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

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Solution

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

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Similar questions

Q. Question 2(iv)
Divide the given polynomial by the given monomial

(x^{3} + 2 x^{2} + 3 x) \div 2 x

Q. Question 2(iv)
Divide the given polynomial by the given monomial

(x^{3} + 2 x^{2} + 3 x) \div 2 x

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Divide the given polynomial by the given monomial:
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. Does it exactly divide the polynomial? State reason

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

x3+2x2+3x=x(x2+2x+3)(x3+2x2+3x)÷2x=x(x2+2x+3)2x=12(x2+2x+3)

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

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