# How do you find the zeros of f(x)=(x^2−3x−4)/(x^2−x−12)?

How do you find the zeros of $f\left(x\right)=\frac{{x}^{2}-3x-4}{{x}^{2}-x-12}$?
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enreciarpv
Factorising and simplifying f(x) as follows.
$f\left(x\right)=\frac{{x}^{2}-3x-4}{{x}^{2}-x-12}=\frac{\overline{)\left(x-4\right)}\left(x+1\right)}{\overline{)\left(x-4\right)}\left(x-3\right)}=\frac{x+1}{x-3}$
with exclusion x ≠ 4
The zeros of f(x) are the values of x which make f(x) equal zero. That is f(x)=0.
For $f\left(x\right)=\frac{x+1}{x+3}$
The denominator of f(x) cannot be zero as this would make f(x) undefined. The numerator is the only part of the rational function that can equal zero.
$⇒x+1=0⇒x=-1\phantom{\rule{1ex}{0ex}}\text{is the zero of f(x)}$