Verified Answer :

  • Divide the polynomial \(x^2 + 4x + 5\) by \(x + 3\) using long division to find the quotient and remainder.
  • The leading term calculation \(x^2 / x = x\) gives the first part of the quotient.
  • Subtracting \(x(x+3)\) from the dividend leaves \(x+5\), and \(x/x = 1\) completes the quotient.
  • The quotient is \(\boxed{x+1}\) and the remainder is \(\boxed{2}\).

Explanation

  1. Problem Analysis

The objective is to perform the polynomial division of \(P(x) = x^2 + 4x + 5\) by the binomial \(D(x) = x + 3\). According to the division algorithm, we can express the polynomial as \(P(x) = D(x)Q(x) + R\), where \(Q(x)\) is the quotient and \(R\) is the constant remainder. Our goal is to determine these two components.

  1. Performing Polynomial Division

To divide \(x^2 + 4x + 5\) by \(x + 3\), we follow the steps of polynomial long division:

  1. Divide the leading term of the dividend (\(x^2\)) by the leading term of the divisor (\(x\)) to get the first term of the quotient: \(\frac{x^2}{x} = x\).
  2. Multiply this term by the divisor: \(x(x + 3) = x^2 + 3x\).
  3. Subtract this product from the original dividend: \((x^2 + 4x + 5) - (x^2 + 3x) = x + 5\).
  4. Now, divide the leading term of the new expression (\(x\)) by the leading term of the divisor (\(x\)) to get the second term of the quotient: \(\frac{x}{x} = 1\).
  5. Multiply this constant by the divisor: \(1(x + 3) = x + 3\).
  6. Subtract this from the current expression: \((x + 5) - (x + 3) = 2\).

The quotient is \(x + 1\) and the remainder is \(2\).

  1. Conclusion and Verification

To confirm our result, we can multiply the divisor by the quotient and add the remainder: \((x + 3)(x + 1) + 2 = (x^2 + x + 3x + 3) + 2 = x^2 + 4x + 5\). Since this matches the original dividend, the calculation is correct. Therefore, the quotient is \(x + 1\) and the remainder is \(2\). The final results are \(\boxed{x+1}\) and \(\boxed{2}\).

Examples

Polynomial division is a fundamental tool in algebra, useful for simplifying complex rational expressions and finding roots of higher-degree polynomials. For example, in engineering or physics, when analyzing the transfer function of a system, you often encounter ratios of polynomials. By dividing the numerator by the denominator, you can identify the long-term behavior (slant asymptote) of the system, represented by the quotient, and the specific impact of the local perturbation, represented by the remainder. This helps in understanding how signals or energy distribute through such linear systems.

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