Question

Find all the real and complex zeros of f(x)=x^3-7x^2+4x-28

Answer 1

For
`x \in\mathbb{C}`

`x = -2i`

`x = 2i`

`x = 7`

Explanation:

`f(x)=x^3-7x^2+4x-28`

For
`f(x) = 0`

`x^3-7x^2+4x-28 = 0`

We can easily factor the expression
`x^3-7x^2+4x-28` by grouping.

`x^3-7x^2+4x-28 = (x^3-7x^2)+(4x-28) = x^2(x-7)+4(x-7) = \boxed{(x^2+4)(x-7) }`

Now we have

`(x^2+4)(x-7) = 0`

Therefore,

`x^2+4 = 0 \text{ or } x-7 = 0`

So,

`x^2 = -4 \implies x = \pm√(-4)`

Once
`√(-1) = i`

`x = \pm√(-4) = \pm√(4)√(-1) = \pm 2i`