Question

guy's mera yh question kro yrr koi ​

Answer 1

Step-by-step explanation:

plz find the answer in attachment

Answer 2

AnswEr :

$\normalsize\bullet\:\sf\ P(x) = x^4 - 3\sqrt{2}x^{3} + 3x^2 + 3\sqrt{2}x - 4$

$\normalsize\bullet\:\sf\ Since, \: two \: zeroes \: are \: \sqrt{2} \: and \: 2\sqrt{2}$

$\because\normalsize\sf\ \left(x - \sqrt{2} \right) \left( x - 2\sqrt{2} \right) = x^2 - 3\sqrt{2}x + 4 \\ \normalsize\sf\quad\ is \: a \: factor \: of \: P(x)$

$\rule{200}1$

$\underline{\dag\:\textsf{According \: to \: question \: now;}}$

$\boxed{\begin{array}{r | l} & \quad\ x^2 - 1 \\ \cline{1-2} x^2 - 3\sqrt{2}x +4 & \: \: x^4 - 3\sqrt{2}x^{3} + 3x^2 + 3\sqrt{2}x - 4 \\ & \: \: x^4 - 3\sqrt{2}x^3 + 4x^2 \\ &- \: \: \: + \: \: \qquad\ - \\ \cline{2-2} & \: \: \:\: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \qquad\ - x^2 + 3\sqrt{2}x - 4 \\ & \: \: \:\: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \qquad\ - x^2 + 3\sqrt{2}x - 4 \\ \cline{2-2} & \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: 0 \end{array}}$

$\rule{200}1$

$\normalsize\dashrightarrow\sf\ x^4 - 3\sqrt{2}x^3 + 3x^2 - 3\sqrt{2}x - 4 \\ \normalsize\sf\ \quad\ = \left( x^2 - 3\sqrt{2}x + 4 \right) \left( x^2 - 1 \right)$

$\scriptsize\sf{ \qquad\dag\ (a^2 - b^2) = (a+b)(a-b)}$

$\normalsize\dashrightarrow\sf\ x^4 - 3 \sqrt{2}x^3 + 3x^2 - 3 \sqrt{2}x - 4 \\ \normalsize\sf\quad\ = \left( x- \sqrt{2}x \right) \left( x - 2\sqrt{2}x \right) \underbrace{\left( x + 1 \right)}_{part1} \underbrace{\left(x - 1 \right)}_{part2}$

$\rule{170}1$

$\normalsize{\dag\:\textsf{Solving \: with \: part \: 1 :}}$

$\normalsize\sf\twoheadrightarrow\ (x - 1) = 0$

$\normalsize\sf\twoheadrightarrow\ x = 0 + 1$

$\normalsize\twoheadrightarrow\sf\red{x = 1}$

$\normalsize{\dag\:\textsf{Solving \: with \: part \: 2 :}}$

$\normalsize\sf\twoheadrightarrow\ (x + 1) = 0$

$\normalsize\sf\twoheadrightarrow\ x = 0 - 1$

$\normalsize\twoheadrightarrow\sf\red{x = -1}$

$\therefore\:\underline{\textsf{Hence, \: the \: other \: two \: zeroes \: are}{\textbf{ 1, -1}}}$