Question

Find all the zeroes of 2x4-3x3-3x2+6x-2 if two of its zeroes are 1 and 1/2

Answer 1

other zeroes are root and -root2

Answer 2

Here your answer goes

Step :- 1

Given ,

p(x) = $2x^4 - 3x^3 -3x^2 +6x-2$

1 and $\frac{1}{2}$ are its two zeroes

$(x-1) and\ ( 2x - 1 )$ is a factor

$2x^2 - x - 2x + 1$

$2x^2 - 3x + 1$

Step :- 2

$\begin{tabular}{r|ll} & x^2-2 \\\cline{2-3} 2x^2-3x+1 & 2x^4-3x^3&-3x^2+6x-2 \\\cline{1-1} & 2x^4-3x^3&+x^2 \\ \cline{2-3} & & -4x^2+6x-2 \\ & & -4x^2+6x-2 \\ \cline{2-3} & & \qquad \quad 0 \end{tabular}$

Step :- 3

$x^2 - 2$ is also  a factor

$x^2 - (\sqrt{2})^2$

$(x - \sqrt{2} ) (x + \sqrt{2} )$

Other zeroes are $\sqrt{2}$ and $-\sqrt{2}$

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Thanks !