How to factorize $x^4+2x^2-x+2$?

Question

look at this: $$x^4+2x^2-x+2$$ How to factorize it?

It should be changed to be in the form of standard factorization formulas.

Answer 1

You can write

$$x^4+2x^2-x+2=(x^2+ax+b)(x^2+cx+d)$$

Then $a+c=0$, $bd=2$, $ad+bc=-1$ and $b+d+ac=2$. If you want integer values for $a,b,c,d$, you must have $b=\pm2$ or $\pm1$, and afters some checks,

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

Here is a more systematic approach. From the preceding equations, you have

$$c=-a$$ $$a(b-d)=1$$ $$b+d=2-ac=2+a^2$$

From $b-d=1/a$ and $b+d=2+a^2$, you get

$$b=\frac12(2+a^2+1/a)$$ $$d=\frac12(2+a^2-1/a)$$

Then, the equation $bd=2$ leads to

$$(2+a^2)^2-1/a^2=8$$

With $t=a^2$,

$$t(2+t)^2-8t-1=0$$

Or

$$t^3+4t^2+4t-8t-1=t^3+4t^2-4t-1=0$$

This one is a reciprocal equation, and it has the obvious solution $t=1$. Then you can choose $a=1$ and $c=-1$, then $b=2$ and $d=1$.

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As DanielV remarks below, this is a general method to solve the quartic equation

$$x^4+Ax^3+Bx^2+Cx+D=0$$

First, let $x=z-A/4$, by expanding you can check that the coefficient of $z^3$ vanishes, and you have to solve

$$z^4+Uz^2+Vz+W=0$$

Then, as above, you write

$$z^4+Uz^2+Vz+W=(z^2+az+b)(z^2+cz+d)$$

And after expanding the RHS and equating coefficients with same power of $z$, you get $a+c=0$, $b+d+ac=U$, $ad+bc=V$, $bd=W$.

Then

$$c=-a$$ $$b+d=U+a^2$$ $$b-d=-V/a$$

Hence

$$b=\frac12(U+a^2-V/a)$$ $$d=\frac12(U+a^2+V/a)$$

Thus

$$4bd=4W=(U+a^2)^2-V^2/a^2$$

And with $t=a^2$, you have to solve

$$t(U+t)^2-4Wt-V^2=0$$

$$t^3+2Ut^2+(U^2-4W)t-V^2=0$$

This is a cubic equation, so you have proved that to solve a quartic, you just have to solve first a cubic, which is supposed to be easier (and there are indeed methods to solve the cubic). Then, there remains two trinomials to solve, and you are done.

This method is also described here, along with others. Look here for solutions of the cubic equation.

Answer 2

The zeroes of this quartic are $$\frac12\pm i\frac{\sqrt 3}{2}$$ and $$-\frac12\pm i\frac{\sqrt 7}{2}$$ Since there are no real zeroes, it can't be factored down to linear factors with only real coefficients.

However, this does mean that the complete factorization is $$\left(x-\tfrac12- i\tfrac{\sqrt 3}{2}\right)\left(x-\tfrac12+ i\tfrac{\sqrt 3}{2}\right)\left(x+\tfrac12- i\tfrac{\sqrt 7}{2}\right)\left(x+\tfrac12+ i\tfrac{\sqrt 7}{2}\right) $$

Addendum: Following the suggestion of @Jean-ClaudeArbaut, this can be partially combined to produce two quadratics with all real coefficients: $$\left(x-\tfrac12- i\tfrac{\sqrt 3}{2}\right)\left(x-\tfrac12+ i\tfrac{\sqrt 3}{2}\right) = x^2 - x + 1$$ and $$\left(x+\tfrac12- i\tfrac{\sqrt 7}{2}\right)\left(x+\tfrac12+ i\tfrac{\sqrt 7}{2}\right) = x^2 + x + 2$$ which gives a factorization into $$(x^2 - x + 1)(x^2 + x + 2)$$ with the quadratics not factorizable with real coefficients.

Answer 3

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

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