# Long Division Polynomials

• Jun 24th 2012, 08:52 PM
waleedrabbani
Long Division Polynomials
(2x^3+5x^2-4x-3) / (2X +1) = 2X^3 + 5X^2-4X-3 Remainder zero

The back of the book says the remainder is -2

I checked over my work a bunch of times, and used google cant find anything help would be much appreciated
• Jun 24th 2012, 08:57 PM
Prove It
Re: Long Division Polynomials
Quote:

Originally Posted by waleedrabbani
(2x^3+5x^2-4x-3) / (2X +1) = 2X^3 + 5X^2-4X-3 Remainder zero

The back of the book says the remainder is -2

I checked over my work a bunch of times, and used google cant find anything help would be much appreciated

The remainder theorem states that when a polynomial \displaystyle \begin{align*} P(x) \end{align*} is divided by \displaystyle \begin{align*} ax + b \end{align*}, then the remainder is equal to \displaystyle \begin{align*} P\left(-\frac{b}{a}\right) \end{align*}.

So here \displaystyle \begin{align*} P(x) = 2x^3 + 5x^2 - 4x - 3 \end{align*} and you are dividing by \displaystyle \begin{align*} 2x + 1 \end{align*}, so the remainder will be equal to \displaystyle \begin{align*} P\left(-\frac{1}{2}\right) \end{align*}.

\displaystyle \begin{align*} P\left(-\frac{1}{2}\right) &= 2\left(-\frac{1}{2}\right)^3 + 5\left(-\frac{1}{2}\right)^2 - 4\left(-\frac{1}{2}\right) - 3 \\ &= 2\left(-\frac{1}{8}\right) + 5\left(\frac{1}{4}\right) + 2 - 3 \\ &= -\frac{1}{4} + \frac{5}{4} - 1 \\ &= 0 \end{align*}

I agree with your answer of 0, so either you have copied down the wrong question, or the book has a typo.