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what are the zeros of the function f(x)=(3x^2-3x-6)/(3x-6) –1 1 2 –2 I am stuck with A or C
Start by factoring the top and the bottom. From the top, all the terms have a 3 as a factor, so you can factor out a 3. From the bottom, both terms also contain 3 as a factor, so you can factor out a 3 from the denominator as well
Ok
Then you need to cancel out the 3 from top and bottom and factor the X^2-x-2 into the form (x + ?) (x + ?)
So you have (x-2) * (x+1) / (x +2)
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Sorry we have (x-2) * (x+1) / (x - 2) So the x-2 can cancel out and leave the x+1 with a hole at 2 So the zero is x = -1
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