# long division using the remainder theorem

• Nov 21st 2005, 11:06 AM
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, 12:30 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

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

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

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

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

(can you see why?)

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

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

RonL