Question

How do you find all the zeros of `g(x) = x^5 – 8x^4 + 28x^3 – 56x^2 + 64x – 32`?

Answer 1

Use the rational root theorem, polynomial division and completing the square to find roots:

`x=2` (with multiplicity `3`)

`x=1+-sqrt(3)i`

Explanation:

`g(x)=x^5-8x^4+28x^3-56x^2+64x-32`

By the rational root theorem, any rational zeros of `g(x)` must be expressible in the form `p/q` for integers `p` and `q` with `p` a divisor of the constant term `32` and `q` a divisor of the coefficient `1` of the leading term.

In addition note that `g(-x) = -x^5-8x^4-28x^3-56x^2-64x-32` has no changes of sign in its coefficients, so there are no negative Real zeros.

Hence the only possible rational zeros are:

`1, 2, 4, 8, 16, 32`

We find:

`g(2) = 32-128+224-224+128-32 = 0`

So `x=2` is a zero and `(x-2)` a factor:

`x^5-8x^4+28x^3-56x^2+64x-32`

`=(x-2)(x^4-6x^3+16x^2-24x+16)`

Let `h(x) = x^4-6x^3+16x^2-24x+16`

`h(2) = 16-48+64-48+16 = 0`

So `x=2` is a zero of `h(x)` and `(x-2)` is a factor:

`x^4-6x^3+16x^2-24x+16 = (x-2)(x^3-4x^2+8x-8)`

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

`k(2) = 8-16+16-8 = 0`

So `x=2` is a zero of `k(x)` and `(x-2)` is a factor:

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

The remaining quadratic factor can be factored by completing the square and using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a = x - 1` and `b=sqrt(3)i`

`x^2-2x+4`

`= x^2-2x+1+3`

`= (x-1)^2-(sqrt(3)i)^2`

`= ((x-1)-sqrt(3)i)((x-1)+sqrt(3)i)`

`= (x-1-sqrt(3)i)(x-1+sqrt(3)i)`

Hence the last two zeros are:

`x = 1+-sqrt(3)i`