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

Sep 2, 2016

Evaluate $f \left(3\right)$ The answer is the remainder.

In this case there is no remainder which means that (x-3) is a factor.

#### Explanation:

$\left(2 {x}^{3} - 3 {x}^{2} - 10 x + 3\right) \div \textcolor{red}{\left(x - 3\right)}$

$\rightarrow \text{make } \textcolor{red}{x - 3 = 0 , \rightarrow x = 3}$

$\text{Let } f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} - 10 x + 3$

$f \left(\textcolor{red}{3}\right)$ will give the remainder.

$f \left(3\right) = 2 {\left(3\right)}^{3} - 3 {\left(3\right)}^{2} - 10 \left(3\right) + 3$

$\textcolor{w h i t e}{\times \times} = 54 - 27 - 30 + 3$

$\textcolor{w h i t e}{\times \times} = 0 \text{ "larr " this is the remainder}$

There is no remainder.

This means that $\left(x - 3\right) \text{ is a factor of } 2 {x}^{3} - 3 {x}^{2} - 10 x + 3$