Question

If the zers of the polynomial x^3-3x^2+x+1 are a-b, a, a+b find 'a' and 'b'

Answer 1

Answer:

a = 1

b = ±√2

Step by step explanation:

=p(x)=x³-3x²+x+1 ..........(1)

We need to find a and b.

Let p,q,r be the zeros of the polynomial.

f(y)=y³+sy²+ts+h, Then

=p+q+r= -s

=pq+qr+rp=t,

pqr= -h

Now taking polynomial (1),

(a-b)+(a)+(a+b)= -(-3)

3a=3

a=1

And then

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

(1-b)1+1(1+b)+(1+b)(1-b)=1

1-b+1+b+1-b²=1

b²=2

∴b=±√2

Thus, the required value of a is 1 and that of b is ±√2.

Hope this helps you!