Question

How do you use the rational root theorem to find the roots of `x^4 - 7x^2 + 12 = 0`?

Answer 1

If `p/q` is a root of `x^4-7x^2+12=0` written in lowest terms, then `p` must be a divisor of the constant term `12` and `q` must be a divisor of the coefficient (`1`) of the highest order term `x^4`.

So `q = +-1`. We might as well choose `q = 1`, since we can choose the sign of `p`.

`p` must be a factor of `12`, i.e. `+-12`, `+-6`, `+-4`, `+-3`, `+-2` or `+-1`.

Since `x^4` grows quite fast with larger positive or negative values of `x`, it's probably better to try the possible smaller values of `x` first.

Trying a few substitutions, we quickly find that `x=+-1` are not solutions and `x=+-2` are solutions. When `|x| = 3`, we find that `x^4 - 7x^2 + 12 = 81 - 63 + 12 = 30`. Larger values of `x` will prduce larger values of `x^4-7x^2+12` as the `x^4` term becomes dominant.

So the only rational roots are `x = 2` and `x = -2`.

Answer 2

George C gives a great answer to find that `+-2` are zeros of the polynomial. (Roots of the equation.)

If you need to find all (real or complex) roots, there a bit more work to do.

Because, `2` and `-2` are roots, we know that
`x-2` and `x+2` are factors of the polynomial on the left..

Divide by the factors using synthetic division or long division. Or divide by the product of these 2 factors, `x^2-4`

You'll get

`x^2-7x+12=(x^2-4)(x^2-3)=0`

So the roots are:

`2, -2, sqrt3, -sqrt3`