# How do you factor `k(x)=2x^3- 5x^2-x+6` and solve for all zeros?

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There are a few ways to factor cubic polynomials. Here, the most effective way will unfortunately be a version of guess and check. The fastest way to find any rational factors will be checking the "p's and q's." `p`'s are all of the factors of the constant term, and `q`'s are all of the factors of the coefficient of `x^3` . All possible rational roots are described by the set of all combinations of `+-p/q`. So, we get all possibilities for rational roots:

Possible rational roots: `+- 6, +-3, +-2, +-3/2, +- 1, +-1/2`

It's generally easiest to throw these numbers into the equation and see for which possibilities results in k(x) = 0 (definition of "roots"). I usually start with 1 and -1 and go from there:

`k(1) = 2`

`k(-1) = 0`

We now have a single root, (x+1). We can use synthetic or long division (see reference link as it's difficult to picture in these answers) to get the following factors from the original equation:

`k(x) = (x+1)(2x^2-7x+6)`

We could continue to run through the possible rational roots above in the same way to get the other two roots, or we could just use the quadratic equation (third reference link) to get them without any guesswork.

`x = (7 +- sqrt(7^2-4(2)(6)))/(2(2)) = 3/2, 2`

Therefore, the factorization of the quadratic function above will be `2(x-3/2)(x-2)`.

Considering we have a fraction, we might want to clean it up by distributing the 2 and the `x-3/2` term to get the **following final factorization for k(x):**

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

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