Question

How do you find all the zeros of `h(x)=x^4+6x^3+10x^2+6x+9`?

Answer 1

`x=-3` (multiplicity 2)

`x=+-i`

Explanation:

`h(x) = x^4+6x^3+10x^2+6x+9`

By the rational root theorem, any rational zeros of `h(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `9` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-3`, `+-9`

In addition, note that all of the coefficients of `h(x)` are positive. So it has no positive zeros. So the only possible rational zeros are:

`-1`, `-3`, `-9`

We find:

`h(-1) = 1-6+10-6+9 = 8`

`h(-3) = 81-162+90-18+9 = 0`

So `x=-3` is a zero and `(x+3)` a factor:

`x^4+6x^3+10x^2+6x+9 = (x+3)(x^3+3x^2+x+3)`

The remaining cubic factors by grouping:

`x^3+3x^2+x+3`

`=(x^3+3x^2)+(x+3)`

`=x^2(x+3)+1(x+3)`

`=(x^2+1)(x+3)`

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

So `x=-3` is a zero again, giving it a total multiplicity `2` and the other zeros are `x=+-i`