Question

if the zero of polynomial x`3-6x'2+x+1are a-b,a,a+b . find a and b​

Answer 1

Answer:

Step-by-step explanation:

Alpha=a-b

Beeta=a

Gama=a+b

p(x)=x^3-6x^2+x+1

On comparing with

ax^3+bx^2+CX+d=0

We get

a=1,b=-6,c=1,d=1

Now,

Alpha+beeta+gama=-b/a

(a-b)+a+(a+b)= -(-6)/1

a-b+a+a+b= 6/1

a+a+a=6

3a=6

a=2

Then,

alpha×beeta×gama=-d/a

(a-b)(a)(a+b)=-1/1

By identity (a+b)(a-b)=a^2-b^2

(a^2-b^2)(a)=-1

a^3-ab^2=-1

Putting a=2

8-2d^2=-1

-2d^2=-1-8

-2d^2=-9

d^2=9/2

d=√9/2

d=3/√2