The correct option is B 2x 1+2x 4/2x^2 1 nbsp;enclose [ 2x 1; 2x^2+0x 1longdiv4x^3 2x^2+0x 3; 4x^3+0x^2 2x ...

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Division of a Polynomial by a Binomial

4 x3-2 x2-3/2...

Question

$\frac{4 x^{3} - 2 x^{2} - 3}{2 x^{2} - 1} =$ ?

2 x + 1 + \frac{2 x - 4}{2 x^{2} - 1}

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2 x - 1 + \frac{2 x - 4}{4 x^{3} - 2 x^{2} - 3}

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2 x - 1 + \frac{2 x - 4}{2 x^{2} - 1}

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2 x + \frac{2 x - 4}{2 x^{2} - 1}

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Solution

The correct option is C $2 x - 1 + \frac{2 x - 4}{2 x^{2} - 1}$

$\begin{matrix} 2 x - 1 2 x^{2} + 0 x - 1 4 x^{3} - 2 x^{2} + 0 x - 3 4 x^{3} + 0 x^{2} - 2 x - + - --------------------- - - 2 x^{2} + 2 x - 3 - 2 x^{2} + 0 x + 1 + - - -------------- - 2 x - 4 \end{matrix}$
$2 x - 1 + \frac{2 x - 4}{2 x^{2} - 1}$

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Similar questions

(4 x^{3} - 2 x^{2} - 3) \div (2 x^{2} - 1)

- x^{6} + 2 x^{4} + 4 x^{3} + 2 x^{2} by \sqrt{2} x^{2}

What must be subtracted from $4 x^{3} - 3 x + 5$ to get $2 x^{2} - 3 x^{3} + 5 x - 2$ ?

f (x) = lim n \to \infty (2 x + 4 x^{3} + \dots + 2^{n} x^{2 n - 1})

x \in (0, \frac{1}{\sqrt{2}}),

\int f (x) d x

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