Question

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

Answer 1

Step-by-step explanation:

Solution:

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.