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What are the zero(s) of the function f(x) =
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\[\frac{ 4x^2-36x }{ x-9 }\] @IMStuck
x = -9 x = 0 x = 9 x = 0 and x = 9
x = 9 is wrong, thought it was that one but got it wrong
Make x=0 what would you answer be?
would it be 0?
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If I were you, I would use long division of polynomials and find the zeros that way. LIke this:
4(0)^2 = 0 36(x) = 0 0 - 9 = -9 0 / -9 = 0? This chapter is confusing to me
Yes, it is 0 when x is 0. Because the numerator is 0 and 0 over anything is always 0. Plus if you factor the numerator, you get 4x(x-9) and that x-9 cancels out with the denominator leaving you with just 4x. 4x=0 when x = 0
Thanks, I got a 100%! :)
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