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

1/2(x²+2x+3)

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1/2

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(x²+2x+3)

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Solution

The correct option is A
1/2(x²+2x+3)

On dividing all the terms of the expression $(x^{3} + 2 x^{2} + 3 x)$ separately by $2 x$ we get:
$x^{3} \div 2 x = \frac{x^{2}}{2}$ ;
$\frac{2 x^{2}}{2 x} = x$ ;
$\frac{3 x}{2 x} = \frac{3}{2}$ .
Therefore $\frac{(x^{3} + 2 x^{2} + 3 x)}{2 x} = \frac{(x^{2} + 2 x + 3)}{2}$

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

The correct option is A 1/2(x2+2x+3) On dividing all the terms of the expression (x3+2x2+3x) separately by 2x we get: x3÷2x=x22; 2x22x=x; 3x2x=32. Therefore (x3+2x2+3x)2x=(x2+2x+3)2

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