Question

Find only the rational zeros of the function. If there are none, state this.

Answer 1

-2, 1 and 2

EXPLANATION :

From the problem, we have the function :

`f(x)=x^5-3x^4+2x^3+8x^2-24x+16`

We can regroup the function :

`f(x)=(x^5-3x^4+2x^3)+(8x^2-24x+16)`

Factor out x^3 from the first group and 8 from the second group :

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

Then rewrite as a product of two polynomials :

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

The first factor is the "sum of two cubes" which can be factored in the form :

`x^3+y^3=(x+y)(x^2-xy+y^2)`

So that will be :

`\begin{gathered} f(x)=(x^3+8)(x^2-3x+2) \\ f(x)=(x+2)(x^2-2x+4)(x^2-3x+2) \end{gathered}`

Factor the last parenthesis :

`f(x)=(x+2)(x^2-2x+4)(x-2)(x-1)`

The zeros of the function is when f(x) = 0

So equate all factors to 0.

x + 2 = 0

x = -2

x - 2 = 0

x = 2

x - 1 = 0

x = 1

x^2 - 2x + 4 = 0

Use the determinant formula to check if it has rational or irrational roots.

`\begin{gathered} b^2-4ac=(-2)^2-4(1)(4) \\ =-12 \end{gathered}`

Since the result is a negative number, it has imaginary roots.

So the real and rational roots are -2, 1, and 2