Question

What are all the possible rational zeros for `y=x^3+x^2-9x+7` and how do you find all zeros?

Answer 1

`x=1` or `x = -1+-2sqrt(2)`

Explanation:

`f(x) = x^3+x^2-9x+7`

By the rational roots 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 `7` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-7`

Since the sum of the coefficients of `f(x)` is zero (i.e. `1+1-9+7 = 0`) then `f(1) = 0`, `x=1` is a zero and `(x-1)` a factor:

`x^3+x^2-9x+7 = (x-1)(x^2+2x-7)`

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

as follows:

`x^2+2x-7 = x^2+2x+1-8`

`color(white)(x^2+2x-7) = (x+1)^2-(2sqrt(2))^2`

`color(white)(x^2+2x-7) = ((x+1)-2sqrt(2))((x+1)+2sqrt(2))`

`color(white)(x^2+2x-7) = (x+1-2sqrt(2))(x+1+2sqrt(2))`

Hence the remaining zeros are:

`x = -1+-2sqrt(2)`