# Thread:

1. ## Polynomial Reduction Help

Reduce to lowest terms:
$\frac {x^3+x^2+x+1}{x^3+3x^3+3x+1}$

I don't see how this can be factored to be reduced. Thanks.
The answer is $\frac{x^2+1}{x^2+2x+1}$

2. ## Re: Polynomial Reduction Help

Just to be sure, is that really $3x^3$ for the second term in the denominator?

3. ## Re: Polynomial Reduction Help

For the numerator:

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

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

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

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

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

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

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

$\frac{x^3+x^2+x+1}{x^3+3x^2+3x+1}=\frac{(x+1)(x^2+ 1)}{(x+1)^3}=\frac{x^2+1}{(x+1)^2}=\frac{x^2+1}{x^ 2+2x+1}$

4. ## Re: Polynomial Reduction Help

grillage - you are correct it is $3x^2$. Thanks for the help markFL2, I didn't know about the rational roots theorem, that will make working with polynomials a whole lot easier.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9256000518798828, "perplexity_flag": "middle"}