Math, asked by educationmaster37, 10 months ago

obtain all the zeroes of the polynomial x^4-3√2x^3+3x^2+3√2x-4, if two of it's zeroes are√2and2√2​

Answers

Answered by Anonymous
20

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}}}

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