Question

17. If the zerors of the polynomial
${x}^{3} - {3x}^{2} + x + 1$
+ x+ 1 are a - b, a, a + b, find a and b.​

Answer 1

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

Let assume that

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

Now, given that

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

We know, In cubic polynomial,

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

So, using this identity we get

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

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

$\bf\implies \:a = 1$

Also, In cubic Polynomial

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$

So, using this we get

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

On substituting the value of a = 1, we get

$\rm :\longmapsto\:(1 - b)(1)(1+ b) = - 1$

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

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

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

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

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

Hence,

$\purple{\boxed{ \bf{ a \: = \: 1 \: \: \: and \: \: \: b \: = \: \pm \: \sqrt{2} }}}$

Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: a {x}^{2} + b {x} + c, \: then}$

$\boxed{ \bf{ \: \alpha + \beta = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta = \dfrac{c}{a}}}$