# How do you use the remainder theorem to find the remainder for the division (3t^3-10t^2+t-5)div(t-4)?

Jul 28, 2018

The remainder is 31.

#### Explanation:

Remainder Theorem :
When we divide a polynomial $f \left(x\right)$ by (x−c) the remainder is $f \left(c\right)$
So to find the remainder after dividing by $\left(x - c\right)$ we don't need to do any division: Just calculate $f \left(c\right) .$

In this case,
$f \left(4\right) = 3 {\left(4\right)}^{3} - 10 {\left(4\right)}^{2} + 4 - 5$
$f \left(4\right) = 31$

Jul 28, 2018

The remainder is 31.

#### Explanation:

Note that the dividend is equal to the divisor times the quotient plus the remainder.

Let Q equal the quotient and R the remainder.

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

Notice that when $t = 4$, the divisor $t - 4$ becomes zero, meaning that anything multiplied to it will still result in zero.

Let's try $t = 4$

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

$\implies 192 - 160 + 4 - 5 = R$

$\implies 31 = R$

$31$
$\text{the remainder when "f(x)" is divided by "(x-a)" is } f \left(a\right)$
$3 {\left(4\right)}^{3} - 10 {\left(4\right)}^{2} + 4 - 5 = 31 \leftarrow \textcolor{b l u e}{\text{remainder}}$