Wait how tf do you do this because my teacher in grade 12 advanced placement math is making me do long division and stuff but bro just whipped through it in 10 seconds?
You can notice that x-2 is a factor here by many methods, one of them is seeing the pattern, the other is by seeing what could be a factor by plugging in x values that make sense.
Try this in general: x^3 + px + q We want to turn this into (x-k)(Q(x) - p) = x^3 - k^3 + px - pk for some quadratic Q. To make that happen, we need to find k such that - k^3 - pk = q ==> k^3 + pk + q = 0 which is exactly the equation we want to solve. In other words, the only way this method works is if you're able to guess a factor right away, in which case you could just divide already.
A cubic polynomial, call it p(x), with smallish coefficients... Just try p(1) (algebraic sum of all coefficients = 0; nope) & p(-1) (sum of odd-term coefficients = sum of even-powered term coeffs; nope) & p(2) (brute-force plug-n-chug; easy b/c small coeffs... p(2)=0, works!). Synthetic div. to get other polynomial factor.. Done! (about 2 mins, MB a tad less; and not dependent on any cleverness, and/or semi-random trying stuff that might/might not work).
Because if something is multiplied to another thing and its equal to 0, then one of those must be equal to 0, and so you can find the roots (srry for bad english im from Brasil)
@@maximofernandez196 Wouldn't they know how to use Wolfram Alpha if they need cubic polynomials factored? This isn't the time of Cardano and mathematical duels.
it's because -8 is a perfect cube. so then he can use that to do perfect cube factoring. also I tried to do it with +8 but then there is no common factor after doing perfect cube rule for that
I remember all the quadratic and cubic identities... and not just this but also over 50 formulae/indenties of Trigonometry as well as of Vector Algebra...
