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How do you write the partial fraction decomposition of #(2x^3-x^2+x+5)/(x(x+1)^2)#?

Answer 1

Elias Ackerman

The answer is #=2+5/(x)-1/(x+1)^2-10/(x+1)#

Since the degree of the numerator is #=# to the degree of the denominator, we must perform a long division

#x(x+1)^2=x(x^2+2x+1)=x^3+2x^2+x#

Now, we perform the long division

#color(white)(aaaa)##2x^3-x^2+ x+5##color(white)(aaaa)##∣##x^3+2x^2+x#

#color(white)(aaaa)##2x^3+4x^2+2x##color(white)(aaaaa)##∣##2#

#color(white)(aaaaaa)##0-5x^2-x+5#

Therefore,

#(2x^3-x^2+ x+5)/(x(x+1)^2)=2+(-5x^2-x+5)/(x(x+1)^2)#

We can make the partial fraction decomposition

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

#=(A(x+1)^2+Bx+Cx(x+1))/((x(x+1)^2)#

So,

#-5x^2-x+5=A(x+1)^2+Bx+Cx(x+1)#

Let #x=0#, #=>#, #5=A#

Let #x=-1#, #=>#, #1=-B#, #=>#, #B=-1#

Coefficients of #x^2#, #=>#,#-5=A+C#, #=>#, #C=-10#

Finally we get

#(2x^3-x^2+ x+5)/(x(x+1)^2)=2+5/(x)-1/(x+1)^2-10/(x+1)#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.