ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 11- Worked Out Examples

Monday 4 August 2014

Example: 1

Expand the following using partial dfractions:

	(a) $\dfrac{{{x^2} - 3x - 4}}{{{x^3} - 6{x^2} + 11x - 6}}$
	(b) $\dfrac{{{x^2} + x + 1}}{{{{(x - 1)}^2}(x - 2)(x - 3)}}$

Solution: 1-(a)

The denominator can be factorised:

${x^3} - 6{x^2} + 11x - 6 = (x - 1)(x - 2)(x - 3)$

The partial dfraction expansion is:

$\dfrac{{(x + 1)(x - 4)}}{{(x - 1)(x - 2)(x - 3)}} = \dfrac{A}{{x - 1}} + \dfrac{B}{{x - 2}} + \dfrac{C}{{x - 3}}$

Cross-multiplying, we obtain

	$(x + 1)(x - 4) = A(x - 2)(x - 3) + B(x - 1)(x - 3) + C(x - 1)(x - 2)$
	Put $x = 1 \Rightarrow\,\,\,\, A = \dfrac{{(2) \times ( - 3)}}{{( - 1) \times ( - 2)}} = - 3$
	Put $x = 2 \Rightarrow\,\,\,\, B = \dfrac{{(3) \times ( - 2)}}{{(1) \times ( - 1)}} = 6$
	Put $x = 3 \Rightarrow\,\,\,\, C = \dfrac{{(4) \times ( - 1)}}{{(2) \times (1)}} = - 2$

Thus, the partial dfraction expansion is

$\dfrac{{ - 3}}{{x - 1}} + \dfrac{6}{{x - 2}} + \dfrac{{ - 2}}{{x - 3}}$

Solution: 1-(b)

As discussed previously, the partial dfraction expansion of this expression would be of the form

$\dfrac{{{x^2} + x + 1}}{{{{(x - 1)}^2}(x - 2)(x - 3)}} = \dfrac{A}{{x - 1}} + \dfrac{B}{{{{(x - 1)}^2}}} + \dfrac{C}{{x - 2}} + \dfrac{D}{{x - 3}}$

Cross-multiplying, we obtain

${x^2} + x + 1 = A(x - 1)(x - 2)(x - 3) + B(x - 2)(x - 3) + C{(x - 1)^2}(x - 3) + D{(x - 1)^2}(x - 2)$

	Put $x = 1 \Rightarrow\,\,\,\, B = \dfrac{{1 + 1 + 1}}{{( - 1) \times ( - 2)}} = \dfrac{3}{2}$
	$x = 2 \Rightarrow\,\,\,\, C = \dfrac{{4 + 2 + 1}}{{{1^2} \times ( - 1)}} = - 7$
	$x = 3 \Rightarrow\,\,\,\, D = \dfrac{{9 + 3 + 1}}{{{2^2} \times (1)}} = \dfrac{{13}}{4}$