Algebra - quadratic polynomial

262

Aditya Bhutra ·Apr 28 '12 at 0:46

same here .

1

even im getting the same,
f(x) = x² - (a+b+c)x + ab+bc+ac

341

Hari Shankar ·Apr 30 '12 at 5:52

P(x) = xf(x) -abc is a cubic with roots a,b,c and hence is equivalent to

$x 3 - x (a + b + c) + x 2 (a b + b c + c a) - a b c$ (bearing in mind that the constant term on both sides is -abc)

so that $f(x) \equiv x^2-x(a+b+c)+(ab+bc+ca)$

Its easily seen that $f (a + b + c) = a b + b c + c a$

1708

man111 singh ·Apr 30 '12 at 9:23

Thanks hsbhatt Sir,Rishab,Aditiya.

quadratic polynomial