# How do you factor 3x^3-12x^2-45x?

Apr 24, 2017

You can write down $3 x \left({x}^{2} - 4 x - 15\right)$

#### Explanation:

If you take a look at your equation, all numeric terms can be divided by 3. Also notice that x is the smallest term found in all terms. Now you can write down:

$3 x \left({x}^{2} - 4 x - 15\right)$.

Apr 24, 2017

$= 3 x \left(x - \frac{4 + 2 \sqrt{19}}{2}\right) \left(x - \frac{4 - 2 \sqrt{19}}{2}\right)$

#### Explanation:

Factorization is determined either by taking common factor or
$\text{ }$
computing $\delta$
$\text{ }$
$3 {x}^{3} - 12 {x}^{2} - 45 x$
$\text{ }$
$= 3 x \left({x}^{2} - 4 x - 15\right)$
$\text{ }$
$\delta = {\left(- 4\right)}^{2} - 4 \left(1\right) \left(- 15\right)$
$\text{ }$
$\delta = 16 + 60$
$\text{ }$
$\delta = 76$
$\text{ }$
Since $\delta > 0$ so ${x}^{2} - 4 x - 15$ admits two roots :
$\text{ }$
The first root is:
$\text{ }$
${x}_{1} = \frac{- \left(- 4\right) + \sqrt{76}}{2}$
$\text{ }$
${x}_{1} = \frac{4 + 2 \sqrt{19}}{2}$
$\text{ }$
The second root is:
$\text{ }$
${x}_{1} = \frac{- \left(- 4\right) - \sqrt{76}}{2}$
$\text{ }$
${x}_{2} = \frac{4 - 2 \sqrt{19}}{2}$
$\text{ }$
Then :
$\text{ }$
${x}^{2} - 4 x - 15 = \left(x - \frac{4 + 2 \sqrt{19}}{2}\right) \left(x - \frac{4 - 2 \sqrt{19}}{2}\right)$
$\text{ }$
Therefore:
$\text{ }$
$3 {x}^{3} - 12 {x}^{2} - 45 x$
$\text{ }$
$= 3 x \left({x}^{2} - 4 x - 15\right)$
$\text{ }$
$= 3 x \left(x - \frac{4 + 2 \sqrt{19}}{2}\right) \left(x - \frac{4 - 2 \sqrt{19}}{2}\right)$