SOLUTION: f(x)=(3x^2+9x+6)/(3x-3)....finde zeros of the function

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Question 611134: f(x)=(3x^2+9x+6)/(3x-3)....finde zeros of the function

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The zeros of a function are the x values that make the function value zero. Your function is a fraction and a fraction is equal to zero only when its numerator is zero (and its denominator is not zero). So all we have to do is figure out the solution(s) to 3x%5E2%2B9x%2B6+=+0 and reject any that would make 3x-3 zero.

3x%5E2%2B9x%2B6+=+0
This is a quadratic equation so we want one side to be zero. The right side is already zero so we can proceed to the next step: Factor (or use the Quadratic Formula). We start factoring, as usual, with the GCF. The GCF here is 3:
3%28x%5E2%2B3x%2B2%29+=+0
And then we factor the rest:
3(x+1)(x+2) = 0
From the Zero Product Property we know that a product can be zero only if one (or more) of the factors is zero. The factor of 3 cannot be zero. But the other two could:
x+1 = 0 or x+2 = 0
Solving these we get:
x = -1 or x = -2

Last of all we need to make sure that these x's do not make the denominator zero. If we replace the x in 3x-3 with -1 we get -6 and if we replace the x in in 3x-3 with -2 we get -9.

So both -1 and -2 are zeros of f(x).