# How do you factor the expression x^4-13x^2+36?

Feb 8, 2016

${x}^{4} - 13 {x}^{2} + 36 = \left(x - 3\right) \left(x + 3\right) \left(x - 2\right) \left(x + 2\right)$

#### Explanation:

We will use the difference of squares identity which can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

First, notice that this quartic is a quadratic in ${x}^{2}$. Note also that $36 = 9 \times 4$ and $13 = 9 + 4$, so we find:

${x}^{4} - 13 {x}^{2} + 36$

$= {\left({x}^{2}\right)}^{2} - 13 \left({x}^{2}\right) + 36$

$= \left({x}^{2} - 9\right) \left({x}^{2} - 4\right)$

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

$= \left(x - 3\right) \left(x + 3\right) \left(x - 2\right) \left(x + 2\right)$

Feb 8, 2016

Use an alternative method to find:

${x}^{4} - 13 {x}^{2} + 36 = \left(x - 3\right) \left(x - 2\right) \left(x + 3\right) \left(x + 2\right)$

#### Explanation:

Another method of factoring ${x}^{4} - 13 {x}^{2} + 36$ goes as follows:

Note that ${x}^{4} = {\left({x}^{2}\right)}^{2}$ and $36 = {6}^{2}$ are both perfect squares.

Note also that there are no terms in ${x}^{3}$ or $x$.

Consider:

$\left({x}^{2} - k x + 6\right) \left({x}^{2} + k x + 6\right) = {x}^{4} + \left(12 - {k}^{2}\right) {x}^{2} + 36$

The opposite middle terms $- k x$ and $+ k x$ cause the ${x}^{3}$ and $x$ terms to cancel out.

If $\left(12 - {k}^{2}\right) {x}^{2} = - 13 {x}^{2}$ then we find ${k}^{2} = 25$ so $k = \pm 5$ and our original quartic factors as:

${x}^{4} - 13 {x}^{2} + 36 = \left({x}^{2} - 5 x + 6\right) \left({x}^{2} + 5 x + 6\right)$

In our example, each of the remaining quadratic factors is reducible to linear factors with Real coefficients:

$\left({x}^{2} - 5 x + 6\right) \left({x}^{2} + 5 x + 6\right) = \left(x - 3\right) \left(x - 2\right) \left(x + 3\right) \left(x + 2\right)$

Note however, that in general we may often be able to factor a quartic with no ${x}^{3}$ or $x$ terms into two quadratics in this way regardless of whether the resulting quadratics have Real roots.

For example:

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