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Question:

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

Solution:

P(x)=x³-3x²+x+1
zeroes are -b, a, a+b
Comparing the given polynomial with px³+qx²+rx+t, we obtain
p=1, q=-3, r=1, t=1
Sum of zeroes = a-b+a+a+b = -q/p
3a = -(-3)/1 = 3
3a = 3
a = 1
Then the zeroes becomes, 1-b, 1, 1+b
Multiplication of zeroes = (1-b) x 1 x (1+b) = -t/p
1-b² = -1
1 = 1-b²
b² = 0
b = 0
Therefore, a=1 and b=0