Question

What are the zeros of the polynomial function f(x) = x3 − 2x2 − 8x?

Answer 1

Our strategy will aim to factor the polynomial as much as possible: once completely factored, the polynomial will become a multiplication of polynomials of lower degree:

`f(x) = f_1(x)\cdot f_2(x) \cdot f_3(x)`

and its zeroes will be the ones of its factors.

Since the polynomial has no constant term, you can factor it as follows:

`x^3 - 2x^2 - 8x = x(x^2 - 2x - 8)`

To continue, we must factor the quadratic expression in the parenthesis. A common way to factor expressions like
`ax^2+bx+c` is to find the two solutions
`x_1` and
`x_2` and write the polynomial as
`ax^2+bx+c=a(x-x_1)(x-x_2)`.

To find the solutions, we can use the quadratic formula

`x_(1,2) = (-b\pm√(b^2-4ac))/(2a)`

and since in our case
`a=1, b=-2,c=-8`, the solving formula becomes

`x_(1,2) = (2\pm√(4+32))/(2) = (2\pm6)/(2) = 1 \pm 3`

So, the two solutions are
`x_1 = -2` and
`x_2 = 4` and we write the polynomial as
`(x+2)(x-4)`.

So, the complete factorization is

`x^3 - 2x^2 - 8x = x(x+2)(x-4)`

So, the zeroes of the cubic polynomial we started with are the zeroes of the three polynomials in the factorization:
`f_1(x)=0` yields a solution for
`x=0`,
`f_2(x)=x+2` yields a solution for
`x=-2` and
`f_3(x)=x-4` yields a solution for
`x=4`.

Answer 2

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