8. If the zeroes of the polynomial x² – 3x2 + x + 1 are a - b, a, a + b, then find the
value of a and b.
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Answer:
Let the given polynomial be:
p(x) = x3 – 3x2 + x + 1
Given,
The zeroes of the p(x) are a – b, a, and a + b.
Now, compare the given polynomial equation with general expression.
px3 + qx2 + rx + s = x3 – 3x2 + x + 1
Here, p = 1, q = -3, r = 1 and s = 1
For sum of zeroes:
Sum of zeroes will be = a – b + a + a + b
-q/p = 3a
Substitute the values q and p.
-(-3)/1 = 3a
Or, a = 1
So, the zeroes are 1 – b, 1, 1 + b.
For the product of zeroes:
Product of zeroes = 1(1 – b)(1 + b)
-s/p = 1 – 2
=> -1/1 = 1 – 2
Or, 2 = 1 + 1 =2
So, b = √2
Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation 3 − 32 + + 1.
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