- For each expression, factor the denominator into irreducible components and use polynomial long division if necessary.
- (a) \(\frac{2}{3(x-1)} + \frac{4x+20}{3(x^2+x+1)}\)
- (b) \(1 - \frac{8}{9(x+8)} + \frac{8}{9(x-1)}\)
- (c) \(\frac{2x-2}{x^2+1} - \frac{1}{x+1}\)
- (d) \(\frac{14x-182}{17(x^2+1)} - \frac{14}{17(x-4)}\)
- (e) \(x-1 + \frac{11}{3(x+2)} + \frac{4}{3(x-1)}\)
- Final answer: The decompositions are \(\boxed{\frac{2}{3(x-1)} + \frac{4x+20}{3(x^2+x+1)}, 1 - \frac{8}{9(x+8)} + \frac{8}{9(x-1)}, \frac{2x-2}{x^2+1} - \frac{1}{x+1}, \frac{14x-182}{17(x^2+1)} - \frac{14}{17(x-4)}, x-1 + \frac{11}{3(x+2)} + \frac{4}{3(x-1)}}\).
Explanation
- Analysis and General Method of Partial Fraction Decomposition
Partial fraction decomposition is a method of breaking down a complex rational expression into simpler fractions that are easier to work with, especially in calculus. The general process involves checking the degrees of the numerator \(P(x)\) and the denominator \(Q(x)\). If the degree of \(P(x)\) is greater than or equal to the degree of \(Q(x)\), we first perform polynomial long division to extract a polynomial part and a proper rational expression (where the degree of the numerator is less than the degree of the denominator). Then, we factor the denominator \(Q(x)\) completely into linear factors \((x-r)\) or irreducible quadratic factors \((x^2+px+q)\). Finally, we express the rational part as a sum of partial fractions, using coefficients to be determined by comparing expressions.
- Solving Part (a)
AFor the expression \(\frac{2x^2+6x-6}{x^3-1}\), the degree of the numerator is 2 and the denominator is 3. Factoring the denominator, we get \(x^3-1 = (x-1)(x^2+x+1)\).
We set up the decomposition as:
\[\frac{2x^2+6x-6}{x^3-1} = \frac{A}{x-1} + \frac{Bx+C}{x^2+x+1}\]
Multiplying by the common denominator: \(2x^2+6x-6 = A(x^2+x+1) + (Bx+C)(x-1)\).
Setting \(x=1\), we find \(2 = 3A\), so \(A = \frac{2}{3}\).
Comparing coefficients of powers of \(x\), we find \(B = \frac{4}{3}\) and \(C = \frac{20}{3}\).
Thus, \(\frac{2x^2+6x-6}{x^3-1} = \frac{2}{3(x-1)} + \frac{4x+20}{3(x^2+x+1)}\).
- Solving Part (b)
BFor \(\frac{x^2+7x}{x^2+7x-8}\), the degree of the numerator is 2 and the denominator is 2, so we divide first:
\[\frac{x^2+7x}{x^2+7x-8} = 1 + \frac{8}{x^2+7x-8} = 1 + \frac{8}{(x+8)(x-1)}\]
Setting up the partial fraction: \(\frac{8}{(x+8)(x-1)} = \frac{A}{x+8} + \frac{B}{x-1}\).
Multiplying by the denominator: \(8 = A(x-1) + B(x+8)\).
Setting \(x=1\), we get \(8 = 9B\), so \(B = \frac{8}{9}\). Setting \(x=-8\), we get \(8 = -9A\), so \(A = -\frac{8}{9}\).
The decomposition is \(1 - \frac{8}{9(x+8)} + \frac{8}{9(x-1)}\).
- Solving Part (c)
CFor \(\frac{x^2-3}{x^3+x^2+x+1}\), the denominator factors by grouping: \(x^2(x+1) + 1(x+1) = (x^2+1)(x+1)\).
Decomposition: \(\frac{x^2-3}{(x^2+1)(x+1)} = \frac{Ax+B}{x^2+1} + \frac{C}{x+1}\).
Multiplying by the denominator: \(x^2-3 = (Ax+B)(x+1) + C(x^2+1)\).
Setting \(x=-1\), we get \(-2 = 2C\), so \(C = -1\).
Expanding and matching coefficients gives \(A=2\) and \(B=-2\).
The result is \(\frac{2x-2}{x^2+1} - \frac{1}{x+1}\).
- Solving Part (d)
DFor \(\frac{42-14x}{x^3-4x^2+x-4}\), the denominator factors: \(x^2(x-4) + 1(x-4) = (x^2+1)(x-4)\).
Decomposition: \(\frac{42-14x}{(x^2+1)(x-4)} = \frac{Ax+B}{x^2+1} + \frac{C}{x-4}\).
Multiplying by denominator: \(42-14x = (Ax+B)(x-4) + C(x^2+1)\).
Setting \(x=4\), we get \(-14 = 17C\), so \(C = -\frac{14}{17}\).
Expanding and equating coefficients results in \(A = \frac{14}{17}\) and \(B = -\frac{182}{17}\).
The result is \(\frac{14x-182}{17(x^2+1)} - \frac{14}{17(x-4)}\).
- Solving Part (e) and Conclusion
EFor \(\frac{x^3+2x+1}{x^2+x-2}\), since the numerator has a higher degree, we use long division to get \(x-1 + \frac{5x-1}{x^2+x-2}\).
Factoring the denominator: \(x^2+x-2 = (x+2)(x-1)\).
Partial fraction for \(\frac{5x-1}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}\).
Solving \(5x-1 = A(x-1) + B(x+2)\) gives \(A = \frac{11}{3}\) and \(B = \frac{4}{3}\).
Final form: \(x-1 + \frac{11}{3(x+2)} + \frac{4}{3(x-1)}\).
Examples
Partial fraction decomposition is fundamental in engineering and physics, specifically in systems analysis. For example, when analyzing the output of an electrical circuit using the Laplace transform, you often encounter complex rational functions in the frequency domain. Decomposing these functions into simpler partial fractions allows you to use inverse transforms to return to the time domain, effectively finding the voltage or current response over time. This helps engineers design stable and efficient circuits.