# How can you tell whether a polynomial is factorable and find whatever factors it has?

Jan 27, 2017

Some thoughts...

#### Explanation:

First let's talk about what is meant by "factorable / not factorable".

If you are given a polynomial with integer coefficients then it may be factorable as a product of simpler polynomials also with integer coefficients.

For example:

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

I think we would agree that that counts as factorable.

The next level of complexity would be irrational coefficients, for example:

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

We could say that ${x}^{2} - 2 x - 2$ is not factorable over the rational numbers. That is it has no factors whose coefficients are rational numbers. However it is factorable if you allow irrational coefficients.

The next level would be Complex coefficients. For example:

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

The quadratic ${x}^{2} - 2 x + 2$ is not factorable over the reals. That is, it has no simpler factors with Real coefficients - only Complex coefficients.

We can tell whether a quadratic $a {x}^{2} + b x + c$ with Real coefficients is factorable over the reals by examining its discriminant:

$\Delta = {b}^{2} - 4 a c$

• If $\Delta > 0$ then $a {x}^{2} + b x + c$ has two distinct Real zeros and is factorable over the Reals. In addition, if $\Delta$ is a perfect square (and $a , b , c$ are rational) then it can be factored over the rationals.

• If $\Delta = 0$ then $a {x}^{2} + b x + c$ is a perfect square trinomial, expressible as ${\left(\sqrt{a} x + \sqrt{c}\right)}^{2}$ or as ${\left(\sqrt{a} x - \sqrt{c}\right)}^{2}$.

• If $\Delta < 0$ then $a {x}^{2} + b x + c$ has two distinct Complex zeros and is not factorable over the reals. It is factorable if you allow Complex coefficients.

You will probably get used to seeing ${x}^{2} + 2$ or similar and recognising that it is not factorable over the reals, but you might then be surprised to find:

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

In theory any polynomial in one variable with real coefficients is factorable as a product of linear and quadratic factors with real coefficients. In practice, those coefficients are not necessarily expressible in terms of $n$th roots. It is also always (in theory) expressible as a product of linear factors with Complex coefficients.

An example of a polynomial whose factors are not expressible in terms of $n$th roots is:

${x}^{5} + 4 x + 2$

Jan 27, 2017

Some more thoughts...

#### Explanation:

Descartes' Rule of Signs

Given any polynomial in one variable with Real coefficients:

$f \left(x\right) = {a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}$

Look at the pattern of the signs of the coefficients and write it as a sequence of pluses and minuses.

For example, given:

$f \left(x\right) = {x}^{3} - 21 x - 90$

we would write:

$+ - -$

Then count the number of changes of sign. In our example, there's one change of sign.

This count tells us the possible number of positive Real zeros, and if the polynomial has fewer positive zeros then it is less by a multiple of $2$. In our example, there has to be exactly $1$ positive zero.

Further, by examining the signs of the coefficients of $f \left(- x\right)$ we can deduce the possible number of negative Real zeros.

In our example:

$f \left(- x\right) = - {x}^{3} + 21 x - 90$

which has coefficients with signs $- + -$.

With $2$ changes of sign, that means that $f \left(x\right)$ has $2$ or $0$ negative Real zeros.

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Rational roots theorem

Given any polynomial in one variable with integer coefficients:

$f \left(x\right) = {a}_{n} {x}^{n} + {a}_{n - 1} {x}^{n - 1} + \ldots + {a}_{1} x + {a}_{0}$

the rational roots theorem tells us that 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 ${a}_{0}$ and $q$ a divisor of the coefficient ${a}_{n}$ of the leading term.

Any such rational zero $x = \frac{p}{q}$ corresponds to a linear factor $\left(q x - p\right)$ of $f \left(x\right)$ with integer coefficients.

If you can find such a factor, then you can divide your starting polynomial by it to get a simpler polynomial to factor.

For example, given:

$f \left(x\right) = {x}^{3} - 21 x - 90$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ where $p$ is a divisor of $90$ and $q$ a divisor of $1$.

So the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 5 , \pm 6 , \pm 9 , \pm 10 , \pm 15 , \pm 18 , \pm 30 , \pm 45 , \pm 90$

We know from Descartes' Rule of Signs that this $f \left(x\right)$ has a positive Real zero, so try the positive values first.

We find:

$f \left(6\right) = {6}^{3} - 21 \cdot 6 - 90 = 216 - 126 - 90 = 0$

So $\left(x - 6\right)$ is a factor:

${x}^{3} - 21 x - 90 = \left(x - 6\right) \left({x}^{2} + 6 x + 15\right)$

Then examining the discriminant of the remaining quadratic factor, we can observe that it is negative, so has no Real zeros and no further linear factors with Real coefficients.