Get rid of the denominator of $\frac{x^4-x^2+1}{x+3}$

Well, I put on Wolfram Alpha the equation $$\dfrac{x^4-x^2+1}{x+3}$$ and it tells me that this simplified is this:$$x^3-3x^2+8x+\frac{73}{x+3}-24$$ But I don't know how did he went there.

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You know how to do synthetic division, yes? – J. M. Apr 27 '12 at 16:45
By the way, you must have entered the expression in incorrectly. – David Mitra Apr 27 '12 at 16:46
How can a third degree polynomial on division by a first degree polynomial yield back a third degree polynomial? – Tomarinator Apr 27 '12 at 16:48
It must have been $\frac{x^4-x^2+1}{x+3}$ – Robert Israel Apr 27 '12 at 16:50

You have

$$\frac{{{x^4} - {x^2} + 1}}{{x + 3}}$$

This can be solved as an usual division:

$1.$ $x^4/x=\color{green}{x^3}$. We now multiply by $(x+3)$ and subtract it from our polynomial. That is

$x^4-x^2+1-x^3(x+3)=\color{red}{-3x^3-x^2+1}$

$2.$ $-3x^3/x=\color{blue}{-3x^2}$. Again, we multiply by $(x+3)$ and subtract it from our last result.

$-3x^3-x^2+1-(-3x^2)(x+3)=8x^2+1$

$3.$ $8x^2/x=\color{orange}{8x}$.

$8x^2+1-8x(x+3)=-24x+1$

$4.$ $-24x/x=\color{violet}{-24}$

$-24x+1-(-24)(x+3)=73$

Now that we have reduced the original expression to a degree lower than the dividend, we rearrange:

$$\color{green}{x^3}\color{blue}{-3x^2}+\color{orange}{8x}\color{violet}{-24}+\frac{73}{x+3}$$

A way cooler algorithm is as follows:

Complete the polynomial to get

$${x^4} + 0{x^3} - {x^2} + 0x + 1$$

Now arrange them in the following table, with the root of the dividend (i.e $-3$)

$\LaTeX$ coding beats me here, so this is what I can give you

This algorithm is very useful when dividing by monic binomial expressions $x-a$, plus, the remainder is $f(a)$, which helps to find the roots of certain polynomials when the remainder is $0$.

It seems english-speaking sites simply call it synthetic division, without mentioning Paolo Ruffini. In Argentina we call it "Ruffini's Rule".

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+1 for the pretty colours – TonyK Apr 27 '12 at 17:47
+1, The Wikipedia entry on Ruffini's rule is this – Américo Tavares Apr 27 '12 at 18:19
I tend to use "synthetic division" and "Horner's method" interchangeably myself; after all, evaluating a polynomial and dividing by a linear divisor is essentially the same problem. – J. M. Apr 27 '12 at 18:59

Generally one may employ the polynomial division algorithm, which is a polynomial analog of the grade-school integer long-division algorithm.

Lacking knowledge of that, one can shift the problem from division by $\rm\: x+3\:$ to the simpler problem of division by $\rm\:x,\:$ by shifting $\rm\:x \to x-3.\:$ Then the problem transforms to

$$\rm \frac{(x\!-\!3)^4\!-(x\!-\!3)^2+1}{x}\: =\: \frac{x^4\!-12x^3+53x^2\!-102x +73}{x}\: =\: x^3\!-12x^2\!+53x\!-\!102 +\!\frac{73}{x}$$ Finally, applying the inverse shift $\rm\:x \to x+3\:$ on the above equation yields $$\rm \frac{x^4\!-x^2+1}{x+3}\: =\: x^3\!-3x^2\!+8x-24 +\frac{73}{x+3}$$

Remark $\$ This is prototypical transformation-based problem solving. See my answer here for further discussion, which mentions as examples the formula for solving a quadratic equation, and Eisenstein's irreducibility criterion. See also the link there to an interesting problem-solving book by Melzak with many further examples.

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I was thinking about something like that to suggest but I guess it gets tedious for higher degree polynomials. However, it is very clever! (+1) – Pedro Tamaroff Apr 27 '12 at 18:01