Question

How do you find all the zeros of `f(x) = x^3 + 11x^2 + 41x + 51`?

Answer 1

Use the rational root theorem to find zero `x = -3`, divide by `(x+3)` then solve the remaining quadratic to find zeros `x = -4+-i`.

Explanation:

`f(x) = x^3+11x^2+41x+51`

by the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p` and `q` where `p` is a divisor of the constant term `51` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational roots are:

`+-1`, `+-3`, `+-17`, `+-51`

In addition, note that all of the coefficients of `f(x)` are positive, so it can only have zeros for negative values of `x`.

That leaves:

`-1`, `-3`, `-17`, `-51`

Trying each in turn we find:

`f(-1) = -1+11-41+51 = 20`

`f(-3) = -27+99-123+51 = 0`

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

`x^3+11x^2+41x+51 = (x+3)(x^2+8x+17)`

Then solve:

`0 = x^2+8x+17`

`= x^2+8x+16+1`

`= (x+4)^2-i^2`

`= ((x+4)-i)((x+4)+i)`

`= (x+4-i)(x+4+i)`

So the remaining zeros are: `x = -4+-i`