Question

What are the possible rational roots of `x^3-5x^2-4x+20=0` and then determine the rational roots?

Answer 1

The possible rational roots are `+-1,+-2,+-4,+-5,+-10,+-20`. The rational roots are `x=-2, x=2, x=5`.

Explanation:

`color(blue)1x^3-5x^2-4x+color(red)(20)=0`

The possible rational roots are the factors of the constant `color(red)20` divided by the factors of the leading coefficient `color(blue)1`. The factors of the constant are called `p` and the factors of the leading coefficient are called `q`.

`p/q=(+-1,+-2,+-4,+-5,+-10,+-20)/(+-1)=`

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

The rational roots of this particular function can be found by factoring.

Factor by grouping.

First, group the first two terms and the second two terms.

`(x^3-5x^2)color(white)a+(-4x+20)=0`

Factor out a GCF from each group.

`x^2(x-5)-4(x-5)=0`

Regroup.

`(x^2-4)(x-5)=0`

Factor `x^2-4` as the difference of squares.

`(x+2)(x-2)(x-5)=0`

Set each factor equal to zero and solve.

`x+2=0color(white)(aa)x-2=0color(white)(aa)x-5=0`

`x=-2color(white)(aaa)x=2color(white)(aaa)x=5`