) \cdot \left(15\right) \mathmr{and} {N}_{1} + {N}_{2} = - 8$
${N}_{1} \cdot {N}_{2} = 15 \mathmr{and} {N}_{1} + {N}_{2} = - 8$

After Trial and Error, we get ${N}_{1} = - 3 \mathmr{and} {N}_{2} = - 5$
$\left(- 3\right) \cdot \left(- 5\right) = 15$ and $\left(- 3\right) + \left(- 5\right) = - 8$

So we can write the expression ${x}^{2} - 8 x + 15$ as
$\textcolor{b l u e}{{x}^{2} - 3 x - 5 x + 15}$
$= x \left(x - 3\right) - 5 \left(x - 3\right)$
$= \left(x - 3\right) \cdot \left(x - 5\right)$

color(green)(x^3-8x^2+15x = x*(x-3)*(x-5)