Question

If f(–2) = 0, what are all the factors of the function f(x)=x^3 -2x^2-68x-120
Use the Remainder Theorem.
A) (x + 2)(x + 60)
B) (x – 2)(x – 60)
C) (x – 10)(x + 2)(x + 6)
D) (x + 10)(x – 2)(x – 6)

Answer 1

f(-2) = 0 ⇒ x+2 is a factor of x^3 -2x^2 -68x -120

Then you can divide x^3 -2x^2 -68x -120 by x + 2.

The quotient of that division is x^2 - 4x -60 [you should know how to divide polynomilas]

Now factor x^2 - 4x -60

x^2 -4x -60 = (x - 10)(x + 6)

Then the factors are (x+2)(x-10)(x+6)

Which is the option C).

Answer 2

The factors of the function are -2 , 6 , 10 .

What is Remainder Theorem ?

If we divide a polynomial by a (x-a)  then the polynomial obtained will have a smaller polynomial with a remainder.

It is given that

(x +2) is a factor of the polynomial given

Then on dividing
x³ -2x²-68x-120 by x+2

we get

x²-4x-60

The factors

x² - 10x+6x -60

x(x-10)- 6(x-10)

(x-6)(x-10)

Therefore the factors are -2 , 6 , 10 .

To know more about Remainder Theorem

https://brainly.com/question/13264870

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