Q. \( \dfrac{3x^3 – x – 2}{x} = \).

Q: How do I simplify \frac{3x^3-x-2}{x} ?

Divide each term by x: \frac{3x^3}{x}-\frac{x}{x}-\frac{2}{x}=3x^2-1-\frac{2}{x}. Domain: x\neq 0.

Q: What is the domain of \frac{3x^3-x-2}{x} ?

All real x except 0, so x \in \mathbb{R}\setminus\{0\}, because division by x is undefined at x=0.

Q: Can the numerator 3x^3-x-2 be factored?

Yes: 3x^3-x-2=(x-1)(3x^2+3x+2). The quadratic has negative discriminant, so only real linear factor is (x-1)..

Q: Can any factor cancel with the denominator x?

No. The numerator has no factor x, so nothing cancels; the expression remains 3x^2-1-\frac{2}{x} with x\neq 0..

Q: Are there holes or vertical asymptotes?

There is a vertical asymptote at x = 0 because the numerator is -2 \neq 0 at x = 0. No removable hole exists.

Q: What is the end behavior as x \to \pm \infty ?

Leading term 3x^2 dominates, so the expression \sim 3x^2 and tends to +\infty as x \to \pm\infty..

Q: What is the derivative of \frac{3x^3-x-2}{x} (for x\neq 0 ?).

Differentiate 3x^2-1-\frac{2}{x}: derivative = 6x+\frac{2}{x^2}, valid for x\neq 0.

Answer

Divide each term by \(x\).

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

Separate the fraction term by term.

\(\frac{3x^3-x-2}{x}=\frac{3x^3}{x}-\frac{x}{x}-\frac{2}{x}\)

Simplify each part.

\(\frac{3x^3}{x}=3x^2\)

\(\frac{x}{x}=1\)

So:

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

Final result: \(3x^2-1-\frac{2}{x}\)

Detailed Explanation

Write the original expression:\( \dfrac{3x^3 – x – 2}{x} \)
Use the property that a sum (or difference) divided by a nonzero number can be written as the sum (or difference) of the quotients. In other words,\( \dfrac{A+B+C}{D} = \dfrac{A}{D} + \dfrac{B}{D} + \dfrac{C}{D} \) for \(D \neq 0\).
Apply this to the numerator:

\( \dfrac{3x^3 – x – 2}{x} = \dfrac{3x^3}{x} – \dfrac{x}{x} – \dfrac{2}{x} \)
Simplify each individual quotient separately (remembering the rule for dividing powers of the same base: \( \dfrac{x^m}{x^n} = x^{m-n} \), valid for \(x \neq 0\)):
- \( \dfrac{3x^3}{x} = 3x^{3-1} = 3x^2 \)
- \( \dfrac{x}{x} = 1 \) (for \(x \neq 0\))
- \( \dfrac{2}{x} \) cannot be simplified further, so it remains \( \dfrac{2}{x} \)
Combine the simplified terms:\( 3x^2 – 1 – \dfrac{2}{x} \)
State the domain restriction that must be observed because of division by x:\( x \neq 0 \)

Final simplified form: \( 3x^2 – 1 – \dfrac{2}{x} \) with \( x \neq 0 \).

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Algebra FAQs

How do I simplify \( \frac{3x^3-x-2}{x} \) ?

Divide each term by \(x\): \(\frac{3x^3}{x}-\frac{x}{x}-\frac{2}{x}=3x^2-1-\frac{2}{x}.\) Domain: \(x\neq 0.\)

What is the domain of \( \frac{3x^3-x-2}{x} \)?

All real \(x\) except \(0\), so \(x \in \mathbb{R}\setminus\{0\}\), because division by \(x\) is undefined at \(x=0\).

Can the numerator \(3x^3-x-2\) be factored?

Yes: \(3x^3-x-2=(x-1)(3x^2+3x+2)\). The quadratic has negative discriminant, so only real linear factor is \((x-1)\)..

Can any factor cancel with the denominator \(x\)?

No. The numerator has no factor \(x\), so nothing cancels; the expression remains \(3x^2-1-\frac{2}{x}\) with \(x\neq 0\)..

Are there holes or vertical asymptotes?

There is a vertical asymptote at \(x = 0\) because the numerator is \(-2 \neq 0\) at \(x = 0\). No removable hole exists.

Solve \( \frac{3x^3-x-2}{x}=0\) ..

What is the end behavior as x \to \pm \infty ?

Leading term \(3x^2\) dominates, so the expression \(\sim 3x^2\) and tends to \(+\infty\) as \(x \to \pm\infty\)..

What is the derivative of \( \frac{3x^3-x-2}{x} \) (for \( x\neq 0 \)?).

Differentiate \(3x^2-1-\frac{2}{x}\): derivative \(= 6x+\frac{2}{x^2}\), valid for \(x\neq 0\).

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