Question

How do you find the roots of `12x^3+31x^2-17x-6=0`?

Answer 1

Use the rational roots theorem, then factor to find:

`x = 2/3" "` or `" "x = -1/4" "` or `" "x = -3`

Explanation:

`f(x) = 12x^3+31x^2-17x-6`

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 `-6` and `q` a divisor of the coefficient `12` of the leading term.

That means that the only possible rational zeros are:

`+-1/12, +-1/6, +-1/4, +-1/3, +-1/2, +-2/3, +-3/4, +-1, +-3/2, +-2, +-3, +-6`

Notice that the signs of the coefficients of `f(x)` are in the pattern `+ + - -`. By Descartes' Rule of Signs, since there is one change of sign, this cubic has exactly one positive Real zero.

So let's look among the positive possibilities:

`f(1) = 12+31-17-6 = 20`

`f(1/2) = 12(1/8)+31(1/4)-17(1/2)-6 = (6+31-34-24)/4 = -21/4`

`f(2/3) = 12(8/27)+31(4/9)-17(2/3)-6 = (32+124-102-54)/9 = 0`

So `x=2/3` is a zero and `(3x-2)` a factor:

`12x^3+31x^2-17x-6 = (3x-2)(4x^2+13x+3)`

To factor the remaining quadratic, use an AC method:

Look for a pair of factors of `AC=4*3=12` with sum `B=13`.

The pair `12, 1` works.

Use this pair to split the middle term and factor by grouping:

`4x^2+13x+3 = (4x^2+12x)+(x+3)`

`color(white)(4x^2+13x+3) = 4x(x+3)+1(x+3)`

`color(white)(4x^2+13x+3) = (4x+1)(x+3)`

So the other two zeros are:

`x = -1/4" "` and `" "x = -3`