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.
Get rid of the denominator of $\frac{x^4-x^2+1}{x+3}$
-
0@ThomasAndrews Indeed, he should be aware of Gauss-Ruffini. – 2012-04-27
2 Answers
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".
-
0I 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. – 2012-04-27
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.
-
0I was thinking about something like that to suggest but I guess it gets tedious for higher degree polynomials. However, it is very clever! (+1) – 2012-04-27