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For What Value(s) of x does the expression below equal 0? (x-2)(x+1)/ x(x+3)
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set the numerator equal to zero and solve
How ?
solve \[(x-2)(x+1)=0\]
what about the denominator
if you have \[\frac{f(x)}{g(x)}=0\] if you multiply by g(x) to both sides you get \[f(x)=0\] so the zeros are determined by the numerator (provided that g(x) is not zero at the same time) also, if you have a product of functions equal to zero set each equal to zero and solve so \[(x−2)(x+1)=0\Rightarrow (x-2)=0\text{ or }(x+1)=0\]
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so thats it or is there more to it?
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