# Find constants of function

I have this equality :

$$f(x)=\frac{9}{(x-1)(x+2)^2}$$

I am required to find the constants A, B and C so that,

$$f(x) = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^{2}}$$

How do we go about solving such a question?

I am not sure on how to solve such questions. Approach and Hints to solve these kinds of questions are welcomed. :)

Thank you!

Edit: Wow, saw all the answers! Didn't know that there were a lot of different ways to solve this question. Math is such a fascinating thing!

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Add the three fractions on the right hand side of your equation

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

I'll leave it to you to simplify the numerator and solve for $A$,$B$ and $C$, such that.

$$A(x+2)^2+B(x-1)(x+2)+C(x-1)=9$$

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Wow...i was thinking it might be much more complicated. Can't believe i missed such a simple thing! Thank you very much :) – Phantom Mar 27 '14 at 18:57
You are most welcome. – Alijah Ahmed Mar 27 '14 at 18:58

Hint Set $\dfrac{9}{(x-1)(x+2)^2} = \dfrac{A}{(x-1)} + \dfrac{B}{(x+2)} + \dfrac{C}{(x+2)^{2}}$ and multiply both sides by $(x-1)(x+2)^2$. Then simplify and solve for $A,B,C$.

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Algebric solution: use Euclide algorithm to solve the Bezout equation $$1 = (X-1)U(X) + (X+2)^2 V(X).$$

Analytic solution:

1. multiply by $x$ and take limit to $\infty$ gives $$A+B=0$$
2. multiply by $x^2$ and take limit to $\infty$ (after plugging $A+B=0$) gives $$0 = 2A-B+C$$
3. multiply by $(x-1)$ and take limit to $1$ gives $$A = \frac 9{3^2}=1.$$

You then get $B=-1, C=-3$.

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Sorry, i just started to learn calculus. I have not learnt this yet :( – Phantom Mar 27 '14 at 18:53
additional analytic proof :) – mookid Mar 27 '14 at 19:00

$$\dfrac{9}{(x-1)(x+2)^2}=\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$$

Multiplying by the GCD(which means multiplying all the terms by $(x-1)(x+2)^2$) $$9=A(x+2)^2+B(x-1)(x+2)+C(x-1)$$ Let $x=-2$; $$9=A(-2+2)^2+B(-2-1)(-2+2)+C(-2-1)$$ $$9=A(0)+B(-3)(0)+C(-3)$$ $$9=-3C$$ $$C=-3$$ Now that we have C,we let x=1; $$9=A(1+2)^2+B(1-1)(1+2)+C(1-1)$$ $$9=A(3)^2+B(0)(3)+C(0)$$ $$9=9A$$ $$\implies A=1$$ then we let x=3 and substitute back A and C $$9=(3+2)^2+B(2)(5)-3(2)$$ $$9=25+10B-5$$ $$9=20+10B$$ $$-11=10B$$ $$B=\dfrac{-11}{10}$$

So,now you have the values of A,B and C. This is how you go about doing it. Try it yourself. There might be a small arithmetic error in this solution. Solve it yourself to find it.

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Why not just use $x = 0$ to find B? The calculation uses smaller numbers, so one is less prone to make the arithmetic error for that value. – RecklessReckoner Mar 27 '14 at 19:12
Good point,that works better. I used 3 to show any arbitrary value other that 1 and -2 can be used. Thank you for the improvement though. Feel free to edit the answer and show that it works with 0 as well – JEET TRIVEDI Mar 27 '14 at 19:16
Well, what you have is fine (except for the intentional "error"). I suggest to students using "easy" numbers for this because they may have to do this "under the gun" on an exam problem. I would fix the one line to read "...substitute back A and C ". – RecklessReckoner Mar 27 '14 at 19:19
I understand what you mean. – JEET TRIVEDI Mar 27 '14 at 19:21

I'd like to bring again one of the below-mentioned proofs, but in a slightly modified form. In fact, I'll use the same words as you can read it in H. Wilfs Generatingfunctionology (section 1.2). The text is for your amusement and the benefit is that you could use these techniques, if you need an answer quickly.

We have the form $$\frac{9}{(x-1)(x+2)^2}=\frac{A}{(x-1)}+\frac{B}{(x+2)}+\frac{C}{(x+2)^2}\qquad\qquad(\ast)$$ and the only problem is how to find the constants $A$, $B$, $C$.

Here's the quick way. First multiply both sides of $(\ast)$ by $(x+2)^2$ and then let $x=-2$. The instant result is that $C=-3$ (don't take my word for it, try it for yourself!). Next multiply $(\ast)$ through by $x-1$ and let $x=1$. The instant result is that $A=1$. The hard one to find is $B$, so let's do that one by cheating. Since we know that $(\ast)$ is an identity, i.e., is true for all values of $x$, let's choose an easy value of $x$, say $x=0$, and substitute that value of $x$ into $(\ast)$. Since we know $A$ and $C$, we find at once that $B=-1$.

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A simple way is

$$A=f(x)(x-1)\bigg|_{x=1}=1$$ $$C=f(x)(x+2)^2\bigg|_{x=-2}=-3$$ $$0=\lim_{x\to\infty}xf(x)=A+B\Rightarrow B=-A=-1$$

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The way this approach (which I was typing out, but won't bother now) is generally taught is a bit of a cheat, because it seems like we are inserting values into the numerator that also make the denominator zero. You show the step that "legitimizes" this, which is that we are removing singularities and taking limits. – RecklessReckoner Mar 27 '14 at 19:15
Why it's a cheat? If we multiply $f(x)$ by $(x-1)$ and cancel we can replace $x$ by $1$ and we find $A$. This's the mean of the expression: $$A=f(x)(x-1)\bigg|_{x=1}=1$$ – user63181 Mar 27 '14 at 19:19
I'm saying the way it is often taught is a cheat because those students are just told to substitute values for $\ x \$ without explaining why it is all right to do that, or why it even gives correct results. – RecklessReckoner Mar 27 '14 at 19:22
Sorry I don't agree with you, explanation of this equality does not require a great thing: it suffices just with eyes watching the two expressions of $f$. – user63181 Mar 27 '14 at 19:29

From $$\frac{9}{(x-1)(x+2)^2} = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^2}$$ it follows, after multiplying each side with $(x-1)(x+2)^2$: $$9=A(x+2)^2+B(x+2)(x-1)+C(x-1)$$ which, after regrouping, gives $$9=(A+B)x^2 + (4A+B+C)x + (4A-2B-C)$$

There are two polynomials, one on each side of this equation, and two polynomials $P_a(x)=\sum a_nx^n$ and $P_b(x)=\sum b_nx^n$ are equal if and only if $a_n=b_n \forall n$.

Therefore, three equations follow from the previous one: \begin{eqnarray} A+B &=& 0 \\ 4A+B+C &=& 0 \\ 4A-2B-C &=& 9 \end{eqnarray}

Solving this system of equations gives the solution: $A=1$, $B=-1$ and $C=-3$.

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