Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I cant figure this out. $$\frac{x-5}{x^3-3x^2+7x-5}$$

I tried by grouping and got $$x^2(x-3)+1(7x-5)$$ for the denominator. I need to use partial fractions on this so I cant use that yet.

share|cite|improve this question
Take $x=1$. ${}{}$ – Git Gud Mar 3 '13 at 19:28
up vote 2 down vote accepted

Let $p(x)=x^3-3x^2+7x-5$. From the rational root test you know that the only possible rational roots of $p(x)$ are $\pm1$ and $\pm5$. Since $p(1)=0$, you know that $1$ is a root and therefore that $x-1$ is a factor, so divide it out to get $p(x)=(x-1)(x^2-2x+5)$. The discriminant of $x^2-2x+5$ is $(-2)^2-4\cdot1\cdot5<0$, so the quadratic factor is irreducible over the real numbers.

share|cite|improve this answer

Clearly (why? Inspection...) , $\,x=1\,$ is a root, thus


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.