Your solution requires guessing (by inspection) a real root, namely 2, which leaves a quadratic solution to be solved, with complex roots. If students know complex numbers, they are surely aware of solving analytically any cubic equation (Cardano's formula), which gives all three roots at once, without guessing.
Cardano's formula isn't really taught at any education level. The solution does not require guessing; you can just use the rational root theorem. All the coefficients are integers, so you can take the leading coefficients (1) and the constant coefficient (-10), then take all the divisors of the latter and divide each by each divisor of the former (so stuff like 10/1, -2/-1, -5/1, etc.) and these are your candidates for rational roots. If the polynomial DOES have rational roots, they MUST all be among these candidates. It's not typical (at any educational level) to give a degree 3 or higher polynomial without a rational root. If they do, it has some trick to it, like being a special kind of polynomial. Essentially solving this type of question requires knowledge on how to solve it AND knowledge on what types of questions you would be given.
Great
Very well explained
Too good
lovely
❤❤
Эта задача решается в уме. Гарвард - это что то вроде заборостроительного техникума?
Nice
Your solution requires guessing (by inspection) a real root, namely 2, which leaves a quadratic solution to be solved, with complex roots. If students know complex numbers, they are surely aware of solving analytically any cubic equation (Cardano's formula), which gives all three roots at once, without guessing.
Cardano's formula isn't really taught at any education level.
The solution does not require guessing; you can just use the rational root theorem. All the coefficients are integers, so you can take the leading coefficients (1) and the constant coefficient (-10), then take all the divisors of the latter and divide each by each divisor of the former (so stuff like 10/1, -2/-1, -5/1, etc.) and these are your candidates for rational roots. If the polynomial DOES have rational roots, they MUST all be among these candidates.
It's not typical (at any educational level) to give a degree 3 or higher polynomial without a rational root. If they do, it has some trick to it, like being a special kind of polynomial.
Essentially solving this type of question requires knowledge on how to solve it AND knowledge on what types of questions you would be given.
So it's a pretend game.
Just freaking write that k=2 found by guessing and divide out that root (k^3+k-10 divide by (k-2) and gonna get k^2+2k+5)
jb k=2 aa hi gya to baki ke solution nialne ki jrurt kyo pdi kyoki imaginary roots h na