How do you factor $x^{4} + x^{2} + 1$ $x^4+x^2+1$ ?

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How do you factor $x^{4} + x^{2} + 1$ $x^4+x^2+1$ ?

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Answer 1 · 2015-05-10T08:13:09

$x^{4} + x^{2} + 1 = (x^{2} + x + 1) (x^{2} - x + 1)$ $x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1)$

To find this, first notice that $x^{4} + x^{2} + 1 > 0$ $x^4 + x^2 + 1 > 0$ for all (real) values of $x$ $x$ . So there are no linear factors, only quadratic ones.

$x^{4} + x^{2} + 1 = (a x^{2} + b x + c) ({d x}^{2} + e x + f)$ $x^4 + x^2 + 1 = (ax^2 + bx + c)(dx^2 + ex + f)$

Without bothering to multiply this out fully just yet, notice that the coefficient of $x^{4}$ $x^4$ gives us $a d = 1$ $ad = 1$ . We might as well let $a = 1$ $a = 1$ and $d = 1$ $d = 1$ .

... $= (x^{2} + b x + c) (x^{2} + e x + f)$ $= (x^2 + bx + c)(x^2 + ex + f)$

Next, the coefficient of $x^{3}$ $x^3$ gives us $b + e = 0$ $b + e = 0$ , so $e = - b$ $e = -b$ .

... $= (x^{2} + b x + c) (x^{2} - b x + f)$ $= (x^2 + bx + c)(x^2 -bx + f)$

The constant term gives us $c f = 1$ $cf = 1$ , so either $c = f = 1$ $c = f = 1$ or $c = f = - 1$ $c = f = -1$ . Let's try $c = f = 1$ $c = f = 1$ .

... $= (x^{2} + b x + 1) (x^{2} - b x + 1)$ $= (x^2 + bx + 1)(x^2 - bx + 1)$

Note that the coefficient of $x$ $x$ will vanish nicely when these are multiplied out.

Finally notice that the coefficient of $x^{2}$ $x^2$ is $(1 - b^{2} + 1) = 2 - b^{2}$ $(1 - b^2 + 1) = 2 - b^2$ , giving us $1 = 2 - b^{2}$ $1 = 2 - b^2$ , thus $b^{2} = 1$ $b^2 = 1$ , so $b = 1$ $b = 1$ or $b = - 1$ $b = -1$ .

Answer 2 · 2017-11-15T07:22:12

$x^{4} + x^{2} + 1 = (x^{2} - x + 1) (x^{2} + x + 1)$ $x^4+x^2+1 = (x^2-x+1)(x^2+x+1)$

Explanation:

This really makes a bit more sense in the complex numbers...

First note that:

$(x^{2} - 1) (x^{4} + x^{2} + 1) = x^{6} - 1$ $(x^2-1)(x^4+x^2+1) = x^6-1$

So zeros of $x^{4} + x^{2} + 1$ $x^4+x^2+1$ are also zeros of $x^{6} - 1$ $x^6-1$ .

What are the zeros of $x^{6} - 1$ $x^6-1$ ?

The real zeros are $1$ $1$ and $- 1$ $-1$ , which are zeros of $x^{2} - 1$ $x^2-1$ , the factor we introduced. So the four zeros of $x^{4} + x^{2} + 1$ $x^4+x^2+1$ are the four complex zeros of $x^{6} - 1$ $x^6-1$ apart from $\pm 1$ $+-1$ .

Here are all $6$ $6$ in the complex plane:

graph{((x-1)^2+(y-0)^2-0.002)((x-1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1/2)^2+(y-sqrt(3)/2)^2-0.002)((x+1)^2+(y-0)^2-0.002)((x+1/2)^2+(y+sqrt(3)/2)^2-0.002)((x-1/2)^2+(y+sqrt(3)/2)^2-0.002) = 0 [-2.5, 2.5, -1.25, 1.25]}

They form the vertices of a regular hexagon.

de Moivre's formula tells us that:

${(cos θ + i sin θ)}^{n} = cos n θ + i sin θ$ $(cos theta + i sin theta)^n = cos n theta + i sin theta$

where $i$ $i$ is the imaginary unit, satisfying $i^{2} = - 1$ $i^2=-1$

For instance we find:

${(cos (\frac{π}{3}) + i sin (\frac{π}{3}))}^{6} = cos 2 π + i sin 2 π = 1 + 0 = 1$ $(cos (pi/3) + i sin (pi/3))^6 = cos 2pi + i sin 2pi = 1 + 0 = 1$

That's the zero $\frac{1}{2} + \frac{\sqrt{3}}{2} i$ $1/2+sqrt(3)/2i$ that we see in Q1.

If $a$ $a$ is a zero of a polynomial then $(x - a)$ $(x-a)$ is a factor.

Hence:

$x^{4} + x^{2} + 1 = (x - \frac{1}{2} - \frac{\sqrt{3}}{2} i) (x - \frac{1}{2} + \frac{\sqrt{3}}{2} i) (x + \frac{1}{2} - \frac{\sqrt{3}}{2} i) (x + \frac{1}{2} + \frac{\sqrt{3}}{2} i)$ $x^4+x^2+1 = (x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)(x+1/2-sqrt(3)/2i)(x+1/2+sqrt(3)/2i)$

$x^{4} + x^{2} + 1 = (x^{2} - x + 1) (x^{2} + x + 1)$ $color(white)(x^4+x^2+1) = (x^2-x+1)(x^2+x+1)$

For example:

$(x - \frac{1}{2} - \frac{\sqrt{3}}{2} i) (x - \frac{1}{2} + \frac{\sqrt{3}}{2} i)$ $(x-1/2-sqrt(3)/2i)(x-1/2+sqrt(3)/2i)$

$=x^2-((1/2+color(red)(cancel(color(black)(sqrt(3)/2i))))+(1/2-color(red)(cancel(color(black)(sqrt(3)/2i)))))x+(1/2+sqrt(3)/2i)(1/2-sqrt(3/2)i)$

$=x^2-x+((1/2)^2-(sqrt(3)/2i)^2)$

$=x^2-x+(1/4-3/4i^2)$

$=x^2-x+(1/4+3/4)$

$=x^2-x+1$

Answer 3 · 2017-11-16T06:48:30

$x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$

Explanation:

Given: $x^4+x^2+1$

The difference of squares identity can be written:

$a^2-b^2 = (a-b)(a+b)$

The difference of cubes identity can be written:

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

The sum of cubes identity can be written:

$a^3+b^3 = (a+b)(a^2-ab+b^2)$

Note that:

$(x^2-1)(x^4+x^2+1) = (x^2-1)((x^2)^2+(x^2)+1)$

$color(white)((x^2-1)(x^4+x^2+1)) = (x^2)^3-1^3$

$color(white)((x^2-1)(x^4+x^2+1)) = x^6-1$

$color(white)((x^2-1)(x^4+x^2+1)) = (x^3)^2-1^2$

$color(white)((x^2-1)(x^4+x^2+1)) = (x^3-1)(x^3+1)$

$color(white)((x^2-1)(x^4+x^2+1)) = (x^3-1^3)(x^3+1^3)$

$color(white)((x^2-1)(x^4+x^2+1)) = (x-1)(x^2+x+1)(x+1)(x^2-x+1)$

$color(white)((x^2-1)(x^4+x^2+1)) = (x^2-1)(x^2+x+1)(x^2-x+1)$

Dividing both ends by $(x^2-1)$ we get:

$x^4+x^2+1 = (x^2+x+1)(x^2-x+1)$

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