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I was wondering how would you factor $$x^4+4x^3+21x^2-20x+25=0\text{ ?}$$ I cannot find a number that allows the expression to equal zero.

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    $\begingroup$ When there aren't any nice roots, there is still a chance of factoring into quadratics $(x^2+ax+b)(x^2+cx+d)$. It takes some observations and guesswork to pull it off. $\endgroup$ – user147263 Jan 1 '16 at 2:51
  • $\begingroup$ There is no quick ways of factorising higher degree polynomial, that's why we install WA on our phone. WA tells me that $(x^2-x+1)(x^2+5x+25)$ with complex roots, so it must be hard to use factor theorem $\endgroup$ – Steve Jan 1 '16 at 2:52
  • $\begingroup$ It is worth mentioning that a generic formula does exist for quartic (degree four) polynomials similar to the ever-familiar $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ however it is quite difficult to memorize and often not very useful to learn in the first place. It is also worth mentioning that one of the big achievements of Galois Theory is that no such general formula could possibly exist for general quintics (degree five). $\endgroup$ – JMoravitz Jan 1 '16 at 3:07
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One of the universal principles in mathematics is the ability to make a good guess and then eventually prove the guess. Note that $-\omega,-\omega^2$, where $\omega$ is the non-real cube roots of unity, satisfies the equation. This means $x^2-x+1$ is a factor. Using this, the polynomial factors as $$(x^2-x+1)(x^2+5x+25)$$

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You really can do this yourself.

A lemma of Gauss says that a polynomial with integer coefficients factors over the integers if it factors over the rationals. in practice, this means there are just four possible products that might work, with integers $A,B$: $$ (x^2 + A x - 5)(x^2 + B x - 5) $$ $$ (x^2 + A x + 5)(x^2 + B x + 5) $$ $$ (x^2 + A x - 1)(x^2 + B x - 25) $$ $$ (x^2 + A x + 1)(x^2 + B x + 25). $$ In all, $A + B = 4$ because of the $4 x^3.$ In the second one, this gives $20 x,$ so the second one is out, but the first one gives $-20x.$ The first then gives $(AB - 10) x^2,$ so $AB = 31. $ As $31$ is prime, we get either $A+B = \pm 32,$ so the first one is out.

In the third we get $(-25 A - B)x,$ so $-25A-B = -20,$ or $25A + B = 20.$ With $A+B = 4,$ we have $24A = 16,$ so $A$ is not an integer.

In the fourth, we get $(25 A + B)x,$ so $25A+B = -20.$ With $A+B = 4,$ we have $24A = -24,$ so $A=-1$ and $B=5.$ We then confirm that $$ (x^2 - x + 1)(x^2 + 5 x + 25) $$ works.

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