1. ## hELP partial fractions....

Resolve into partial...
4.) (3x^3 + 8x^2 + 6x - 2) / (x+2)(x+1)(x^2+2)
And the rule
= A / (x +2) + B ( x+1) + (Cx + D)/(x^2 + 2)

the only part i cant seem to find is by solving A,B,C and D
3 = A + B + C
8 = A + 2B + 3C + D
6 = 2A + 2B + 2C + 3D
-2 = 2A + 4B + 2D
can u check if thiss correct?

Resolve into partial...
4.) (3x^3 + 8x^2 + 6x - 2) / (x+2)(x+1)(x^2+2)
And the rule
= A / (x +2) + B ( x+1) + (Cx + D)/(x^2 + 2)

the only part i cant seem to find is by solving A,B,C and D
3 = A + B + C
8 = A + 2B + 3C + D
6 = 2A + 2B + 2C + 3D
-2 = 2A + 4B + 2D
can u check if thiss correct?
These give the correct solution, so yes they are correct.

RonL

3. cud u help me answer please that is the part where i got stuck

Resolve into partial...
4.) (3x^3 + 8x^2 + 6x - 2) / (x+2)(x+1)(x^2+2)
And the rule
= A / (x +2) + B ( x+1) + (Cx + D)/(x^2 + 2)

the only part i cant seem to find is by solving A,B,C and D
3 = A + B + C
8 = A + 2B + 3C + D
6 = 2A + 2B + 2C + 3D
-2 = 2A + 4B + 2D
can u check if thiss correct?
Look at equations 1 and 3, one is twice the other except for an extra
3D, so D must be zero.

So the equations now become (after discarding the duplicate)

3 = A + B + C
8 = A + 2B + 3C
-2 = 2A + 4B

multiply equation 2 by 2 and we have:

16=2A+4B+6C

but from wuation 3 we have

-2=2A+4B,

so C=18/6=3.

so now we have:

A+B=0
2A+4B=-2

so 2B=-2, B=-1, A=1.

Now you will need to check that this is correct as my algebra comes with
no guarantees

RonL

RonL