# Factoring a fourth degree polynomial

How would I go about factoring $$x^4-x^2-12=0$$

I'm supposed to find the roots of this without the use of calculators (well we do have TI's but some don't so I suppose they're assuming we can solve this manually). But I can't see how I'd factor this simply?

• As an intermediate step, write it as a polynomial in $y = x^2$. Commented Apr 13, 2014 at 18:41
• Sorry, I don't understand what you mean? Do you mean I write it as: $$x^2 = x^4-12$$? Commented Apr 13, 2014 at 18:42
• No, $y^2-y-12$. Commented Apr 13, 2014 at 18:44

Let $t=x^2$, then you have $t^2-t-12=0$ which is a lot easier to factor. Once you're done that, resubstitute $x^2$ back in.

Write $y=x^2$, giving $y^2-y-12=0=(x^2+3)(x^2-4)$. I'll leave the steps in the factorization of the quadratic $y^2-y-12=0$ to you.

• Ah right, that's how you did it. I thought I had to pull $12$ down by a factor of $x^2$ as well for some reason. Commented Apr 13, 2014 at 18:43
• @user3200098 No, doing that would be redundant and wrong.
– user122283
Commented Apr 13, 2014 at 18:44

Change the variable so you have $y=x^2$ and so solve for $$y^2 - y -12 =0$$

You will find $y = -3$ or $y = 4$, then solve $x^2 = -3$ and $x^2 = 4$. You will have multiple roots which are $$x_1 = -2$$

$$x_2 = 2$$ $$x_3 = -i\sqrt{3}$$ $$x_4 = i\sqrt{3}$$

You will end up with

$$x^4 - x^2 -12 = (x-2)(x+2)(x-i\sqrt{3})(x+i\sqrt{3})$$

Forget $x_3$ and $x_4$ if you are in $\mathbb{R}$ :

$$x^4 - x^2 -12 = (x-2)(x+2)(x^2+3)$$

Let $t=x^2$

Now we have

$t^2-t-12=0$

$\Delta=1+48=49$

$t_1=-3$, $t_2=4$

go back to x

$x_1=\sqrt-3$, $x_2=-\sqrt-3$, $x_3=2$, $x_4=-2$

If you had complex numbers in school, you can easily compute that $x_1=i\sqrt3$ and $x_2=-i\sqrt3$

If not, just write that there are no real roots of negative numbers.

Note that $x^4 - x^2 -12 = (x^2 - 4)(x^2 + 3)$; this factorization was guesswork on my part, based on the observation that $(x^2 - \mu_1)(x^2 - \mu_2) = x^4 - (\mu_1 + \mu_2)x^2 + \mu_1 \mu_2$; so I looked for integers $\mu_1$, $\mu_2$ such that $\mu_1 + \mu_2 = 1$ and $\mu_1 \mu_2 = -12$; it was easy to make such a guess. But this factorization could have been done systematically by setting $y = x^2$, as several others have suggested, and then realizing the roots of the quadratic $y^2 - y -12 = 0$ are $4, -3$; this of course can be done with aid of the quadratic formula. Once we have $x^4 - x^2 -12 = (x^2 - 4)(x^2 + 3)$, we can factor further by using $x^2 - 4 = (x + 2)(x - 2)$; thus $x^4 - x^2 -12 = (x + 2)(x - 2)(x^2 + 3)$; we can't go further over the reals since $x^2 + 3$ has no real zeroes.

That's how I'd do it, in any event.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

This "quartic" equation is actually a quadratic in $x^2$. Notice that $$x^4 - x^2 - 12 \equiv \left(x^2\right)^2 - \left(x^2\right) - 12$$ If we re-name $x^2$ as $u$ then we have $$x^4 - x^2 - 12 = 0 \iff u^2 - u - 12 = 0$$ We are able to factorise $u^2-u-12$. Indeed $u^2-u-12 \equiv (u+3)(u-4)$. It follows that $$u^2-u-12=0 \iff u=-3,\, 4$$ Since $u$ is used in place of $x^2$ we have $x^2=-3, \, 4$. It follows that $$x^4-x^2-12=0 \iff x = \pm\mathrm{i}\sqrt{3}, \, \pm 2$$ If you are working over the real numbers then $x= \pm 2$ are the only valid solutions.

$$x^4-x^2-12=0\,\,\text{can be written as: }\color{blue}{(x^2)}^2-\color{blue}{(x^2)}-12=0.$$ which with a change of variable is a quadratic equation, that can be solved using the quadratic formula to get: $$x^2=\dfrac{1\pm\sqrt{1+4\cdot12}}{2}.$$ And so, you will get $2$ roots, and recall that: If $\alpha$ is a root of $\mathrm P(x)$, then there exist one polynomial $\mathrm Q(x)$ such that $\mathrm P(x)=(x-\alpha)\cdot\mathrm Q(x)$ and so you'll have to use polynomial long division.

After doing all of those steps, you will find that: $$x^4-x^2-12=(x-2 ) (2 + x) (3 + x^2).$$