# How do you use the remainder theorem to find the remainder for each division (4x^3+4x^2+2x+3)div(x-1)?

Apr 9, 2017

There are a few ways we can find the remainder. The two most common are long division and synthetic division. I prefer synthetic division, so I'll be using that.

$1 | 4 \textcolor{w h i t e}{. .} 4 \textcolor{w h i t e}{. .} 2 \textcolor{w h i t e}{. .} 3$
$\textcolor{w h i t e}{1} | \textcolor{w h i t e}{\ldots .4} 4 \textcolor{w h i t e}{. .} 8 \textcolor{w h i t e}{. .} 10$
$\textcolor{w h i t e}{1 |} \textcolor{b l a c k}{- - - - - -}$
$\textcolor{w h i t e}{1} \textcolor{w h i t e}{|} 4 \textcolor{w h i t e}{. .} 8 \textcolor{w h i t e}{. .} 10 \textcolor{w h i t e}{.} 13$

That leaves us with $4 {x}^{2} + 8 x + 10$ and a remainder of $13$. To write this as an equation, we need to that our remainder and place it over the divisor ($x - 1$). That means our final solution is $y = 4 {x}^{2} + 8 x + 10 + \frac{13}{x - 1}$.

Apr 15, 2017

$x - 1$ is not a factor. The remainder will be 13.

#### Explanation:

The remainder theorem is a quick and useful way to determine whether an expression as a factor before you actually go ahead with the whole dividing process.

Let: $f \left(\textcolor{b l u e}{x}\right) = 4 {x}^{3} + 4 {x}^{2} + 2 x + 3$

The divisor is $\left(x - 1\right)$ . Set it equal to 0 and solve for $x$
$x - 1 = 0 \text{ } \rightarrow \textcolor{b l u e}{x = 1}$

$f \textcolor{b l u e}{\left(1\right)} = 4 {\textcolor{b l u e}{\left(1\right)}}^{3} + 4 {\textcolor{b l u e}{\left(1\right)}}^{2} + 2 \textcolor{b l u e}{\left(1\right)} + 3 = 13$

If f(x) =0, then the expression is a factor.
Here, $f \left(1\right) = 13 , s o \left(x - 1\right)$ is NOT a factor, and the remainder will be $13$