Question

If the zeros of the polynomial x^3 - 3x^2 + x + 1 are (a - b), a, (a + b), find the sum of all values of b
(A) 1
(B) 0
(D) None of these
(C) 3​

Answer 1

$\large\underline{\sf{Solution-}}$

Let we assume that

$\rm :\longmapsto\:f(x) = {x}^{3} - {3x}^{2} + x + 1$

Now,

Given that

$\rm :\longmapsto\:a - b, \: a, \: a + b \: are \: zeroes \: of \: f(x)$

We know, that

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}$

$\rm :\longmapsto\:a - b + a + a + b = - \: \dfrac{( - 3)}{1}$

$\rm :\longmapsto\:3a = 3$

$\bf\implies \:a = 1 - - - (1)$

Also,

We know that,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{3}}}}$

$\rm :\longmapsto\:(a - b) \times a \times (a + b) = - \: \dfrac{1}{1}$

$\rm :\longmapsto\:a( {a}^{2} - {b}^{2}) = - 1$

$\rm :\longmapsto\:1 \times ( {1}^{2} - {b}^{2}) = - 1$

$\rm :\longmapsto\:1- {b}^{2} = - 1$

$\rm :\longmapsto\:- {b}^{2} = - 1 - 1$

$\rm :\longmapsto\:- {b}^{2} = - 2$

$\rm :\longmapsto\: {b}^{2} = 2$

$\bf\implies \:b \: = \: \pm \: \sqrt{2}$

So,

$\rm :\longmapsto\:Sum \: of \: values \: of \: b \: = \sqrt{2} - \sqrt{2} = 0$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: Option \: (B) \: is \: correct}}}$

Additional Information :-

1. Quadratic polynomial

$\rm :\longmapsto\: \alpha,\beta \: are \: zeroes \: of \: {ax}^{2} + bx + c, \: \: then$

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

and

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

2. Useful Identities

$\green{ \boxed{ \bf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta }}$

$\green{ \boxed{ \bf \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta( \alpha + \beta ) }}$

$\green{ \boxed{ \bf \: {( \alpha - \beta )}^{2} = {( \alpha + \beta) }^{2} - 4 \alpha \beta }}$