Question

How do you find all the zeros of `f(x)=x^4+7x^3-x^2-67x-60`?

Answer 1

Zeros: `-1, 3, -4, -5`

Explanation:

`f(x)=x^4+7x^3-x^2-67x-60`

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-60` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros are:

`+-1, +-2, +-3, +-4, +-5, +-6, +-10, +-12, +-15, +-20, +-30, +-60`

We find:

`f(-1) = 1-7-1+67-60 = 0`

So `-1` is a zero and `(x+1)` is a factor:

`x^4+7x^3-x^2-67x-60 = (x+1)(x^3+6x^2-7x-60)`

Let `g(x) = x^3+6x^2-7x-60`

We find:

`g(3) = 27+54-21-60 = 0`

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

`x^3+6x^2-7x-60 = (x-3)(x^2+9x+20)`

To factor the remaining quadratic, note that `4+5 = 9` and `4 xx 5 = 20`, so we find:

`x^2+9x+20 = (x+4)(x+5)`

Hence remaining two zeros: `-4` and `-5`.