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Mathematics 15 Online
OpenStudy (anonymous):

What are the zero(s) of the function f(x) =

OpenStudy (anonymous):

\[\frac{ 4x^2-36x }{ x-9 }\] @IMStuck

OpenStudy (anonymous):

x = -9 x = 0 x = 9 x = 0 and x = 9

OpenStudy (anonymous):

x = 9 is wrong, thought it was that one but got it wrong

OpenStudy (anonymous):

Make x=0 what would you answer be?

OpenStudy (anonymous):

would it be 0?

OpenStudy (imstuck):

If I were you, I would use long division of polynomials and find the zeros that way. LIke this:

OpenStudy (anonymous):

4(0)^2 = 0 36(x) = 0 0 - 9 = -9 0 / -9 = 0? This chapter is confusing to me

OpenStudy (imstuck):

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

OpenStudy (anonymous):

Thanks, I got a 100%! :)

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