Mathematics

If $$\displaystyle \frac{3x^3-8x^2+10}{(x-1)^4} = \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3} + \frac{D}{(x-1)^4}$$, then $$A+B+C+D$$ is equal to


ANSWER

2


SOLUTION
$$\dfrac { 3x^{ 3 }-8x^{ 2 }+10 }{ (x-1)^{ 4 } } =\dfrac { A }{ x-1 } +\dfrac { B }{ (x-1)^{ 2 } } +\dfrac { C }{ (x-1)^{ 3 } } +\dfrac { D }{ (x-1)^{ 4 } } $$

Multiply both sides by $$(x-1)^{ 4 }$$

$$3x^{ 3 }-8x^{ 2 }+10=A(x-1)^{ 3 }+B(x-1)^{ 2 }+C(x-1)+D$$

Expand and collect in terms of power of $$x$$

$$3x^{ 3 }-8x^{ 2 }+10=Ax^{ 3 }+\left( B-3A \right) { x }^{ 2 }+\left( 3A-2B+C \right) x-A+B-C+D$$

Gives

$$A=3$$

$$ B-3A=-8\Rightarrow B=-8+3(3)=1$$

$$ 3A-2B+C=0\Rightarrow C=2B-3A=2-9=-7$$

$$ -A+B-C+D=10\Rightarrow D=10+C-B+A=10-7-1+3=5$$

$$\therefore A=3,B=1,C=-7,D=5$$

Hence $$A+B+C+D=2$$
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