Question

How do you find all the zeros of `f(x)=12x^3+31x^2-17x-6`?

Answer 1

`f(x)` has zeros: `2/3`, `-1/4`, `-3`

Explanation:

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

By the rational root 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.

So the only possible rational zeros are:

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

That's rather a lot of possibilities to check, so let's search by approximately binary chop:

`f(0) = -6`

`f(1) = 12+31-17-7 = 19`

`f(1/2) = 3/2+31/4-17/2-6 = (6+31-34-24)/4 = -21/4`

`f(3/4) = 12(27/64)+31(9/16)-17(3/4)-6 = (81+279-204-96)/16 = 15/4`

`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)` is a factor:

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

To factor `4x^2+13x+2` we can 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 = 4x(x+3)+1(x+3) = (4x+1)(x+3)`

So the remaining zeros are `x=-1/4` and `x=-3`