Question

What are all the zeroes of `f(x)= x^3 +2x^2 -3x +20`?

Answer 1

`x = -4`

`x = 1 + 2i`

`x = 1 - 2i`

Explanation:

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` in lowest terms, where `p, q in ZZ`, `p` a divisor of the constant term `20` and `q` a divisor of the coefficient `1` of the leading term.

So the possible rational roots are the divisors of `20`:

`+-1`, `+-2`, `+-4`, `+-5`, `+-10`, `+-20`

After a bit of trial and error, find:

`f(-4) = -64+32+12+20 = 0`

so `x=-4` is a zero and `(x+4)` is a factor of `f(x)`.

`x^3+2x^2-3x+20=(x+4)(x^2-2x+5)`

Find the Complex zeros of the remaining quadratic factor using the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a) = (2+-sqrt(4-20))/2 = (2+-sqrt(-16))/2`

`= (2+-4i)/2 = 1 +-2i`