Question

a cubic equation of the form x^3+bx^2+cx+d=0 has solutions x=3, x=4 and x=5. What are the valves of b, c and d?

Answer 1

There are a couple of ways to do this. The way i like the most is:\[(x-r_1)(x-r_2)(x-r_3)=x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_1r_3+r_2r_3)x-(r_1r_2r_3)\]So -b is the sum of the roots, c is the sum of the roots taken two at a time, and -c is the product of the roots.

Answer 2

Some stuff got cut off on my page, i'll retype:\[(x-r_1)(x-r_2)(x-r_3)\]\[=x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_1r_3+r_2r_3)x-(r_1r_2r_3)\]

Answer 3

thx

Answer 4

Ah, Viete's formulas. Never got around to memorizing them.