Question

find the zero of the polynomial f(x) = x3 - 3x2-16x+48 if it is given that 2 zeros are equal on magnitude but opposite in sign​

Answer 1

Answer:

f(x)=0

x3-3x2-16x+48=0

let the zeroes be a, b, c and ATQ, say a = -b,

then, a+b+c = -(-3)/1, ab+bc+ca = -16/1 and abc = -48/1

Putting a = -b we get

-b+b+c = 3, -b.b+bc+(-bc) = -16 and -b.b.c= -48

which gives c = 3, b = +4 or -4

Hence, a = -4 or +4

So the zeroes of the polynomial are -4,4,3 or 4,-4,3.