Q. Simplify \(\dfrac{x^3+x^2+x+2}{x^2-1}\)?

Answer

\[
\text{Divide: } \frac{x^3+x^2+x+2}{x^2-1}.
\]
Leading-term division: \(x^3/(x^2)=x\). Multiply and subtract: \(x^3+x^2+x+2-(x^3-x)=x^2+2x+2\).
Next term: \(x^2/(x^2)=1\). Multiply and subtract: \(x^2+2x+2-(x^2-1)=2x+3\).
Final result:
\[
\frac{x^3+x^2+x+2}{x^2-1}=x+1+\frac{2x+3}{x^2-1}.
\]

Detailed Explanation

Step-by-step solution

Divide the polynomial \(x^3 + x^2 + x + 2\) by \(x^2 – 1\) using polynomial long division.
- First term: divide \(x^3\) by \(x^2\) to get \(x\).
- Multiply divisor by \(x\): \(x(x^2 – 1) = x^3 – x\).
- Subtract: \((x^3 + x^2 + x + 2) – (x^3 – x) = x^2 + 2x + 2\).
Continue the division.
- Divide \(x^2\) by \(x^2\) to get \(1\).
- Multiply divisor by \(1\): \(1(x^2 – 1) = x^2 – 1\).
- Subtract: \((x^2 + 2x + 2) – (x^2 – 1) = 2x + 3\).
Write the quotient and remainder form.

\(x^3 + x^2 + x + 2\) divided by \(x^2 – 1\) equals \(x + 1 + \frac{2x + 3}{x^2 – 1}\).
Factor the denominator for partial fractions: \(x^2 – 1 = (x – 1)(x + 1)\). Decompose the remainder fraction:
\[ \frac{2x+3}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \]
Multiply through by \((x-1)(x+1)\) to get:
\[ 2x + 3 = A(x+1) + B(x-1) \]
Equate coefficients:

\(A + B = 2\) (coefficient of \(x\)) and \(A – B = 3\) (constant term).

Solve the system:

\(A = \frac{5}{2}\) and \(B = -\frac{1}{2}\).
Substitute the partial fractions back:
\[ \frac{2x+3}{x^2-1} = \frac{5/2}{x-1} – \frac{1/2}{x+1} \]
Therefore the full simplified form is:

\(x + 1 + \dfrac{5/2}{x-1} – \dfrac{1/2}{x+1}\).

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FAQs

Q1: How do I simplify \(\dfrac{x^3+x^2+x+2}{x^2-1}\)?

A1: Perform polynomial division: \(\dfrac{x^3+x^2+x+2}{x^2-1}=x+1+\dfrac{2x+3}{x^2-1}.\)

Q2: What are the quotient and remainder?

A2: -Quotient \(=x+1\). Remainder \(=2x+3\). So \( \dfrac{N}{D}= \text{quotient} + \dfrac{\text{remainder}}{D}.\)

Q3: Can this fraction be reduced?

A3: No. Denominator factors as \((x-1)(x+1)\). Numerator has no common factor with those, so no cancellation.

Q4: What is the partial fraction decomposition?

A4: \(\dfrac{2x+3}{x^2-1}=\dfrac{5/2}{x-1}-\dfrac{1/2}{x+1}.\) So full form: \(x+1+\dfrac{5/2}{x-1}-\dfrac{1/2}{x+1}.\)

Q5: What is the domain?

A5: -All real \(x\) except where denominator zero: \(x\neq 1\) and \(x\neq -1\).

Q6: -Are there vertical asymptotes?

A6: Yes. Vertical asymptotes at \(x=1\) and \(x=-1\) (no cancellation yields infinite behavior).

Q7: Is there a horizontal or oblique asymptote?

A7: Oblique asymptote \(y=x+1\) (degree numerator = degree denominator + 1). No horizontal asymptote.

Q8: -Are there any holes in the graph?

A8: No holes, because numerator and denominator do not share factors to cancel.

Q9: How to perform the division quickly?

A9: Use polynomial long division: divide leading terms, multiply, subtract, repeat. For linear factors you can use synthetic division; for quadratic divisor use long division.

Q10: What is the remainder when dividing the numerator by \(x-1\) or \(x+1\)?

A10: By the Remainder Theorem: \(f(1)=1+1+1+2=5\) (remainder for \(x-1\)). \(f(-1)=-1+1-1+2=1\) (remainder for \(x+1\)).

Q11: Does the function behave like \(x+1\) for large \(|x|\)?

A11: Yes. -As \(x\to\pm\infty\), \(\dfrac{x^3+x^2+x+2}{x^2-1}=x+1+o(1)\), so it approaches the line \(y=x+1\).

Q12: How can I check my division result?

A12: Multiply divisor by quotient and add remainder: \((x^2-1)(x+1)+(2x+3)=x^3+x^2+x+2\). If equal to original numerator, the division is correct.

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