# Another Partial fractions Q

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#1
*Sigh* .

a.) Find the values of A, B and C.

x(4x+5)/(x-1)(x+2)² = A/(x-1) + B/(x+2) + C/(x+2)²

I've done;
When x=1, A=1/3

I don't know what to do from here as if x=-2, it makes the RHS 0.

How do I find out the values of B&C?

b.) Show that the tangent to the curve y=f(x) at the point where x=-1 has the equation;

3x-4y+5=0  0
15 years ago
#2
so your deterministic function is the following:
x(4x+5)/(x-1)(x+2)² = A/(x-1) + B/(x+2) + C/(x+2)²

x(4x+5)/(x-1)(x+2)² =[A(X+2)^2+B(X-1)(X+2)+C(X-1)=X(4X+5)]/(x-1)(x+2)²

A(X+2)^2+B(X-1)(X+2)+C(X-1)=X(4X+5)

X=1 --> 9A=9 --> A=1
X=-2 --> -3C=6 --> C=-2

(X+2)^2+B(X-1)(X+2)-2(X-1)=X(4X+5)

X=0 --> 4-2B+2=0 --> B=3

So... A=1, B=3, C=-2
0
#3
Many thanks! 0
15 years ago
#4
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