Question

(2x^4-3x^3-3x^2+6x-2) divided by (x^2-2) hint: polynomial ☺☺☺☺​

Answer 1

Answer:

2x²−3x+1

Step-by-step explanation:

(2x^4-3x^3-3x^2+6x-2)/(x^2-2)

(x−1)(2x−1)(x²−2)/x²-2

=2x²−3x+1

Answer 2

Division of polynomials :
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$\begin{array}{r|l|r} {x}^{2} - 2& 2 {x}^{4} - 3 {x}^{3} - 3 {x}^{2} + 6x + 2&2 {x}^{2} - 3x - 1 \\ \cline{1-1} \cline{3-3} \: & 2 {x}^{4} \qquad \: \: \: - 2 {x}^{2} \quad & \: \\ \: & - \qquad \qquad + & \: \\ \cline{2-2} \: &0 \quad - 3 {x}^{3} - {x}^{2} + 6x + 2& \: \\ \: & \qquad - 3 {x}^{3} \qquad \: + 6x \quad & \: \\ \: & \qquad + \qquad \qquad - \qquad & \: \\ \cline{2-2} \: & \quad \qquad 0 \: \: \: - {x}^{2} \qquad \: \: + 2& \: \\ \: & \qquad \: \: \quad \: \: \: \: - {x}^{2} \qquad \: \: \: + 2 & \\ \: & \qquad \: \qquad + \qquad \qquad - & \: \\ \cline{2-2} \: & \qquad \qquad \quad 0 \qquad \qquad 0& \: \\ \end{array}$

Hence,
$\underline{\boxed{ \frac{2 {x}^{4} - 3 {x}^{3} - 3 {x}^{2} + 6x + 2}{ {x}^{2} - 2} = 2 {x}^{2} - 3x - 1 \\ }}$