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

Question

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.

Thanks in advance.

score 6 · Accepted Answer · answered Oct 31 '14 at 05:47

6

HINT:

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

answered Oct 31 '14 at 05:47

lab bhattacharjee

274,582

ahh...the left side confuses me a bit, but you completed the square...thanks. wait a minute. – drawnonward Oct 31 '14 at 05:54
oh I see, and then you just grouped it. But what tipped you off to go in that direction? – drawnonward Oct 31 '14 at 06:03
Well regardless, I will keep it in mind in the future. Cheers. – drawnonward Oct 31 '14 at 06:09
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. – MT_ Oct 31 '14 at 06:23
agreed, that was the conclusion I came too. – drawnonward Oct 31 '14 at 06:27
how did you get -4x^2? – Lily Mar 21 '18 at 20:40
@Lily, $$(x^2+4)^2=?$$ – lab bhattacharjee Mar 22 '18 at 02:06

score 2 · Answer 2 · answered Oct 31 '14 at 05:56

2

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:

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

answered Oct 31 '14 at 05:56

JimmyK4542

54,331

Thanks, but the question is how to factor $x^4 + 4x^2 + 16$ – drawnonward Oct 31 '14 at 06:00
2

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 06:03
Thanks again Jimmy, but you've missed the point of this question. – drawnonward Oct 31 '14 at 06:06
3

@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 06:27
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 06:34
http://math.stackexchange.com/questions/899109/problems-that-become-easier-in-a-more-general-form/900398#900398 – Kim Jong Un Oct 31 '14 at 07:27

score 2 · Answer 3 · answered Oct 31 '14 at 06:25

2

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.

answered Oct 31 '14 at 06:25

MT_

19,603
9
40
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Interesting....thanks for that, I will research this identity in the near future. – drawnonward Oct 31 '14 at 06:42

Axoren · Answer 4 · 2014-10-31T06:39:39.763

0

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

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

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

Then, use the quadratic formula:

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

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.

edited Oct 31 '14 at 06:39

answered Oct 31 '14 at 06:11

Axoren

2,303

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 06:14
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 06:24
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 06:27
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 06:27
looks like we will be getting some fourth roots of unity or something? – drawnonward Oct 31 '14 at 06:30
@drawnonward I corrected the mistake, but when it's correct it looks really inelegant. However, if a student does give you this answer, you know that they used this method rather than the one you expect. – Axoren Oct 31 '14 at 06:41
Haha....that's what I thought it might look like....thanks for the work anyways :) I still think you are making a sign mistake though....right at the beginning you said -4/2 = 2 should be $x^2 = -2 \pm 2i\sqrt{3}$ – drawnonward Oct 31 '14 at 06:44
Yeah, although this isn't the most elegant way to do it, it's what I'd expect a student in precalculus to be able to do. Groupings like the one provided by lab bhattacharjee tend to work really well, but finding them isn't easy. Using the quadratic formula is guaranteed to work (if you use it right, unlike myself). – Axoren Oct 31 '14 at 06:49

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

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