# Thread: Dividing equations and finding quotiant.

1. ## Dividing equations and finding quotiant.

i)Find the quotient and the remainder when $\displaystyle 3x^3-2x^2+x+7$ is divided by $\displaystyle x^2-2x+5$#

ii)Hence, or otherwise, determine the values of the constants $\displaystyle a$ and $\displaystyle b$ such that, $\displaystyle 3x^3-2x^2+ax+b$ is divded by $\displaystyle x^2-2x+5$, there is no remainder.

Been looking at this for ages and have forgotten how to dot his kind of quiestion.
Thanks

2. Originally Posted by George321
i)Find the quotient and the remainder when $\displaystyle 3x^3-2x^2+x+7$ is divided by $\displaystyle x^2-2x+5$#

ii)Hence, or otherwise, determine the values of the constants $\displaystyle a$ and $\displaystyle b$ such that, $\displaystyle 3x^3-2x^2+ax+b$ is divded by $\displaystyle x^2-2x+5$, there is no remainder.

Been looking at this for ages and have forgotten how to dot his kind of quiestion.
Look at the leading term. $\displaystyle x^2$ divides into $\displaystyle 3x^3$ 3x times. Now multiply the entire divisor, $\displaystyle x^2- 2x+ 5$ by that and subtract:
$\displaystyle \begin{array}{cccc}3x^3& - 2x^2& + x& + 7 \\ \underline{3x^3}& \underline{- 6x^2}& \underline{+ 15x} & \\ & 4x^2 & -14x & 7\end{array}$
Now, $\displaystyle x^2$ will divide into $\displaystyle 4x^2$ 4 times. Multiply $\displaystyle x^2- 2x+ 5$ by 4 and subtract:
$\displaystyle \begin{array}{ccc}4x^2 & -14x & 7 \\\underline{4x^2}& \underline{-8x} & \underline{20}\\ & -6x& - 13\end{array}$.
That is, $\displaystyle x^2- 2x+ 5$ divides into $\displaystyle 3x^3- 2x^2+ x+ 7$ 3x+ 4 times with remainder -6x- 13.