# How do you factor F(x) = x^4+6x^3+2x^2-3 ?

Mar 1, 2017

This quartic has no rational or simple irrational factors, but:

${x}^{4} + 6 {x}^{3} + 2 {x}^{2} - 6 x - 3 = \left(x - 1\right) \left(x + 1\right) \left(x + 3 - \sqrt{6}\right) \left(x + 3 + \sqrt{6}\right)$

#### Explanation:

Given:

$F \left(x\right) = {x}^{4} + 6 {x}^{3} + 2 {x}^{2} - 3$

By the rational roots theorem, any rational zeros of $F \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 3$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 3$

None of these work. Hence $F \left(x\right)$ has no rational zeros.

It is a fairly typically nasty quartic with two real irrational zeros and two complex ones.

I suspect there may be a missing term $- 6 x$ in the question, since:

${x}^{4} + 6 {x}^{3} + 2 {x}^{2} - 6 x - 3 = \left(x - 1\right) \left(x + 1\right) \left(x + 3 - \sqrt{6}\right) \left(x + 3 + \sqrt{6}\right)$