# Factoring $x^4 + 4x^2 + 16$

I was putting together some factoring exercises for my students, and came across one that I am unsure of how to factor.

I factored $x^6 - 64$ as a difference of squares, and then tried it as a difference of cubes, but was left with $(x^2 - 4)(x^4 + 4x^2 + 16)$ is there a general method for factoring $x^4 + 4x^2 + 16$?It factors into two irreducible quadratic trinomials, which is where I think the problem is stemming from.

HINT:

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

• ahh...the left side confuses me a bit, but you completed the square...thanks. wait a minute. – drawnonward Oct 31 '14 at 5:54
• oh I see, and then you just grouped it. But what tipped you off to go in that direction? – drawnonward Oct 31 '14 at 6:03
• Well regardless, I will keep it in mind in the future. Cheers. – drawnonward Oct 31 '14 at 6:09
• I wouldn't call this factorization technique common, but I have seen it before a couple of times. I think that the intuition in this case just comes from experience. – MCT Oct 31 '14 at 6:23
• agreed, that was the conclusion I came too. – drawnonward Oct 31 '14 at 6:27

Here is another method that might generalize to more situations.

Using DeMoivre's Theorem, the roots of $x^6-64 = 0$ are $x = 2e^{ik\pi/3}$ for $k = 0,1,2,3,4,5$.

So, the polynomial is the product of the following complex linear factors:

$(x-2)(x+2)(x-2e^{i\pi/3})(x-2e^{i5\pi/3})(x-2e^{i2\pi/3})(x-2e^{i4\pi/3})$

Now pair up factors that are complex conjugates, and simplify to get:

$(x-2)(x+2)(x^2-(4\cos\frac{\pi}{3})x+4)(x^2-(4\cos\frac{2\pi}{3})x+4)$

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

Alternatively, you can first use the difference of squares factorization, and then use the sum of cubes and difference of cubes factorizations:

$x^6-64 = (x^3+8)(x^3-8) = (x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$

• Thanks, but the question is how to factor $x^4 + 4x^2 + 16$ – drawnonward Oct 31 '14 at 6:00
• Well, once you've factored $x^6-64$, then you can use $x^4+4x^2+16 = \dfrac{x^6-64}{(x-2)(x+2)}$. – JimmyK4542 Oct 31 '14 at 6:03
• Thanks again Jimmy, but you've missed the point of this question. – drawnonward Oct 31 '14 at 6:06
• @drawnonward JimmyK4542 didn't miss anything. Read again what he has taken time to type up for you. – Kim Jong Un Oct 31 '14 at 6:27
• perhaps, but why would I want to consider a 6th degree polynomial to factor a 4th degree one? Thats like saying to factor 9, consider the prime factorization of 27....makes no sense. – drawnonward Oct 31 '14 at 6:34

In general, if we have $x^4 + (2a - b^2)x^2 + a^2$, we can factor it as $(x^2 + a)^2 - (bx)^2 = (x^2 - bx + a)(x^2 + bx + a)$.

On a related note, this factorization is the intuition behind the Sophie-Germain Identity:

$$x^4 + 4y^4 = (x^2 + 2y^2 + 2xy)(x^2 + 2y^2 - 2xy)$$

Indeed, plug in $a = 2y^2, b = 2y$ to get the result.

• Interesting....thanks for that, I will research this identity in the near future. – drawnonward Oct 31 '14 at 6:42

$$x^4 + 4x^2 + 16$$

Let $y = x^2$, the polynomial is then equal to

$$y^2 + 4y + 16$$

$$y = \frac{-4 \pm \sqrt{4^2 - 4(1)(16)}}{2(1)}$$ $$y = \frac{-4 \pm \sqrt{-48}}{2}$$ $$y = 2 \pm 2i\sqrt{3}$$

Return to $x$

$$x^2 = 2 \pm 2i\sqrt{3}$$

So, we can now convert $x^4 + 4x^2 + 16$ into factors.

$$x^4 + 4x^2 + 16 = \left(x^2 - {2 + 2i\sqrt{3}}\right)\left(x^2 - {2 - 2i\sqrt{3}}\right)$$

Repeat quadratic formula for each factor.

$$x_{left} = \frac{0 \pm \sqrt{0^2 - 4(1)(2 + 2i\sqrt{3})}}{2}$$ $$x_{left} = \pm \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}$$

$$x_{right} = \frac{0 \pm \sqrt{0^2 - 4(1)(2 - 2i\sqrt{3})}}{2}$$ $$x_{right} = \pm \frac{\sqrt{8i\sqrt{3} - 8}}{2}$$

So, we can now convert $x^4 + 4x^2 + 16$ into factors again.

$$x^4 + 4x^2 + 16 \\ = \left(x - \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}\right)\left(x + \frac{\sqrt{-8 - 8i\sqrt{3}}}{2}\right)\\ \left(x - \frac{\sqrt{8i\sqrt{3} - 8}}{2}\right)\left(x + \frac{\sqrt{8i\sqrt{3} - 8}}{2}\right)$$

None of this is too outlandish for a student to know how to do.

Edit: Corrected the mistake in the quadratic formula.

Edit: Fixed the mistake with naively square-rooting the y-roots.

• don't think so....should be b^2-4ac, but i dont think you can use the quad formula in this case. – drawnonward Oct 31 '14 at 6:14
• Lets say this does work, than $(x^2 + 2 - 2i\sqrt{3})(x^2 + 2 + 2i\sqrt{3})$ should expand into our original equation, but it doesn't. Unless I'm making a mistake. – drawnonward Oct 31 '14 at 6:24
• Something's still wrong. When those factors combine, it turns into $x^4 -4x^2 + 16$. I'm trying to find my logical error. Regardless, I don't think there's any restriction on using the quadratic formula. – Axoren Oct 31 '14 at 6:27
• you lost your minus b sign. yeah my galois theory is rusty, so couldn't tell you, but when I tried it this way, I wasn't getting the desired result....when you cancelled the denominator. – drawnonward Oct 31 '14 at 6:27
• looks like we will be getting some fourth roots of unity or something? – drawnonward Oct 31 '14 at 6:30