# Thread: Finding the ordered Triplet

1. ## Finding the ordered Triplet

Find the ordered triplet (A,B,C) if:

(6x^2 - 8x -20) / [(x+2)(x-3)(x-2)] = (A/(x+2)) + (B/(x-3)) + (C/(x-2))

**hint: mulitply both sides by (x+2)(x-3)(x-2)

2. Originally Posted by Mr_Green
Find the ordered triplet (A,B,C) if:

(6x^2 - 8x -20) / [(x+2)(x-3)(x-2)] = (A/(x+2)) + (B/(x-3)) + (C/(x-2))

**hint: mulitply both sides by (x+2)(x-3)(x-2)
Multiply through and you get,

$6x^2-8x-20=A(x-3)(x-2)+B(x+2)(x-2)+C(x+2)(x-3)$
Now this is a partial fraction decomposition which can always be achived. Thus this is true for all $x$.

If $x=2$
$6(2)^2-8(2)-20=A(2-3)(0)+B(2+2)(0)+C(2)(-1)$ you can solve for $C$

If $x=3$
$6(3)^3-8(3)-20=A(0)(3-2)+B(3+2)(3-2)+C(5)(0)$ you can solve for $B$

If $x=-2$
$6(-2)^2-8(-2)-20=A(-5)(-4)+B(0)(-2-2)+C(0)(-5)$ you can solve for $A$

3. ## what i got.

i got

A = 1

B = 23.6

C = 6

Could someone check that i got this right?

4. Nvm, i got the answer and checked it...

A = 1

B = 2

C = 3

thanks everyone!

5. Originally Posted by Mr_Green
i got
A = 1
B = 23.6
C = 6
Could someone check that i got this right?
Hello, Mr Green,

In

If x=3
$6(3)^3-8(3)-20=A(0)(3-2)+B(3+2)(3-2)+C(5)(0)$ you can solve for B
the exponent of the first 3 should be 2

Then you'll get: B = 2

EB

thanks