# How do you know if a polynomial is not factorable?

Feb 8, 2015

The most reliable way I can think of to find out if a polynomial is factorable or not is to plug it into your calculator, and find your zeroes. If those zeroes are weird long decimals (or don't exist), then you probably can't factor it. Then, you'd have to use the quadratic formula.

If you don't have a calculator, then I don't know of any definite way in which you can tell if a polynomial is factorable or not, but you should just try playing with the numbers (factor out stuff, etc), and see if it works.

Hope that helped :)

Oct 21, 2016

Any polynomial in one variable with Real coefficients factors as a product of linear and quadratic factors with Real coefficients, but those coefficients may be hard to find...

#### Explanation:

It is a common error (and one that I have made myself) to think that just because a polynomial has no Real zeros (and therefore no linear factors with Real coefficients), that it has no factors with Real coefficients. Such is not the case, as we shall observe.

Restricting the question to the case of polynomials in one variable with Real coefficients, note that in theory any such polynomial can be factored as a product of quadratic and linear factors with Real coefficients. We can tell whether any quadratic factors can be reduced to linear ones as follows...

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$f \left(x\right) = a {x}^{2} + b x + c$

It is possible to factor this into linear factors with Real coefficients if and only if its discriminant $\Delta = {b}^{2} - 4 a c$ is non-negative.

In the following examples we see polynomials that can be factored with different complexities of coefficients...

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Example 1

$f \left(x\right) = {x}^{4} + 4$

Note that for any Real value of $x$ we have ${x}^{4} \ge 0$ and hence $f \left(x\right) \ne 0$.

We can deduce that $f \left(x\right)$ has no linear factors with Real coefficients, but it is possible to factor it as a product of quadratics:

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

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Example 2

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

Again this has no linear factors, but we can find quadratic factors with irrational coefficients:

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

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Example 3

$h \left(x\right) = {x}^{5} + 4 x + 2$

This quintic has $5$ distinct zeros, of which one is Real and the other $4$ occur as two pairs of Complex conjugates.

Unfortunately for us, none of these zeros can be expressed in terms of elementary functions, including $n$th roots, trigonometric, exponential or logarithmic functions. There is a factorisation of the form:

${x}^{5} + 4 x + 2 = \left(x + a\right) \left({x}^{2} + b x + c\right) \left({x}^{2} + \mathrm{dx} + e\right)$

where $a , b , c , d$ and $e$ are Real, but about the best we can do is find numerical approximations to them.