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

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

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Answer 1 · 2017-03-01T00:28:03

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

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

Explanation:

Given:

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

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

That means that the only possible rational zeros are:

$\pm 1, \pm 3$ $+-1, +-3$

None of these work. Hence $F (x)$ $F(x)$ 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 $- 6 x$ $-6x$ in the question, since:

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

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

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Explanation:

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How do you factor F(x) = x^4+6x^3+2x^2-3F(x)=x4+6x3+2x2−3F(x) = x^4+6x^3+2x^2-3 ?

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Explanation:

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