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

Mar 14, 2017

$P \left(3\right) = 168$

#### Explanation:

According to remainder theorem when we divide a polynomial $P \left(x\right)$ (here $P \left(x\right) = 6 {x}^{3} - {x}^{2} + 4 x + 3$), by a binomial of degree one say $\left(x - a\right)$, (here $\left(x - 3\right)$ as we desire $P \left(a\right)$ wih $a = 3$)

the remainder is $P \left(a\right)$

Normally to find $P \left(a\right)$, one substitute $x$ with $a$, but as we have to use Remainder theorem, we should divide $P \left(x\right)$ by $\left(x - a\right)$ and then the remainder would be $P \left(a\right)$. For this let us use synthetic division to divide $6 {x}^{3} - {x}^{2} + 4 x + 3$ by $\left(x - 3\right)$.

$3 | \textcolor{w h i t e}{X} 6 \text{ "color(white)(X)-1color(white)(XXX)4" "" } 3$
$\textcolor{w h i t e}{1} | \text{ } \textcolor{w h i t e}{X X \times} 18 \textcolor{w h i t e}{X X X} 51 \textcolor{w h i t e}{X X} 165$
" "stackrel("—————————————)
$\textcolor{w h i t e}{x} | \textcolor{w h i t e}{X} \textcolor{b l u e}{6} \textcolor{w h i t e}{X \times x} \textcolor{b l u e}{17} \textcolor{w h i t e}{X X X} \textcolor{b l u e}{55} \textcolor{w h i t e}{X X} \textcolor{red}{168}$

i.e. while quotient is $6 {x}^{2} + 17 x + 55$, remainder is $168$.