Question

What are the roots of `6x^4-11x^3-28x^2-15x+18=0` ?

Answer 1

The roots of `6x^4-11x^3-28x^2color(red)(+)15x+18=0` are:

`1, 3, -2/3, -3/2`

Explanation:

Assuming that this problem is supposed to be reasonably tractable, there is probably a typo in the question, with the correct quartic being:

`6x^4-11x^3-28x^2color(red)(+)15x+18=0`

First note that the sum of the coefficients is `0`, that is:

`6-11-28+15+18 = 0`

So `x=1` is a root and `(x-1)` a factor:

`6x^4-11x^3-28x^2+15x+18 = (x-1)(6x^3-5x^2-33x-18)`

By the rational roots theorem, any rational zeros of the remaining cubic are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-18` and `q` a divisor of the coefficient `6` of the leading term.

Further note that since `6` has only `2` prime factors, then if all of the zeros are rational then at least one is an integer factor of `18`.

Trying a few possibilities we find `x=3` is a zero and `(x-3)` a factor:

`6^3-5x^2-33x-18 = (x-3)(6x^2+13x+6)`

To factor the remaining quadratic use an AC method:

Find a pair of factors of `AC = 6*6 = 36` with sum `B=13`.

The pair `9,4` works.

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

`6x^2+13x+6 = (6x^2+9x)+(4x+6)`

`color(white)(6x^2+13x+6) = 3x(2x+3)+2(2x+3)`

`color(white)(6x^2+13x+6) = (3x+2)(2x+3)`

Hence the other two zeros are:

`x = -2/3" "` and `" "x = -3/2`