Harvard College Interview Question

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[@billcook4768]

Not a Harvard College interview question.

[@wobaguk]

A way to avoid the division: a generic expansion of a cubic (with roots a,b,c) gives constant abc=2961 or ab=141 when you sub in 21. the x^2 coef is a+b+c=0 ie. a+b+21=0, you can sub in b=141/a, simplify, and you have a familiar quadratic to solve for the complex roots.

[@mikeeisler6463]

The long division is tedious. We know that if we multiply (x-21) by the quadratic, (x^2 + kx + 141) that we get the cubic. Thus -300x = k * -21x + 141x. So cancel out x, solve for k. k = 21

[@prime423]

After the root 21 is found, synthetic division avoids the rest of the slog. The observation any mathlete would make is the sum of the roots is zero. This would also lead to the analysis.

[@reeb3687]

would it be cheap to pull out a simplified cubic formula?

[@oshaya]

Instead of « pulling out » a simplified cubic formula, one just needs a bit of experience with the Tartaglia solution to the depressed cubic.

[@reeb3687]

@@oshaya is it depressed because there is no x^2 term or is it because the x^3 term has a coefficient of 1?

[@oshaya]

@@reeb3687 Because there is no quadratic term.

[@samueldeandrade8535]

​@@oshaya are you sure about that? Could you explain?

[@prollysine]

by faktoring , x^3-21x^2 +21x^2 -441x +141x-2961=0 , (x-21)(x^2+21x+141)=0 , x-21=0 , x=21 ,
test , 21^3-300*21=9261-6300 , 9261-6300=2961 , for complex , x^2+21x+141=0 ,
solu., x = 21, (-21+i*sqrt123)/2 , (-21-i*sqrt123)/2 ,

[@KyyTyy]

This is an very easy question, i don't believe it's a Harvard interview question.

[@samueldeandrade8535]

This video ... is bad. Very bad.