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One x-intercept of the graph of the of the cubic function f(x) = x^3 – 2x^2 – 111x – 108 is –9.

What are the other zeros?

User Westlee
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1 Answer

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Answer:

The zeros of the given function are -9,-1 and 12.

Explanation:

The given function is


f(x)=x^3-2x^2-111x-108

It is given that the x-intercept of the graph is -9, therefore -9 is a zero of the function.

Since -9 is the zero of the function, therefore (x+9) is a factor of f(x). Use synthetics method or long division method to divide the function by (x+9).


f(x)=(x+9)(x^2-11x-12)=0


f(x)=(x+9)(x^2-12x+x-12)=0


f(x)=(x+9)(x(x-12)+(x-12))=0


f(x)=(x+9)(x-12)(x+1)=0

Use zero product property and quote each factor equal to 0.


x=-9,-1,12

Therefore zeros of the given function are -9,-1 and 12.

One x-intercept of the graph of the of the cubic function f(x) = x^3 – 2x^2 – 111x-example-1
User Andrsnn
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