Question

How do you find the rational roots of `x^4+5x^3+7x^2-3x-10=0`?

Answer 1

Use the rational roots theorem to find rational roots `x=1` and `x=-2`.

Explanation:

Let `f(x) = x^4+5x^3+7x^2-3x-10`

The rational roots theorem tells us that all rational roots of `f(x) = 0` must be of the form `p/q` where `p` and `q` are integers, `q != 0`, `p` is a divisor of the constant term `10` and `q` is a divisor of the coefficient `1` of the highest order term `x^4`.

So the possible rational roots are:

`+-1`, `+-2`, `+-5` and `+-10`

Note that the sum of the coefficients of `f(x)` is `0`:

`1+5+7-3-10 = 0`

So `f(1) = 0`.

That is `x=1` is a root of `f(x) = 0` and `(x-1)` is a factor of `f(x)`

We also find

`f(-2) = 16-40+28+6-10 = 0`

So `x=-2` is a root of `f(x) = 0` and `(x+2)` is a factor of `f(x)`

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

We can long divide `f(x)` by `x^2+x-2` to find:

`f(x)/(x^2+x-2) = x^2+4x+5`

A quick check of the discriminant of this quotient polynomial finds:

`Delta = 4^2-(4xx1xx5) = 16-20 = -4`

So there are no more real roots and no more linear factors with real coefficients.