Question

What are all the possible rational zeros for `y=x^4-2x^3-21x^2+22x+40` and how do you find all zeros?

Answer 1

All the zeros are for `x=-1, 2, -4, 5`

Explanation:

Let `f(x)=x^4-2x^3-21x^2+22x+40`
By trial and error
`f(-1)=1+2-21-22+40=0`
And `f(2)=16-16-84+44+40=0`
Therefore, `(x+1)(x-2)` is a factor
We have tto make a long division
`x^4-2x^3-21x^2+22x+40` `color(white)(aaaa)` `∣` `x^2-x-2`
`x^4-x^3-2x^2` `color(white)(aaaaaaaaaaaaaaaa)` `∣` `x^2-x-20`
`color(white)(a)` `0-x^3-19x^2+22x`
`color(white)(aaa)` `-x^3+x^2+2x`
`color(white)(aaaa)` `0-20x^2+20x+40`
`color(white)(aaaaaa)` `-20x^2+20x+40`
So `x^2-x-20=(x+4)(x-5)` these are factors
all the zeros are for `x=-1, 2, -4, 5`

Answer 2

The "possible" rational zeros are:

`+-1, +-2, +-4, +-5, +-8, +-10, +-20, +-40`

The actual zeros are: `-1`, `2`, `5` and `-4`.

Explanation:

`color(white)()`
`f(x) = x^4-2x^3-21x^2+22x+40`

By the rational roots 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 `40` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-5, +-8, +-10, +-20, +-40`

Start by trying each in turn:

`f(1) = 1-2-21+22+40 = 40`

`f(-1) = 1+2-21-22+40 = 0`

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

`x^4-2x^3-21x^2+22x+40 = (x+1)(x^3-3x^2-18x+40)`

Let `g(x) = x^3-3x^2-18x+40`

`g(-1) = -1-3+18+40 = 54`

`g(2) = (2)^3-3(2)^2-18(2)+40 = 8-12-36+40 = 0`

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

`x^3-3x^2-18x+40 = (x-2)(x^2-x-20)`

To factor the remaining quadratic note, find a pair of factors of `20` which differ by `1`. The pair `5, 4` works, so we find:

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

Hence the other two zeros are `color(blue)(x = 5)` and `color(blue)(x = -4)`.