If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
Given, 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;
∴ ax3+bx2+ cx + d = x3 – 3x2+ x + 1
a = 1, b = -3, c = 1 and d = 1
Sum of zeroes = a – b + a + a + b
-b/a = 3a
Putting the values b and a
- (-3)/1 = 3a
a = 1
Thus, the zeroes are 1 - b, 1, 1 + b.
Now, product of zeroes = 1(1 – b) (1 + b)
d/a = 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
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