# How do you factor 3t^3-21t^2-12t?

Jun 9, 2018

$3 t \left(t - \frac{1}{2} \cdot \left(7 - \sqrt{65}\right)\right) \left(t - \frac{1}{2} \cdot \left(7 + \sqrt{65}\right)\right)$

#### Explanation:

We can write
$3 t \left({t}^{2} - 7 t - 4\right)$

to get the other factors we must solve the quadrat6ic

${t}_{1 , 2} = \frac{7}{2} \pm \sqrt{\frac{49}{4} + \frac{16}{4}}$
this is

${t}_{1 , 2} = \frac{7}{2} \pm \frac{\sqrt{65}}{2}$

Jun 9, 2018

$3 {t}^{3} - 21 {t}^{2} - 12 t = 3 t \left(t - \frac{7}{2} - \frac{\sqrt{65}}{2}\right) \left(t - \frac{7}{2} + \frac{\sqrt{65}}{2}\right)$

#### Explanation:

Given:

$3 {t}^{3} - 21 {t}^{2} - 12 t$

Note that all of the terms are divisible by $3 t$. So we can separate that out as a monomial factor:

$3 {t}^{3} - 21 {t}^{2} - 12 t = 3 t \left({t}^{2} - 7 t - 4\right)$

The remaining quadratic is in the form:

$a {t}^{2} + b t + c$

with $a = 1$, $b = - 7$ and $c = - 4$

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(\textcolor{b l u e}{- 7}\right)}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{- 4}\right) = 49 + 16 = 65$

Since $\Delta > 0$ this quadratic does have real zeros and hence linear factors with real coefficients, but since $\Delta$ is not a perfect square those zeros and coefficients are irrational.

We can use the quadratic formula to find them:

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{t} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{t} = \frac{7 \pm \sqrt{65}}{2}$

Hence:

${t}^{2} - 7 t - 4 = \left(t - \frac{7}{2} - \frac{\sqrt{65}}{2}\right) \left(t - \frac{7}{2} + \frac{\sqrt{65}}{2}\right)$

Putting it all together:

$3 {t}^{3} - 21 {t}^{2} - 12 t = 3 t \left(t - \frac{7}{2} - \frac{\sqrt{65}}{2}\right) \left(t - \frac{7}{2} + \frac{\sqrt{65}}{2}\right)$