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The question pretty much says it all, though I have a specific example in mind. In the mid-1500s while working on his Ars Magna Cardano asked Tartaglia to solve the cubic $x^3=9x+10$. Using Tartaglia's formula results in having to take the square root of a negative number, which was undefined at the time; neither Tartaglia nor Cardano could solve the equation.

However, with a little work the above equation can be factored into $(x+2)(x^2-2x-5)$, and the three roots readily found using nothing more complicated than the quadratic formula which was known at the time. Clearly neither Tartaglia nor Cardano did this, as they had no solution to the original cubic.

All that aside, when was it realized that such higher-order equations could be factored?

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  • $\begingroup$ Well, for polynomials of degree > 4, the answer is "never" :-) $\endgroup$ – Carl Witthoft Dec 17 '18 at 13:23
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    $\begingroup$ @CarlWitthoft: It is possible to factor many polynomials of high degree, we just can't do it for all such polynomials using radicals. $\endgroup$ – Rory Daulton Dec 18 '18 at 1:06
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I found no trace of factoring polynomials being done before Descartes, in his La Géométrie, written in 1637. In it, he wrote:

It is evident from the above that the sum of an equation having several roots [that is, the polynomial itself] is always divisible by a binomial consisting of the unknown quantity diminished by the false roots. In this way, the degree of an equation can be lowered.

This is how Descartes stated what we call today the factor theorem.

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  • $\begingroup$ And the existence of roots from Descartes' Rule of Signs, which I cited in my dissertation. $\endgroup$ – Andrew Lazarus Jan 18 '19 at 20:53

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