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I have this:

$f(x) = 0$


$f(x) := \cfrac{3x^2 - 5x + 2}{x + 2}$

How do I solve that?

Do I multiply by $(x + 2)$ and solve $3x^2 - 5x + 2=0$ or solve $3x^3 + x^2 - 8x + 4=0$ with Horner method?

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Yes, you solve for the roots of the numerator of your rational function. You do not solve for the roots of $3x^3 + x^2 - 8x + 4$ since that is more complicated, and one of the roots of that polynomial is not a root of your original rational function. – J. M. Aug 9 '12 at 10:43
In plain words, do I solve it with Horner? so it's like a/b = 0 => a*b = 0? – Rrjrjtlokrthjji Aug 9 '12 at 10:44
The denominator doesn't enter into finding the roots, so just take the polynomial on top and solve for its roots. – J. M. Aug 9 '12 at 10:45
so I solve 3x^2 - 5x + 2 = 0? – Rrjrjtlokrthjji Aug 9 '12 at 10:46
Yes.$\phantom{}$ – J. M. Aug 9 '12 at 10:48
up vote 8 down vote accepted

In plain words, if $a/b$ is defined and equal to $0$, then $a=0$. Of course, this does imply that $a\cdot b=0,$ but that isn't relevant. You need only determine the solutions of $3x^2-5x+2=0$.

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I solved it as a*b = 0 and got 1 and -2 as roots – Rrjrjtlokrthjji Aug 9 '12 at 10:52
@Nick, $-2$ is not a valid solution. Try substituting it into your original expression. – J. M. Aug 9 '12 at 10:55
There should be another solution to that quadratic. Also, what is $f(-2)$? It's worth noting that $a\cdot b=0$ does not imply that $a/b=0$. It implies that $a/b=0$ or is undefined. – Cameron Buie Aug 9 '12 at 10:56
@Nick: Forget Horner; Horner's method has absolutely nothing to do with this problem. To make it extremely clear: $$\frac{a}{b} = 0 \Leftrightarrow a=0$$ Now $\frac{a}{b} = 0 \Rightarrow ab=0$, but $ab=0 \not \Rightarrow \frac{a}{b} = 0$. In this problem we only care about when $a=0$. – Clive Newstead Aug 9 '12 at 11:02
That is, $\frac{a}{b}=0$ if and only if $a=0$ and $b\neq 0$. – Cameron Buie Aug 9 '12 at 11:03

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