Question

If anybody know so plz get help me to find out..​

Answer 1

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

Given polynomial is

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

Further given that,

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

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

$\rm :\longmapsto\:\bigg(x - \dfrac{1}{ \sqrt{2} } \bigg) \bigg(x + \dfrac{1}{ \sqrt{2} } \bigg) \: is \: factor \: of \: f(x)$

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

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

$\rm :\longmapsto\:{2x}^{2} - 1 \: is \: factor \: of \: f(x)$

So, by using long division, we have

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

Now, we know that,

Dividend = Divisor × Quotient + Remainder

Here,

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

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

$\rm :\longmapsto\:Quotient = {x}^{2} - 3x + 2$

$\rm :\longmapsto\:Remainder = 0$

So,

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

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

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

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

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

So,

Remaining zeroes are 2 and 1