Question

find the all zeros of polynomial x⁴-3x³-x²+9x-6. If 2 of its zeros are √3 and -√3​

Answer 1

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

Let assume that

$\rm :\longmapsto\:f(x) = {x}^{4} - {3x}^{3} - {x}^{2} + 9x - 6$

Given that

$\rm :\longmapsto\: - \sqrt{3} \: and \: \sqrt{3} \: are \: zeroes \: of \: f(x)$

$\rm :\longmapsto\:(x - \sqrt{3}) \: and \: (x + \sqrt{3}) \: are \: factors \: of \: f(x)$

$\rm :\longmapsto\:(x - \sqrt{3}) \: (x + \sqrt{3}) \: is \: factor \: of \: f(x)$

$\rm :\longmapsto\:( {x}^{2} - 3) \: \: is \: factor \: of \: f(x)$

So, By using Long Division, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: \: \: \: {x}^{2} - 3x + 2 \: \: \:}}}\\ {{\sf{ {x}^{2} - 3}}}& {\sf{\: {x}^{4} - {3x}^{3} - {x}^{2} + 9x - 6 \:}} \\{\sf{}}&\underline{\sf{\: \: \: { - x}^{4} + \: \: \: \: \: \: \: \: \: 3{x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: {3x}^{3} + {2x}^{2} + 9x - 6 \:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: - {3x}^{3} \: \: \: \: \: \: \: \: \: - 9x \: \: \: \: \: \: \: \: \: \:\:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {2x}^{2} - 6\:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: { - 2x}^{2} + 6 \:\:}}\\{\sf{}}&\underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}$

We know that,

Dividend = Divisor × Quotient + Remainder

So,

$\rm :\longmapsto\: {x}^{4} - {3x}^{3} - {x}^{2} + 9x - 6$

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

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

$\rm \: = \: \: ( {x}^{2} - 3) \bigg(x(x - 2) - 1(x - 2) \bigg)$

$\rm \: = \: \: ( {x}^{2} - 3) \bigg((x - 2)(x - 1) \bigg)$

So, zeroes of

$\rm :\longmapsto\:{x}^{4} - {3x}^{3} - {x}^{2} + 9x - 6 \: are \: - \sqrt{3}, \sqrt{3},1,2$

Additional Information :-

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

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

and

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

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

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

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

and

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