Question

Find the partial fraction decompostion
-10x/(2x^2-3x-2)

Answer 1

how to applybthecfunction of sin cos tan in an angle

Answer 2

\[{-10x \over 2x^2-3x-2}={-10x \over (2x+1)(x-2)}={A \over 2x+1}+{B \over x-2}\]
Multiplying both sides by \((2x+1)(x-2)\) gives \(A(x-2)+B(2x+1)=-10x\).
Pluging x=2 gives \(5B=-20 \implies B=-4\) and \(x=-1/2\) gives \(-5/2 A=-5 \implies A=2\). So, \(\frac{-10x}{2x^2-3x-2}=\frac{2}{2x+1}-\frac{4}{x-2}\).

Answer 3

Whoops! A should be -2.

Answer 4

thanku

Answer 5

d/dx f[x]

Answer 6

I don't understand what you're saying santhosh!

Answer 7

i am not fluent in english sorry

Answer 8

i want to know more about trignometric calculations

Answer 9

how to apply the equations

Answer 10

anwar

Answer 11

a+b\[\sqrt[?]{?}\]

Answer 12

OK.. Just write your question in a new post, and I'll help you :)

Answer 13

\[\sqrt[?]{?}\]