# How do you use the remainder theorem to determine the remainder when the polynomial (2t-4t^3-3t^2)/(t-2)?

Mar 31, 2017

The remainder is $- 40$

#### Explanation:

According to remainder theorem, when a polynomial function $f \left(x\right)$ is divided by $\left(x - a\right)$, the remainder is $f \left(a\right)$.

Here we have the function $f \left(t\right) = - 4 {t}^{3} - 3 {t}^{2} + 2 t$, which is divided by $t - 2$ and hence the remainder is

$f \left(2\right) = - 4 \times {2}^{3} - 3 \times {2}^{2} + 2 \times 2$

$= - 4 \times 8 - 3 \times 4 + 4$

$= - 32 - 12 + 4 = - 40$

One can check it too

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

= $- 4 {t}^{2} \left(t - 2\right) - 8 {t}^{2} - 3 {t}^{2} + 2 t = - 4 {t}^{2} \left(t - 2\right) - 11 {t}^{2} + 2 t$

(we have subtracted $8 {t}^{2}$ to compensate for $- 4 {t}^{2} \times \left(- 2\right) = 8 {t}^{2}$)

= $- 4 {t}^{2} \left(t - 2\right) - 11 t \left(t - 2\right) - 22 t + 2 t = - 4 {t}^{2} \left(t - 2\right) - 11 t \left(t - 2\right) - 20 t$

= $- 4 {t}^{2} \left(t - 2\right) - 11 t \left(t - 2\right) - 20 \left(t - 2\right) - 40$

= $\left(- 4 {t}^{2} - 11 t - 20\right) \left(t - 2\right) - 40$

i.e. quotient is $\left(- 4 {t}^{2} - 11 t - 20\right)$ and remainder is $- 40$