The correct option is D √(2)x^2 Quotient 2x^3/√(2) x = ( 2/√(2)) ( x^3/x) = √(2)×√(2) x^3/√(2) x = √(2) x^2 ...

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Dividing a Monomial by a Monomial

Divide 2x3 ...

Question

Divide $2 x^{3}$ by $\sqrt{2} x$

\sqrt{5} x^{3}

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\sqrt{3} x^{3}

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\sqrt{2} x^{2}

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\sqrt{2} x

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Solution

The correct option is D $\sqrt{2} x^{2}$
Quotient
$\frac{2 x^{3}}{\sqrt{2} x}$
$= (\frac{2}{\sqrt{2}}) (\frac{x^{3}}{x})$
$= \frac{\sqrt{2} \times \sqrt{2} x^{3}}{\sqrt{2} x} = \sqrt{2} x^{2}$

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Similar questions

p (x) = x^{4} - 3 x^{3} + 2 x^{2} + 1

2 x - 1

3 x^{3} + 2 x^{2} - x + 4

3 + 2 x

\frac{- 1}{8}

p(x)= 2x⁴ - 3x³ + 2x² + 2x - 1 is divided by (x - 2) and q(x) = 3x³ - 2x² + x - 1 is divided by (x - 1). So, twice the sum of the remainders is:

The remainder when $3 x^{3} + 2 x^{2} - x + 4$ is divided by $3 + 2 x$ is $\frac{- 1}{8}$ .

When $3 x^{3} + 2 x^{2} - x + 4$ is divided by $3 + 2 x$ the remainder is $\frac{- 1}{8}$ .

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