Question

find the zeroes of the polynomial p(x)=(x+2) (x_3)​

Answer 1

(x+2)(x-3) = x^2 -3x +2x -6

=x^2 -x -6

now, alpha + beta = -b/a

and alpha(beta) = c/a

alpha + beta = 1. (1)

alpha(beta) = -6. (2)

alpha = 1-beta

in 2

(1-beta)(beta) = -6

beta^2 -beta - 6 = 0

solving further.......

we get 2 values for beta,

beta = -2 and 3

taking beta = 3

from 1

alpha = -2

thus, alpha = -2

beta = 3