How can we factorize $x^4 - 2x^2 + 49$ with coefficients in $\mathbb{R}$? [duplicate]

How can we factorize ${x^4} - 2{x^2} + 49$ with coefficients in $\mathbb{R}$?

A problem would be easier if it was a quadratic equation - we could simply find the roots and get the linear factors.

Moreover, the polynomial $x^4 - 2x^2 + 49$ does not have a real root which would be easy to guess. (If we have one root, we could divide by linear factor determined by this root.)

WolframAlpha says that this can be factorized as $$x^4 - 2x^2 + 49 = (x^2-4x+7)(x^2+4x+7).$$ But how can we get to this factorization?

Factorization of this polynomial also appears as an example in an answer tp another question: Does the Rational Root Theorem ever guarantee that a polynomial is irreducible?

marked as duplicate by Namaste, Stefan Mesken, Bill Dubuque, астон вілла олоф мэллбэрг, R_DNov 26 '16 at 5:40

• @costrom - If ${x^4} - 2{x^2} + 49 = 0$ then, this equation is not the answer in $\mathbb{R}$. – Under sky Nov 25 '16 at 20:12

The idea is try to complete square using the terms $x^4$ and $49$ so let's reorganize the expression:

$x^4+14x^2+49-16x^2=(x^2+7)^2-(4x)^2=(x^2-4x+7)(x^2+4x+7)$

• - Thanks. ......... – Under sky Nov 25 '16 at 20:23
• If the downvoters could please tell me why I deserve that I could improve this and the futures anwers. – Arnaldo Jan 8 '17 at 11:24
• I think we may all have been downvoted without comment. :-/ – Cameron Buie Jan 8 '17 at 15:04
• @Cameron Buie: I also think so. Sadly, that behavior doesn't help anybody. Neither we nor the OP or even the downvoter. – Arnaldo Jan 8 '17 at 15:20

There are a few possible forms for such a factorization:

• as a product of four (not necessarily distinct) linear factors with real coefficients,
• as a product of two quadratic factors with real coefficients, neither of which has real roots,
• as a product of two linear factors and a quadratic factor, all with real coefficients, such that the quadratic factor has no real roots.

Can you see why there are no other possibilities?

Now, note that $$x^4-2x^2+49=(x^2)^2-2x^2+1+48=(x^2-1)^2+48\ge 48$$ for all real $x,$ so that the polynomial has no real roots. This rules out the possibility of any linear factors with real coefficients, so the only possibility remaining is that we can factor it as a product of quadratics with real coefficients.

Set $$x^4-2x^2+49=(px^2+qx+r)(sx^2+tx+u)$$ with $p,s$ non-zero. We must have $ps=1.$ (Why?) Without loss of generality, we can assume that $p=s=1,$ because if not, we can factor out $p$ from the first quadratic and $s$ from the second to get $$x^4-2x^2+49=\left(x^2+\frac qpx+\frac rp\right)\left(x^2+\frac tsx+\frac us\right),$$ which is of the form $$x^4-2x^2+49=(x^2+q'x+r')(x^2+t'x+u').$$

Now, starting from $$x^4-2x^2+49=(x^2+qx+r)(x^2+tx+u),$$ expand the right-hand side and gather like terms. This will yield a system of $4$ equations in $4$ variables, which you should try to solve. If it has real solutions, we've succeeded! If not, it isn't possible.

With $t=x^2$,

$$t^2-2t+49$$ has the roots

$$t=1\pm i\sqrt{48}$$ and

$$x=\pm\sqrt{1\pm i\sqrt{48}}=\pm a\pm ib.$$

Then

$$(x+a+ib)(x-a+ib)(x+a-ib)(x-a-ib)=\\ (x^2+2ax+a^2+b^2)(x^2-2ax+a^2+b^2)$$

$$x^4-2x^2-49 = x^4+14x^2+49-16x^2$$ =$$(x^2+7)^2-16x^2$$ =$$(x^2-4x+7)(x^2+4x+7)$$.

And we are done