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I'm supposed to show that:

$$y=\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$

and the required answers are: $$ P=5, Q=4, R=-4 $$

I tried to solve this with partial fractions like so:

$$5(x-1)(x+2) = A(x+3) + B(x-2)$$

$\implies$ $A$=4, $B$=-4
$\implies$ $Q$=4, R=-4

But where does $P$=5 come from?

Or should I have first multiplied out the numerator and denominator and then used long division to solve?

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    $\begingroup$ You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. $\endgroup$ Mar 6, 2013 at 10:51
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    $\begingroup$ Yes, you should have done long division first. Look at that last displayed equation of yours --- quadratic on the left, linear on the right --- that can't possibly be right. $\endgroup$ Mar 6, 2013 at 12:13

3 Answers 3

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$$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$ gives
$$5(x-1)(x+2)=P(x-2)(x+3)+Q(x+3)+R(x-2)........(1)$$
Now put $x=2, x=-3,x=0$ respectively on both sides of $(1)$ to get, $Q=4,R=-4,P=5$ respectively. Hence you can reach the desired result.

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  • $\begingroup$ Thanks... I wasn't sure if P was in the equation then what its denominator would be, but now it is clear I should have put P into the partial fractions calculations. $\endgroup$
    – Ozzy
    Mar 6, 2013 at 13:35
  • $\begingroup$ @Ozzy You are welcome. $\endgroup$
    – user52976
    Mar 6, 2013 at 13:37
  • $\begingroup$ I would like to add that comparing the coefficients of x-squared terms gives P=5 directly $\endgroup$
    – Ozzy
    Mar 6, 2013 at 13:53
  • $\begingroup$ @Ozzy Yes.You are right. Nice observation. $\endgroup$
    – user52976
    Mar 6, 2013 at 13:54
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You are given the wrong expression. After some calculation I figure out that in order for these two expressions to be equivalent,

$$\frac{5(x-1)(x-2)}{(x-2)(x+3)} = 5 + \frac{4}{(x-2)} + \frac{-4}{(x+3)}$$

instead of having $\frac{5(x-1)(x-2)}{(x-2)(x+3)}$, it should be

$$\frac{5(x-1)(x+2)}{(x-2)(x+3)}$$

Now solve it again it would work well.

My approach to solving it is

$$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = \frac{5x^2+5-10}{(x-2)(x+3)} =\frac{\frac{5x^2+5-10}{(x-2)}}{(x+3)}$$

$$\frac{\frac{5x^2+5-10}{(x-2)}}{(x+3)} = \frac{5x+15+\frac{40}{(x-2)}}{(x+3)}$$

$$\frac{5x+15+\frac{40}{(x-2)}}{(x+3)} = 5+\frac{40}{(x-2)(x+3)}$$

$$5+\frac{40}{(x-2)(x+3)}=5 + \frac{4}{(x-2)} + \frac{-4}{(x+3)}$$

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  • $\begingroup$ I'm sorry that was a typo, but you are correct that it was (x+2) not (x-2). I can't really follow your calculation steps... I think you are doing the long division and partial fractions at the same time? $\endgroup$
    – Ozzy
    Mar 6, 2013 at 13:39
  • $\begingroup$ @Ozzy I expand it first then do two long devisions before doing a simple partial fraction expansion. $\endgroup$ Mar 6, 2013 at 14:48
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You don't need to do any long division. In the expression

$$y = \frac{N(x)}{D(x)} = \frac{5(x-1)(x+2)}{(x-2)(x+3)}$$

The numerator $N(x) = 5(x-1)(x+2)$ and denominator $D(x) = (x-2)(x+3)$ are polynomials of degree $2$. Since the roots of $D(x)$ are simple, $y$ can be rewritten in the form:

$$y = P(x) + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$

where $P(x)$ is a polynomial of degree $\deg N(x) - \deg D(x) = 2 - 2 = 0$, i.e. a constant. To evaluate the 3 coefficients, you can evaluate both side at 3 different values of $x$: the two roots of $D(x)$ and $\infty$:

$$\begin{align} P &= \lim_{x\to\infty} \frac{N(x)}{D(x)} = 5\\ Q &= \lim_{x\to 2} (x-2)\frac{N(x)}{D(x)} = \lim_{x\to 2}\frac{5(x - 1)(x+2)}{x+3} = \frac{5(2 - 1)(2+2)}{2+3} = 4\\ R &= \lim_{x\to -3}(x+3)\frac{N(x)}{D(x)} = \lim_{x\to-3}\frac{5(x-1)(x+2)}{x-2} = \frac{5(-3-1)(-3+2)}{-3-2} = -4 \end{align}$$

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  • $\begingroup$ Loving it.. This is how it should be done. $\endgroup$ Mar 6, 2013 at 14:50
  • $\begingroup$ Nice... wasn't taught this method $\endgroup$
    – Ozzy
    Mar 8, 2013 at 6:53

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