How do you go from (3w-1)/(w+2) and get: 3-(7/(w+2))?

[adelaide87]

How do you go from

(3w-1)/(w+2)

and get:

3-(7/(w+2))?

 

[Mark44]

How do you go from

(3w-1)/(w+2)

and get:

3-(7/(w+2))?

Add 0 to the numerator like this: 3w - 1 = 3w + 6 - 6 - 1 = 3(w + 2) - 7
Now split the rational expression into two parts.

 

[HallsofIvy]

Another way to do the same thing: divide 3w- 1 by w+ 2. w, alone, divides into 3w 3 times, of course. (3w- 1)/(w+ 2)= 3+ ?. Multiplying both sides by w+ 2, 3w- 1= 3w+ 6- ?(w+2). ?(w+ 2)= (3w+ 6)- (3w- 1)= 7 so that ?= 7/(w+ 2).

(3w- 1)/(w+ 2)= 3+ 7/(w+ 2).

 

[adelaide87]

What about this one?

2x/((x+3)(x+1))

Simplified to

3/(x+3) - 1/(x+1)

This is probably one of the harder parts of algebra for me for some reason. I don't understand the steps.

 

[Mark44]

This is an example of rational function decomposition. In this case you need to rewrite the original expression as the sum of two simpler rational expressions.

\frac{2x}{(x + 3)(x + 1)} = \frac{A}{x + 3} + \frac{B}{x + 1}

The equation above needs to be identically true -- true for all reasonable values of x (the ones for which the denominators aren't zero).

Multiply both sides of the equation above by (x + 3)(x + 1) and solve for A and B.

 

[adelaide87]

This one is a bit trickier

1/[(u^2)(u-1)(u+1)]


to get


(1/2)/(u-1) - 1/(u^2) - (1/2)/(u+1)



I tried doing it they way you just showed me but I get stuck.

 

[Mark44]

Here you have a repeated factor, so the decomposition has to look like this:
\frac{1}{u^2(u - 1)(u + 1)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{u - 1} + \frac{D}{u + 1}

As before, multiply both sides by u²(u - 1)(u + 1) and solve for A, B, C, and D.

There should be some examples of decomposition in your book...