+116 Fill in the blanks, to make a true equation:
(8x^3 + 24x^2 + 15x + 1)/((x^2 - 1)(x^2 + 3x)) = ___/(x - 1) + ___/(x + 3) + ___/x + ___/(x + 1)
+2 I just typed out how to do it and then my post was deleted 
I'm just gonna give the summary:
fill in a,b,c,d for the blanks.
Then, find a common denominator for all variables on the right side. At this point it should look like the image below:
\(\frac{8x^3 + 24x^2 + 15x + 1}{(x+1)(x-1)(x)(x+3)} = \frac{a(x+1)x(x+3)}{(x+1)(x-1)(x)(x+3)} + \frac{b(x+1)x(x-1)}{(x+1)(x-1)(x)(x+3)} + \frac{c(x+1)(x-1)(x+3)}{(x+1)(x-1)(x)(x+3)} + \frac{d(x-1)x(x+3)}{(x+1)(x-1)(x)(x+3)} \)
You can cancel the denominator on both sides and foil the polynomials on the right side (but do not multiply by a,b,c or d):
\(8x^3 + 24x^2 + 15x + 1 = a(x^3 + 4x^2 + 3x) + b(x^3 + x^2 - x - 1) + c(x^3 + 2x^2 - 3x) + d(x^3 - 2x^2 + 1) \)
\(x^3\) can only be paired with \(x^3\), \(x^2\) with \(x^2\), and so on. This means \(ax^3+bx^3+cx^3+dx^3=8x^3\), or \(a+b+c+d=8\). We can apply this rule to \(x^3, x^2, x,\) and the constant. So we get:
\(\begin{aligned} a + b + c + d &= 8 \\ 4a + b + 2c - 3d &= 24 \\ 3a - b - 3c &= 15 \\ -c + d &= 1 \end{aligned} \)
You may use any method to solve this system of equations. The first step is to substitute \(d=c+1\) into every equation. I used matrices to solve from there, but you may want to add/subtract equations depending on your knowledge. You should get the final answer:
| \(a = \frac{46}{9}, \, b = 5, \, c = \frac{-14}{9}, \, d = \frac{-5}{9} \) |
p.s: you would probably have more success in getting replies if you typed your equation into the website's built-in LaTeX editor. Just sayin'
