Decompose into Partial Fraction. Image Added. [duplicate]

I really had no idea how to write these questions out without copying and pasting them onto here, so I am sorry for that..I hope adding a picture is fine. I would appreciate any kind of help, and if you dont mind suggesting a useful website, or video that can explain these problems to me. Thank you

marked as duplicate by user147263, Harish Chandra Rajpoot, Claude Leibovici, Mario Carneiro, user91500Oct 18 '15 at 10:02

It can be decomposed into the form $$\frac{x^2}{(x+\alpha)(x-\beta)^2} = \frac{\lambda}{x+\alpha} + \frac{\mu}{x-\beta} + \frac{\nu}{(x-\beta)^2}$$ which means $$\lambda(x-\beta)^2 + \mu(x+\alpha)(x-\beta) + \nu(x+\alpha) = x^2$$ It can be written as a set of equations \begin{align*} \lambda + \mu &= 1 \\ -2\beta\lambda + (\alpha - \beta)\mu + \nu &= 0 \\ \beta^2\lambda - \alpha\beta\mu + \alpha\nu &= 0 \end{align*} Solving for $\lambda, \mu, \nu$ will give you the correct decomposition. (I used CAS here) $$\left[\begin{array}{ccc|c} 1 & 1 & 0 & 1 \\ -2\beta & \alpha-\beta & 1 & 0 \\ \beta^2 & -\alpha\beta & \alpha & 0 \end{array}\right] \rightarrow \left[\begin{array}{ccc|c} 1 & 0 & 0 & \frac{\alpha^2}{(\alpha+\beta)^2} \\ 0 & 1 & 0 & -\frac{\beta^2+2\alpha\beta}{(\alpha+\beta)^2} \\ 0 & 0 & 1 & \frac{\beta^2}{\alpha+\beta} \end{array}\right]$$ Hence $$\lambda = \frac{\alpha^2}{(\alpha+\beta)^2}, \mu = -\frac{\beta^2+2\alpha\beta}{(\alpha+\beta)^2}, \nu = \frac{\beta^2}{\alpha+\beta}$$
• @user281364 $x$ is not involved in the three linear equations. These three equations are used to solve for the coefficients $a, b, c$ in $ax^2 + bx + c$. – Henricus V. Oct 20 '15 at 23:14