Thank you mrh, again! Oh, yes brings me back to 2nd Half of The Calculus l. Lessons requiring Proper and impeccable use of Algebraic operations is first and foremost in Decomposition of polynomial fractions. Followed some months later with the treatment of The Conics, Geometrically and Algebraically, via the General Form of the Equations of Geometric plain figures. Math IS fun!
We can calculate the values of A, B and C from the above lines i.e., 5x + 7 = A (x+1)(x+2) + B (x-1)(x+2) + C (x-1)(x+1) in an easier way. If we put x = 1, we get 12 = A . 2. 3, that is, A = 2. Same way if put x = -2, we get -3 = C . (-3). (-1), that is C = -1 and putting x = -1, we get 2 = B. (-2). (1), that is B = -1.
It was so long ago that I took Integral calculus, I don't remember if we covered partial fractions or not. I am wondering if decomposition works with rational expressions where the denominator is not factorable? Very nice presentation.
For a denominator factor such as (x^2 - 2), this has no rational factors, but it is factorable as (x - sqrt(2))*(x + sqrt(2)). Irrational factors are preferred when necessary over the alternative, for applications of partial fractions. You may also get a factor with no real roots, such as (x^2 + 4). It is factorable with complex factors, and you can take a complex number detour if you prefer. But it usually is easier to set up the irreducible quadratic term, with an arbitrary linear numerator. Instead of just a constant, it is (B*x + C). Example: (x - 9)/((x + 1)*(x^2 + 4)) Setup: A/x + (B*x + C)/(x^2 + 4) Solution: -2/x + (2*x + 1)/(x^2 + 4) For an irreducible cubic, there'd be a quadratic on top, with 3 unknowns. In general, the polynomial on top is of degree one less than the polynomial on bottom. This form isn't very useful for most applications, so usually you'll get at least one rational root to help you. If you had an "irreducible" cubic in an application, you'd need a cubic formula to reduce it to at least one linear factor.
Mr. H should have shown first how he got the three different denominators for the right hand side of the equation. He probably factored the original polynomial using Synthetic Division? Perhaps next time make it a two-part solution showing ALL the steps. So, 3 out of 5 stars...
After setting this up, and multiplying to eliminate the fractions, you end up with: 5*x + 7 = A*(x + 1)*(x + 2) + B*(x - 1)*(x + 2) + C*(x - 1)*(x + 1) You can see that there are no x^2 terms on the left. This means, all x^2 terms on the right, have to add up to zero. Each of the three products on the right, will have an x^2 term, after they expand, each multiplied by the coefficient. When expanding, we get: 5*x + 7 = A*x^2 + 3*A*x + 2*A + B*x^2 + B*x - 2*B + C*x^2 - C Gather coefficients on like-terms of x: 5*x + 7 = (A + B + C)*x^2 + (3*A + B)*x + (2*A - 2*B - C) And we can equate coefficients, to produce a 3x3 system of equations to solve.
There is a shortcut, that is my preferred method, called Heaviside coverup. For simple linear factors, and the highest denominator power of repeated linear factors, Heaviside coverup allows you to more directly get at the answer. The trick works by covering up the factor in the original expression, that corresponds to each partial fraction denominator. Then, figure out what you can make the variable be, so that what is under your hand is equal to zero. Evaluate what remains, and this tells you the value of the unknown coefficient. For this example: (5*x + 7)/[(x + 1)*(x + 2)*(x - 1)] = A/(x - 1) + B/(x + 1) + C/(x + 2) A = (5*x + 7)/[(x + 1)*(x + 2)*covered], evaluated at x=+1 A = (5*1 + 7)/[(1 + 1)*(1 + 2)] = 2 B = (5*x + 7)/[covered*(x + 2)*(x - 1)], evaluate at x = - 1 B = (5*-1 + 7)/[(-1 + 2)*(-1 - 1)] = -1 C = (5*x + 7)/[(x + 1)*covered*(x - 1)], evaluate at x = -2 C = (5*-2 + 7)/[(-2 + 1)*(-2 - 1)] = -1 The underlying idea of why this works, is that near the pole of x=+1, the A/(x - 1) term will dominate the behavior of the equation, and the other terms will be insignificantly small. So you are ultimately taking the limit as x approaches the problem point of -1 and the other poles, and matching the behavior on both sides. This is related to a concept in complex analysis, called the residue of a pole.
The simple case is when you have all distinct linear factors. The setup is a simple unknown constant on top of each one of them. There is a shortcut to find that constant, called Heaviside coverup. Repeated linear terms require us to either build-the-power, or my preference, to descend the power. So if (x+1)^2 were one of your denominator factors, then A/(x+1)^2 + B/(x + 1) would be the setup associated with it. For (x + 1)^3 as a denominator factor, you'd have 3 terms, A/(x+1)^3 + B/(x+1)^2 + C/(x+1). I like descending the power instead of building the power, so I can find the first constant with Heaviside coverup first. Quadratic factors with no real roots, require an arbitrary linear term on top, such as (B*x + C)/(x^2 + 1). Cubics and beyond, would start with an arbitrary polynomial, that is one degree less than the denominator. They aren't that useful for most applications of this process.
...and all this came already shortened - because you came with the polynomial already factored out. Otherwise one needs to do synthetic division first.
This also has use in Laplace transforms, which are methods to solve differential equations. Laplace transforms are methods of converting time-domain calculus into s-domain algebra, for purposes of solving differential equations. Applications in real life of this method, are vibrations, control systems, and electric filter & amplifier circuits.
