Question

find all the zeros of 2x^4 - 3x^3 - 3x^2 +6x -2 if you know that two of its zeros are 1/2 and 1​

Answer 1

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

Given that

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

and

$\rm :\longmapsto\:\dfrac{1}{2} \: and \: 1 \: are \: zeroes \: of \: f(x)$

$\rm :\implies\:(x - 1) \: and \: x - \dfrac{1}{2} \: are \: factors \: of \: f(x)$

$\rm :\implies\:(x - 1)\bigg(x - \dfrac{1}{2} \bigg) \: is \: a \: factor \: of \: f(x)$

$\rm :\implies\:(x - 1)\bigg(\dfrac{2x - 1}{2} \bigg) \: is \: a \: factor \: of \: f(x)$

$\rm :\implies\:\bigg(\dfrac{2 {x}^{2} - 3x + 1}{2} \bigg) \: is \: a \: factor \: of \: f(x)$

$\rm :\implies\:2 {x}^{2} - 3x + 1\: is \: a \: factor \: of \: f(x)$

So,

By using long division method, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\: \: \: \: \: \: \: {x}^{2} \: - \: 2 \: \: \: \: \:\:}}}\\ {{\sf{ {2x}^{2} - 3x + 1}}}& {\sf{\: {2x}^{4} - {3x}^{3} - {3x}^{2} +6x - 2 \:}} \\{\sf{}}&\underline{\sf{\: \: \: { - 2x}^{4} + {3x}^{3} -{x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}}\\{\sf{}}&{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { - 4x}^{2} + 6x - 2 \:\:}}\\{\sf{}}&\underline{\sf{\:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { 4x}^{2} - 6x + 2 \:\:}}\\{\sf{}}&\underline{\sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0\:\: \: \: \: \: \: \: \: \: }}\end{array}\end{gathered}\end{gathered}\end{gathered}$

Now,

We know that

Dividend = Divisor × Quotient + Remainder

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

$\: \rm \: = \:( {2x}^{2} - 3x + 1)( {x}^{2} - 2) + 0$

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

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

$\: \: \: \: \: \: \: \red{\bigg \{ \because \: {x}^{2} - {y}^{2} = (x + y)(x - y) \bigg \}}$

$\bf\implies \:Remaining \: zeroes \: of \: f(x) = \sqrt{2} \: and \: - \: \sqrt{2}$

Hence,

$\boxed{ \rm{ \: Zeroes \: of \: f(x) \: are \: 1, \: \dfrac{1}{2}, \: \sqrt{2} \: and \: - \: \sqrt{2}}}$

Additional Information :-

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

$\boxed{ \sf{ \: \alpha + \beta = - \frac{b}{a}}}$

and

$\boxed{ \sf{ \: \alpha\beta = \frac{c}{a}}}$

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

$\boxed{ \sf{ \: \alpha + \beta + \gamma = - \frac{b}{a}}}$

$\boxed{ \sf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}}}$

and

$\boxed{ \sf{ \: \alpha \beta \gamma = - \frac{d}{a}}}$