# long division using the remainder theorem

• Nov 21st 2005, 12:06 PM
emydawn
long division using the remainder theorem
we have just learnt this topic but i dont understand how i would do this:

2x^4 + 3x^3 – 2x^2 + 4x – 6

divided by

x^2 + x – 2

using the remainder theorem
(aka in the form (Ax^2 + Bx + C)(x^2 + x – 2) + Dx + E

please help asap I have no idea how to do it, because there is a Dx and E at the end, and the power of 4 in the equn!

thanksss
xx
• Nov 22nd 2005, 01:30 PM
CaptainBlack
Quote:

Originally Posted by emydawn
we have just learnt this topic but i dont understand how i would do this:

2x^4 + 3x^3 – 2x^2 + 4x – 6

divided by

x^2 + x – 2

using the remainder theorem
(aka in the form (Ax^2 + Bx + C)(x^2 + x – 2) + Dx + E

please help asap I have no idea how to do it, because there is a Dx and E at the end, and the power of 4 in the equn!

thanksss
xx

$2x^4+3x^3-2x^2+4x-6\ =$ $(Ax^2+Bx+C)(x^2+x-2)+Dx+E$

Now we may factor $x^2+x-2$ as follows:

$x^2\ +\ x\ -\ 2\ =\ (x-1)(x+2)$

Then when $x=1,$
$2x^4 + 3x^3-2x^2 + 4x-6\ =\ D\ +\ E$

(can you see why?)

and when $x=-2,$
$2x^4 + 3x^3-2x^2 + 4x-6\ =\ -2D\ +\ E$

This gives a pair of simultaneous equations in $D$ and $E$, which I will leave for you to solve.

RonL