Reduce to lowest terms:

$\displaystyle \frac {x^3+x^2+x+1}{x^3+3x^3+3x+1}$

I don't see how this can be factored to be reduced. Thanks.

The answer is $\displaystyle \frac{x^2+1}{x^2+2x+1}$

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- Dec 10th 2012, 08:58 PMKhanDisciplePolynomial Reduction Help
Reduce to lowest terms:

$\displaystyle \frac {x^3+x^2+x+1}{x^3+3x^3+3x+1}$

I don't see how this can be factored to be reduced. Thanks.

The answer is $\displaystyle \frac{x^2+1}{x^2+2x+1}$ - Dec 10th 2012, 10:30 PMgrillageRe: Polynomial Reduction Help
Just to be sure, is that really$\displaystyle 3x^3$ for the second term in the denominator?

- Dec 11th 2012, 01:15 AMMarkFLRe: Polynomial Reduction Help
For the numerator:

$\displaystyle f(x)=x^3+x^2+x+1$

You could use the rational roots theorem, which states that if the given polynomial has a rational root, it will come from the list $\displaystyle x=\pm1$. We can see that $\displaystyle x=+1$ cannot work, hence we find:

$\displaystyle f(-1)=-1+1-1+1=0$

So, we know $\displaystyle x+1$ is a factor of $\displaystyle f(x)$. Use of division finds:

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

Now, for the denominator, we should recognize the binomial coefficients arising from the cube of a binomial, i.e.:

$\displaystyle x^3+3x^2+3x+1=(x+1)^3$ and so we may state:

$\displaystyle \frac{x^3+x^2+x+1}{x^3+3x^2+3x+1}=\frac{(x+1)(x^2+ 1)}{(x+1)^3}=\frac{x^2+1}{(x+1)^2}=\frac{x^2+1}{x^ 2+2x+1}$ - Dec 11th 2012, 05:27 AMKhanDiscipleRe: Polynomial Reduction Help
grillage - you are correct it is $\displaystyle 3x^2$. Thanks for the help markFL2, I didn't know about the rational roots theorem, that will make working with polynomials a whole lot easier.