Given the function f(x)=x^3-5x^2-12x+36 and given that f(6)=0, what are the roots of the function?

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Answer 1 · 2015-09-28T23:18:01

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Hence #f(6)=0# that means that #x-6# is a factor of the polynomial

#x^3-5x^2-12x+36 # hence we can phrase it as

#x^3-5x^2-12x+36=(x-6)*(x^2+bx+c)+d#

Doing same basic algebraic calculation we find that

#x^3-5x^2-12x+36=(x-6)*(x^2+x-6)+0=(x-6)(x^2+x-6)#

but #x^2+x-6=(x-2)*(x+3)# hence we have that

#x^3-5x^2-12x+36=(x-6)(x-2)(x+3)#

The roots are #6,2,-3#