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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.

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Take $x=1$. ${}{}$ –  Git Gud Mar 3 '13 at 19:28
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2 Answers

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.

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

$$x^3-3x^2+7x-5=(x-1)(x^2-2x+5)$$

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