Question

How do you find all the zeros of `(3x^6+x^2-4)(x^3+6x+7)`?

Answer 1

This product has zeros:

• `1`

• `-1` (with multiplicity `2`)

• `+-(sqrt((4sqrt(3)-3)/12))+-(sqrt((4sqrt(3)+3)/12))i`

• `1/2+-(3sqrt(3))/2 i`

Explanation:

The zeros of `(3x^6+x^2-4)(x^3+6x+7)` are all the values of `x` which are zeros of `(3x^6+x^2-4)` or `(x^3+6x+7)`.

`color(white)()`
Zeros of `bb(3x^6+x^2-4)`

Note that all of the degrees are even and the sum of the coefficients is `0`.

Hence this sextic has zeros `+-1` and quadratic factor:

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

We find:

`3x^6+x^2-4 = (x^2-1)(3x^4+3x^2+4)`

Treating the remaining quartic factor as a quadratic in `x^2` and using the quadratic formula, we find:

`x^2 = (-3+-sqrt(3^2-4(3)(4)))/(2*3)`

`=(-3+-sqrt(9-48))/6`

`=(-3+-sqrt(-39))/6`

`=-1/2+-sqrt(39)/6i`

The square roots of `a+-bi` are:

`+-(sqrt((sqrt(a^2+b^2)+a)/2)) +- (sqrt((sqrt(a^2+b^2)-a)/2))i`

(see https://socratic.org/s/aw38evei)

So, putting `a=-1/2` and `b=sqrt(39)/6` we find:

`a^2+b^2 = (-1/2)^2+(sqrt(39)/6)^2 = 1/4+39/36 = 48/36 = 4/3`

So:

`sqrt(a^2+b^2) = sqrt(4/3) = sqrt(4/9*3) = (2sqrt(3))/3`

Hence:

`x = +-(sqrt(((2sqrt(3))/3-1/2)/2))+-(sqrt(((2sqrt(3))/3+1/2)/2))i`

`=+-(sqrt((4sqrt(3)-3)/12))+-(sqrt((4sqrt(3)+3)/12))i`

`color(white)()`
Zeros of `bb(x^3+6x+7)`

Notice that `-1` is a zero, so `(x+1)` is a factor and we find:

`x^3+6x+7 = (x+1)(x^2-x+7)`

The remaining quadratic factor has zeros given by the quadratic formula:

`x = (1+-sqrt((-1)^2-4(1)(7)))/(2*1)`

`=(1+-sqrt(-27))/2`

`=1/2+-(3sqrt(3))/2 i`