Divide the polynomial by monomial
$6 x^{2} + 5 x^{2} - 12 x + 10; 3 x - 2$

(3 x - 2) (2 x^{2} + 3 x - 5) + 0

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(3 x - 2) (2 x^{2} + 3 x - 5) + 1

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(3 x - 2) (2 x^{2} + 3 x - 5) + 2

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(3 x - 2) (2 x^{2} + 3 x - 5) + 3

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Solution

The correct option is A $(3 x - 2) (2 x^{2} + 3 x - 5) + 0$
$6 x^{2} + 5 x^{2} - 21 x + 10 \div 3 x - 2$
$2 x^{2} + 3 x - 5$
$3 x - 2$ $6 x^{2} + 5 x^{2} - 21 x + 10$
$6 x^{3} - 4 x^{2}$
- +
$+ 9 x^{2} - 21 x$
$9 x^{2} - 6 x$
- +
$- 15 x + 10$
$- 15 x + 10$
+ -

$0$
Therefore,
$6 x^{2} + 5 x^{2} - 21 x + 10 = (3 x - 2) (2 x^{2} + 3 x - 5) + 0$

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a^{2}

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