question about partial fractions why $\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}$

Question

can anyone tell me why

$$\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$

I don't undestand the $\frac{C}{(x-2)^2}$ and also what is wrong according to basic math rules if I write the equation in the following way

$$\frac{6x^2+19x+15}{(x-1)(x-2)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-2}$$

Thanks

Answer 1

Two things wrong with what you propose.

1. Those last two terms could be combined into one, $${B\over x-2}+{C\over x-2}={D\over x-2}$$

2. It just doesn't work ---- what would you get for $B$ and $C$ if you tried to do $${1\over(x-2)^2}={B\over x-2}+{C\over x-2}$$

Answer 2

This is one of the things that is to be just learned. n-roots in the denominator will be decomposed into more than 1 term and you may need all the $\frac{A_i}{(x-x_0)^i}$-terms where $i \in \{i\}_{i=1}^n$. Here $x_0=2$ and $n=2$. It wasn't shown in studybooks.

Answer 3

An easy way to see why your proposal doesn't work is the following: Multiply both sides of your equation with $(x-1)(x-2)^2$ and extend the formula continuously to $x=1$ and $x=2$. We now have $$6x^2+19x+15 = A(x-2)^2+B(x-1)(x-2)+C(x-1)(x-2).$$ If we now enter $x=2$, we get $24+38+15 = A\cdot 0^2+B\cdot 1\cdot 0 + C\cdot 1\cdot 0 = 0$, which is obviously false. On the other hand $$6x^2+19x+15 = A(x-2)^2 + B(x-1)(x-2)+C(x-1)$$ would have reduced to $24+38+15 = C$, which is perfectly fine.

Answer 4

It is easy to prove that if no denominator on the RHS has a factor of $\,(x\!-\!2)^2\,$ then neither does the LHS, i.e. a factor of $\,x\!-\!1$ can be cancelled out of the LHS, since

$\quad\dfrac{f(x)}{(x\!-\!2)^2 g(x)}\, =\, \dfrac{a(x)}{(x\!-\!2)b(x)} + \dfrac{c_1(x)}{d_1(x)}+\cdots+\dfrac{c_n(x)}{d_n(x)}\ \ \bigg\lbrace\ \begin{eqnarray} x\!-\!2\nmid b(x),\,d_i(x),\ \ {\rm i.e.}\\ b(2)\ne 0,\ \ d_i(2)\ne 0\quad\ \ \end{eqnarray}$

$\quad \Rightarrow\ f(x)\, =\, (x\!-\!2) g(x) \left(\dfrac{a(x)}{b(x)} + \dfrac{(x\!-\!2)c_1(x)}{d_1(x)}+\cdots + \dfrac{(x\!-\!2)c_n(x)}{d_n(x)}\right)$

$\quad \stackrel{\large x=2}\Rightarrow\ f(2) = 0\ \Rightarrow\ f(x) = (x\!-\!2) h(x)\ \Rightarrow\ \dfrac{f(x)}{(x\!-\!2)^2 g(x)}\, =\, \dfrac{h(x)}{(x\!-\!2)g(x)}$