Divide the given polynomial by the given monomial.
$(x^{3} + 2 x^{2} + 3 x) \div 2 x$

\frac{1}{2} (x^{2} + 2 x + 3)

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

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\frac{1}{2} (x^{2} - 2 x + 3)

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\frac{1}{2} (x^{2} + 2 x - 3)

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Solution

The correct option is A $\frac{1}{2} (x^{2} + 2 x + 3)$
$(x^{3} + 2 x^{2} + 3 x) \div 2 x$
$= \frac{(x^{3} + 2 x^{2} + 3 x)}{2 x}$
$= \frac{x (x^{2} + 2 x + 3)}{2 x}$
$= \frac{(x^{2} + 2 x + 3)}{2}$

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Divide the given polynomial by the given monomial.(x3+2x2+3x)÷2x

The correct option is A 12(x2+2x+3) (x3+2x2+3x)÷2x=(x3+2x2+3x)2x=x(x2+2x+3)2x=(x2+2x+3)2

Divide the given polynomial by the given monomial.
$(x^{3} + 2 x^{2} + 3 x) \div 2 x$

The correct option is A $\frac{1}{2} (x^{2} + 2 x + 3)$
$(x^{3} + 2 x^{2} + 3 x) \div 2 x$
$= \frac{(x^{3} + 2 x^{2} + 3 x)}{2 x}$
$= \frac{x (x^{2} + 2 x + 3)}{2 x}$
$= \frac{(x^{2} + 2 x + 3)}{2}$