Q. 48 - Precalculus Enhanced with Graphing Utilities

Question

Use the division algorithm to rewrite each improper fraction as the sum of a quotient and proper fraction. Find the partial fraction decomposition of the proper fraction. Finally, express the improper fraction as the sum of a quotient and the partial fraction decomposition.

$\frac{x^3-3x^2+1}{x^2+5x+6}$

Step-by-Step Solution

Answer

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

Step 1: Perform Polynomial Long Division

Divide \(x^3 - 3x^2 + 1\) by \(x^2 - 2x + 1 = (x-1)^2\):
\(x^3 - 3x^2 + 1 = (x-1)(x^2 - 2x + 1) + (-2x^2 + x + 1) + \ldots\)
Actually: \(\frac{x^3 - 3x^2 + 1}{(x-1)^2}\). Dividing: \(x^3 - 3x^2 + 1 \div (x^2 - 2x + 1)\).
Quotient: \(x - 1\), remainder: \(x^3 - 3x^2 + 1 - (x-1)(x^2-2x+1)\).
\((x-1)(x^2-2x+1) = x^3 - 3x^2 + 3x - 1\)
Remainder: \(x^3 - 3x^2 + 1 - x^3 + 3x^2 - 3x + 1 = -3x + 2\)

Step 2: Decompose the Proper Fraction

\(\frac{-3x + 2}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2}\)
\(-3x + 2 = A(x-1) + B\)
Setting \(x = 1\): \(-3 + 2 = B\), so \(B = -1\).
Comparing coefficients of \(x\): \(-3 = A\), so \(A = -3\).

Step 3: Write the Final Answer

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