Question

How do you use the rational root theorem to find the roots of `3x^4+2x^3-9x^2-12x-4=0`?

Answer 1

Use the rational root theorem to identify the possible roots, try some and narrow down the search to find roots `-2/3`, `-1`, `-1` and `2`.

Explanation:

`f(x) = 3x^4+2x^3-9x^2-12x-4`

By the rational root theorem any rational roots of `f(x)=0` are expressible in lowest terms as `p/q` for some integers `p` and `q`, where `p` is a divisor of the constant term `4` and `q` is a divisor of the coefficient `3` of the leading term.

So the possible rational roots are:

`+-1/3`, `+-2/3`, `+-1`, `+-4/3`, `+-2`, `+-4`

Try:

`f(1/3) = 1/27+2/27-1-4-4 = 1/9-9 = -80/9`

`f(-1/3) = 1/27-2/27-1+4-4 = -28/27`

`f(2/3) = 16/27+16/27-4-8-4 = 32/27`

`f(-2/3) = 16/27 - 16/27-4+8-4 = 0`

So `x=-2/3` is a root. This 'uses up' one of the possible factors of `2` for `p` and the factor of `3` for `q`. So the only other possible rational roots are:

`+-1`, `+-2`

Try:

`f(1) = 3+2-9-12-4 = -20`

`f(-1) = 3-2-9+12-4 = 0`

`f(2) = 48+16-36-24-4 = 0`

`f(-2) = 48-16-36+24-4 = 16`

So we have identified `3` rational roots out of `4` roots, viz `-2/3`, `-1` and `2`. The fourth root must also be rational and having 'used up' the factor of `2` for `p`, it must also be `-1`.

To check, we can multiply out the corresponding factors:

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

`=(3x^2+5x+2)(x^2-x-2)`

`=3x^4+(5-3)x^3-(6+5-2)x^2-(10+2)x-4`

`=3x^4+2x^3-9x^2-12x-4`