3. If the zeroes of the polynomial x3-3x2+x+1 are a – b, a, a + b, find a and b.

Solution: We are given with the polynomial here, p(x) = x3-3x2+x+1 And zeroes are given as a – b, a, a + b Now, comparing the given polynomial with general expression, we get; ∴px3+qx2+rx+s = x3-3x2+x+1 p = 1, q = -3, r = 1 and s = 1 Sum of zeroes = a – b + a + a + b -q/p = 3a Putting the values q and p. -(-3)/1 = 3a a=1 Thus, the zeroes are 1-b, 1, 1+b. Now, product of zeroes = 1(1-b)(1+b) -s/p = 1-b2 -1/1 = 1-b2 b2 = 1+1 = 2 b = √2 Hence,1-√2, 1 ,1+√2 are the zeroes of x3-3x2+x+1.