Asked by Adebola
I need help finding the zeros of a function algebraically. And here is one of them f(x)= x^3-4x^2-9x+36
Answers
Answered by
Damon
try writing as:
(x^3 - 9x) -(4x^2-36) =0
x(x^2-9) -4 (x^2-9)=0
(x^2-9)(x-4) = 0
(x-3)(x+3)(x-4) = 0
3 , -3 , 4
(x^3 - 9x) -(4x^2-36) =0
x(x^2-9) -4 (x^2-9)=0
(x^2-9)(x-4) = 0
(x-3)(x+3)(x-4) = 0
3 , -3 , 4
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