Divide the polynomial by monomial:
$3 x^{5} - 4 x^{4} + 3 x^{3} + 2 x; x^{2} - 3$

(x^{2} - 3) (3 x^{3} - 4 x^{2} + 12 x - 12) + 38 x - 36

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(x^{2} - 3) (x^{3} - 4 x^{2} + 2 x - 2) + 3 x - 6

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(2 x^{2} - 7) (3 x^{3} + 4 x^{2} + x - 1) + 47 x - 109

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(2 x^{2} + 7) (3 x^{3} - 4 x^{2} + 12 x - 12) + 38 x - 36

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Solution

The correct option is A $(x^{2} - 3) (3 x^{3} - 4 x^{2} + 12 x - 12) + 38 x - 36$
$3 x^{5} - 4 x^{4} + 3 x^{3} + 2 x \div x^{2} - 3$
$3 x^{3} - 4 x^{2} + 12 x - 12$
$x^{2} - 3$ $3 x^{5} - 4 x^{4} + 3 x^{3} + 2 x$
$3 x^{5} - 9 x^{3}$
- +
$- 4 x^{4} + 12 x^{3}$
$4 x^{4} + 12 x^{2}$
- -
$+ 12 x^{3} - 12 x^{2} + 2 x$
$+ 12 x^{3} + 36 x$
- -
$- 12 x^{2} + 38 x$
$- 12 x^{2} + 36$
+ -
$38 x - 36$

Therefore,
$3 x^{3} - 4 x^{2} + 12 x - 12 = (x^{2} - 3) (3 x^{3} - 4 x^{2} + 12 x - 12) + 38 x - 36$

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Similar questions

x^{4} - 4 x^{2} + 12 x + 9

x^{2} + 2 x - 3

\begin{matrix} {lim}_{x \to 1} \frac{x^{2} - 3 x + 2}{x^{2} - 4 x + 3} {lim}_{x \to 2} \frac{x^{3} + x^{2} - 12}{x^{3} - x^{2} - x - 2} {lim}_{x \to \sqrt{2}} \frac{x^{2} + x \sqrt{2} - 4}{x^{2} - 3 x \sqrt{2} + 4} {lim}_{x \to 3} \frac{x^{3} - 3 x^{2} - x + 3}{x^{3} - 6 x^{2} + 9 x} \end{matrix}

2 (x - 1)^{2} = 4 x^{2} - 2 x + 1

2 x - x^{2} = x^{2} + 5

{(\sqrt{2} x + \sqrt{3})}^{3} = 3 x^{2} - 5 x

(x^{2} + 2 x)^{2} = x^{4} + 3 + 4 x^{2}

\frac{(4 x + 1)^{2} + (2 x + 3)^{2}}{4 x^{2} + 12 x + 9} = \frac{61}{36}

\lim_{x \to \infty} \frac{3 x^{3} - 4 x^{2} + 6 x - 1}{2 x^{3} + x^{2} - 5 x + 7}

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