Question

Verify that 3,-1,-1/3 are the zeroes of the cubic polynomial p(x) = 3x^2 – 5x^2 – 11x – 3 and then verify the relationship between the zeroes and the coefficients.

Answer 1

Step-by-step explanation:

let p(x)=3

p(3)=3(3)^3-5(3)^2-11(3)-3=0

=81-45-33-3=0

=81-81=0

0=0 hence it is a zero of the polynomial

let p(x)=-1

p(-1)=3(-1)^3-5(-1)^2-11(-1)-3

= -3-5+11-3=0

0=0 hence it is a zero of the polynomial

letp(x)=-1/3

p(-1/3) = 3(-1/3)^3-5(-1/3)^2-11(-1/3)-3

= -1/9-5/9+11/3-3

= -2/3+2/3=0

0=0 hence it is a zero of the polynomial

so,

alpha+beeta+gaama=3+(-1)+(-1/3)=2-1/3=5/3=-(-5)/3=-b/a

alpha×beeta+beeta×gaama+gaama×alpha=3(-1)+(-1)(-1/3)+(-1/3)(3)= -3+1/3-1= -11/3= c/a

alpha×beeta×gaama=3×(-1)×(-1/3)=1=-(-3)/3= -d/a

Hence verified

Hope it helps

Answer 2

Answer:

let p(x)=3,

Then,

p(3)=3(3)^3 -5(3)^2 -11(3) -3=0

=> 81-45-33-3=0

=> 81-81=0

=> 0=0

Hence 3 is a zero of the polynomial

let p(x)=-1,

Then,

p(-1)=3(-1)^3 -5(-1)^2 -11(-1) -3

=> -3-5+11-3=0

=> 0=0

Hence -1 is a zero of the polynomial

Let p(x)=-1/3

Then

p(-1/3) = 3(-1/3)^3 -5(-1/3)^2 -11(-1/3) -3

=> -1/9-5/9+11/3-3

=> -2/3+2/3=0

=> 0=0

Hence -1/3 is a zero of the polynomial

To verify the relationship between the zeroes and the coefficients,

alpha+beta+gamma=3+(-1)+(-1/3)=2-1/3

=5/3

= -(-5)/3=-b/a

alpha×beta+beta×gamma+gamma×alpha=3(-1)+(-1)(-1/3)+(-1/3)(3)

= -3+1/3-1

= -11/3= c/a

alpha×beta×gamma=3×(-1)×(-1/3)=1

=-(-3)/3= -d/a

Hence verified

hope this helps:)