- #1

chwala

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- Homework Statement
- Find the values of ##A,B## and ##C## if the expression

##Ax(x-2)(x+3)+Bx(x-2)+Cx(x+3)+ (x-2)(x+3)##

has a constant value for all values of ##x##.

- Relevant Equations
- Partial fractions

My working follows here,

Let

##(x-2)[Ax(x+3)+Bx] + (x+3)[Cx +(x-2)]=0##

##(x-2)(x+3) \left[Ax+\dfrac{Bx}{(x+3)}\right]+(x+3)(x-2)\left[\dfrac{Cx}{(x-2)} +1\right]##=0

##⇒Ax+\dfrac{Bx}{(x+3)} = -\left[\dfrac{Cx}{(x-2)} +1\right]##

##Ax(x+3)(x-2)+Bx(x-2) = -[Cx(x+3)+(x-2)(x+3)]##

On comparing coefficients we end up with;

##A=0##

...and the simultaneous equation;

##B+C=-1##

##-2B+3C=-1##

Giving me

##B=-\dfrac{2}{5}## and ##C=-\dfrac{3}{5}##

There could be a better approach...And also your insight on my steps are welcome. Cheers!

Let

##(x-2)[Ax(x+3)+Bx] + (x+3)[Cx +(x-2)]=0##

##(x-2)(x+3) \left[Ax+\dfrac{Bx}{(x+3)}\right]+(x+3)(x-2)\left[\dfrac{Cx}{(x-2)} +1\right]##=0

##⇒Ax+\dfrac{Bx}{(x+3)} = -\left[\dfrac{Cx}{(x-2)} +1\right]##

##Ax(x+3)(x-2)+Bx(x-2) = -[Cx(x+3)+(x-2)(x+3)]##

On comparing coefficients we end up with;

##A=0##

...and the simultaneous equation;

##B+C=-1##

##-2B+3C=-1##

Giving me

##B=-\dfrac{2}{5}## and ##C=-\dfrac{3}{5}##

There could be a better approach...And also your insight on my steps are welcome. Cheers!

Last edited: