# Math Help - Dividing equations and finding quotiant.

1. ## Dividing equations and finding quotiant.

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

ii)Hence, or otherwise, determine the values of the constants $a$ and $b$ such that, $3x^3-2x^2+ax+b$ is divded by $x^2-2x+5$, 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 $3x^3-2x^2+x+7$ is divided by $x^2-2x+5$#

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

Been looking at this for ages and have forgotten how to dot his kind of quiestion.
Thanks
Essentially, the same as long division of numbers:
Look at the leading term. $x^2$ divides into $3x^3$ 3x times. Now multiply the entire divisor, $x^2- 2x+ 5$ by that and subtract:
$\begin{array}{cccc}3x^3& - 2x^2& + x& + 7 \\ \underline{3x^3}& \underline{- 6x^2}& \underline{+ 15x} & \\ & 4x^2 & -14x & 7\end{array}$

Now, $x^2$ will divide into $4x^2$ 4 times. Multiply $x^2- 2x+ 5$ by 4 and subtract:
$\begin{array}{ccc}4x^2 & -14x & 7 \\\underline{4x^2}& \underline{-8x} & \underline{20}\\ & -6x& - 13\end{array}$.

That is, $x^2- 2x+ 5$ divides into $3x^3- 2x^2+ x+ 7$ 3x+ 4 times with remainder -6x- 13.

For the second problem, do the same thing but with "ax+ b" rather than "x+ 7". The remainder will involve a and b- set the coefficient of x and the constant term equal to 0 and solve for a and b.