Question

Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³+3x²-x-3 and check the relationship between zeroes and the coefficients.​

Answer 1

Step-by-step explanation:

let p (x) = x^3 + 3x^2-x-3

x = 1 = P(1) = 1^3+3×1^2-1-3

= 1+3-1-3=4-4=0

X = -1 = P(-1) = (-1)^3 + 3×(-1)^2-(-1)-3

= -1+3+1-3=4-4=0

X = -3 = P(13) = (-3)^3+3(-3)^2-(-3)-3

= -27+27+3-3=0

since, 1,-1 & -3 are the zeroes of the given

cubic polynomial

let, a = alfa and b = beta,

a = 1 , b = -1, r = -3

zeroes. coefficents

a= 1. A= 1

b=-1. B=3

r=-3. C=-1

D=-3

a+b+r= -B/A = 1+(-1)+(-3) = -3/1

= 1-1-3=-3= -3 = -3

ab + br + ar = C/A

= 1(-1)+(-1)(-3)+(1)(-3) = -1/1

= -1+3-3 = -1 = -1

abr = -B/A = 1(-1)(-3) = -(-3)/1 = 3

= 3