# How do you simplify (t^5 - 3t^2 - 20) (t-2)^-1?

Jun 25, 2018

$\left({t}^{5} - 3 {t}^{2} - 20\right) {\left(t - 2\right)}^{- 1} = {t}^{4} + 2 {t}^{3} + 4 {t}^{2} + 5 t + 10$

#### Explanation:

We can write $\left({t}^{5} - 3 {t}^{2} - 20\right) {\left(t - 2\right)}^{- 1}$ as

$\frac{{t}^{5} - 3 {t}^{2} - 20}{t - 2}$

Now observe that for the numerator $f \left(t\right) = {t}^{5} - 3 {t}^{2} - 20$,

$f \left(2\right) = {2}^{5} - 3 \cdot {2}^{2} - 20 = 32 - 12 - 20 = 0$ and from factor theorem $\left(t - 2\right)$ is a factor of ${t}^{5} - 3 {t}^{2} - 20$

and hence we can divide ${t}^{5} - 3 {t}^{2} - 20$ by $t - 2$ and

${t}^{5} - 3 {t}^{2} - 20$

= ${t}^{4} \left(t - 2\right) + 2 {t}^{3} \left(t - 2\right) + 4 {t}^{2} \left(t - 2\right) + 5 t \left(t - 2\right) + 10 \left(t - 2\right)$

= (t-2)(t^4+2t^3+4t^2+5t+10

and hence $\left({t}^{5} - 3 {t}^{2} - 20\right) {\left(t - 2\right)}^{- 1} = {t}^{4} + 2 {t}^{3} + 4 {t}^{2} + 5 t + 10$