Question

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

Answer 1

`f(x)` has zeros `x=2` and `x=-3/2+-sqrt(13)/2`

Explanation:

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `2` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-2`

We find:

`f(2) = 8+4-14+2 = 0`

So `x=2` is a zero and `(x-2)` a factor:

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

We can factor the remaining quadratic expression by completing the square. I will multiply by `4` first to cut down on the fractions involved:

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

`=4x^2+12x-4`

`=(2x+3)^2-9-4`

`=(2x+3)^2-(sqrt(13))^2`

`=((2x+3)-sqrt(13))((2x+3)+sqrt(13))`

`=(2x+3-sqrt(13))(2x+3+sqrt(13))`

Hence `x = -3/2+-sqrt(13)/2`