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Why must the constant of the divisor be negated in synthetic division?

For example, if one was dividing a polynomial by $x+a$, $a$ must be changed to $-a$. Why is this?

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2 Answers 2

Synthetic division is best thought of as hijacking an efficient way to evaluate a polynomial (Horner's scheme) to determine a quotient and remainder.

Any polynomial $p(x)$ can be decomposed like so:

$$p(x)=(x-a)Q(x)+R$$

where $Q(x)$ is the quotient and $R$ is the remainder. If you let $x=a$ in both sides of the equation, you find that getting the remainder of $\frac{p(x)}{x-a}$ is the same as evaluating the polynomial $p(x)$ at $x=a$. The negation in synthetic division, then, amounts to solving for $x$ in the equation $x-a=0$.

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Synthetic division is based on the fact that evaluating a polynomial at $x=a$ is the same finding the remainder when dividing by $x-a$. (The reason why it's $x-a$ and not $x+a$ is that $x=a$ if and only if $x-a=0$, so there is a connection between $x-a$ and $x=a$).

So if you want to divide by $x+a = x-(-a)$, you evaluate at $-a$, which is what synthetic division is essentially doing.

To see what I mean when I say that synthetic division is computing the quotient by evaluating, consider the problem of dividing $x^4 - 2x^2 + 2x - 7$ by $x-3$. The way synthetic division works is by writing out the coefficients of the polynomial, and evaluating at $3$; then you add the result to the next coefficient: $$\begin{align*} &\begin{array}{r|rrrrr} &1 & 0 & -2 & 2 & -7\\ 3& & & & &\\ \hline & 1 \end{array}\\ &\begin{array}{r|rrrrr} &1 & 0 & -2 & 2 & -7\\ 3& & 3 & & &\\ \hline & 1& 3 \end{array}\\ &\vdots\\ &\begin{array}{r|rrrrr} &1 & 0 & -2 & 2 & -7\\ 3& & 3 & 9 &21& 69 &\\ \hline & 1 & 3 & 7 & 23 & 62 \end{array} \end{align*}$$ What we are really doing is computing the value of the polynomial $x^4 -2x^2 + 2x - 7$ at $x=3$ by looking at the polynomial as $$ x\Biggl(x\Bigl(x\bigl(x+0\bigr) -2\Bigr)+2\Biggr)-7$$ so we are evaluating "inside out". First doing $3+0$; then multiplying by $3$ and subtracting $2$; then multiplying by $3$ and adding $2$; then multiplying by $3$ and subtracting $7$: $$\begin{align*} 3+0 &\longrightarrow 3(3+0) - 2 \\ &\longrightarrow 3\Bigl(3\bigl(3+0\bigr)-2\Bigr) + 2\\ &\longrightarrow 3\Biggl(3\Bigl(3\bigl(3+0\bigr)-2\Bigr)+2\Biggr)-7 \end{align*}$$ which is the same as evaluating $p(x)$ at $x=3$, which is the same as finding the remainder when you divide $p(x)$ by $x-3$.

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Whoops, you beat me. :) I'll delete my answer in a bit... –  J. M. May 4 '11 at 19:53
    
@J.M. No, no; you have Horner's scheme there mentioned explicitly; I just added it ad hoc-like. I say keep yours as well. –  Arturo Magidin May 4 '11 at 20:04

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