Question

if the zeroes of the polynomial x3-3x2+x+1 are (a-b),a and (a+b), find the value of a

Answer 1

$\alpha + \beta + \gamma = - b \div a$
a-b+a+a+b=3
3a=3
a=1

product =-d/a
$\alpha \beta \gamma$
(a-b)(a)(a+b)=-1
put the vale of a :-1
(1-b)(1)(1+b) =-1
1-b2=-1
2=b2
b=+-√2

Answer 2

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.