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what are the zeros of the function f(x)=(3x^2-3x-6)/(3x-6)
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–1 1 2 –2
WILL GIVE MEDAL!!!!
@thomaster @Juliette32801
well, you have one there already, you just need to solve the quadratic factor there
What do you mean?
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3x^2 - 3x - 6 = 0 x^2 - x - 2 = 0 can u solve this?
is it this: 2x^2-2x-4=0?
just solve x^2 - x - 2 = 0 by factoring then check if the roots are valid
you need to check because the equation is fractional - the denominator cannot be zero
(x -2)(x + 1) = 0 so x = 2 . -1 which one is valid?
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x=-1,2
@cwrw238 so is it A OR C?
@uri
it cannot be 2 because the denominator 3x - 6 gives 3(2) - 6 = 0
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