# How do you use the remainder theorem to find P(c) p(x)=x^4+3x^3-4x^2+6x-31; 2?

Jul 16, 2015

Divide $P \left(x\right)$ by $x - 2$, the remainder is $P \left(2\right)$

#### Explanation:

$P \left(x\right) = {x}^{4} + 3 {x}^{3} - 4 {x}^{2} + 6 x - 31$.

The Remainder Theorem tells us that if we divide $P \left(x\right)$, by $x - c$, then the remainder is $P \left(c\right)$.

So we'll divide $P \left(x\right)$ by $x - 2$. You could use long division, but synthetic division requires less writing and we can use it for any linear divisor with leading coefficient $1$, so:

$\text{2} | |$ $\text{1 " " " "3 " " " "-4 " " " "6" " " } - 31$
$+$ $\text{ " " " " "" "2" " " " 10" " " "12" " " " " "36}$
$\text{ " }$$- - - - - - - - -$
$\text{ " " " 1" " " " 5" " "" " "6" " " "18" " " "||" " 5}$

The remainder is $5$, so $P \left(2\right) = 5$

If you use long division, it will look something like:

$\text{ " " " " " }$ ${x}^{3}$ $+ 5 {x}^{2}$$+ 6 x$ $+ 18$
$\text{ " " }$$- - - - - - - - - - -$
x-2 ) ${x}^{4}$ $+ 3 {x}^{3}$ $- 4 {x}^{2}$ $+ 6 x$ $- 31$
$\text{ " " }$ ${x}^{4}$ $- 2 {x}^{3}$
$\text{ " " }$$- - - - - - - - - - -$
$\text{ " " " " "" }$ $5 {x}^{3}$$\text{ }$ $- 4 {x}^{2}$ $+ 6 x$ $- 31$
$\text{ " " " " "" }$ $5 {x}^{3}$$\text{ }$ $- 10 {x}^{2}$
$\text{ " " }$$- - - - - - - - - - -$
$\text{ " " " " " " " " "" " " " " }$ $6 {x}^{2}$ $+ 6 x$ $- 31$
$\text{ " " " " " " " " "" " " " " }$ $6 {x}^{2}$ $- 12 x$
$\text{ " " }$$- - - - - - - - - - -$
$\text{ " " " " " " " " "" " " " " " " " " }$ $18 x$ $- 31$
$\text{ " " " " " " " " "" " " " " " " " " }$ $18 x$ $- 36$
$\text{ " " }$$- - - - - - - - - - -$
$\text{ " " " " " " " " "" " " " " " " " " " " " " " " }$ $5$

Again, the remainder is $5$, so $P \left(2\right) = 5$