Question

plz solve it fast......​

Answer 1

Answer:

Explanation:

zeros of the polynomial  $x^{3} - 3x^{2} + x +1$ are a-b, a, a+b

Sum of zeros of a cubic polynomial = $\frac{-b}{a}$

(a-b) + (a+b) + (a) = 3a

$\frac{-b}{a}$ = 3a

$\frac{-(-3)}{1}$ = 3a

$a = 1$

we also know that,

Product of zeros of polynomial = $\frac{-d}{a}$

(a+b)(a-b)(a) = $\frac{-d}{a}$

$(a^{2} -b^{2} )a$ = $\frac{-d}{a}$

since a = 1,

$(1^{2} - b^{2})1 = \frac{-d}{a}$

$1 - b^{2} = \frac{-1}{1}$

$-b^{2} = -1-1$

$b^{2} = 2$

$b = \sqrt{2}$

hence proved