0
$\begingroup$

Can someone correct this for me? I thought I got it right, but no. First I factored the bottom to get (x-1)^2 and then I put it into A and B form.

image

  • 0
    Notice that the degree in the numerator is $\geq$ the degree in the denominator. – 2017-02-11

1 Answers 1

2

Hint: Try doing long division on $$\frac{x^4-2x^3+x^2+9x-5}{x^2-2x+1}$$ Then you should get $$\frac{x^4-2x^3+x^2+9x-5}{x^2-2x+1}=x^2+\frac{\dots}{\dots}$$ Then you can decompose the unknown fraction into partial fractions.

  • 0
    Correct, the polynomial part is missing! – 2017-02-11
  • 0
    Ah, I did not realize that part would matter in this question. I thought all I had to put was the A+B+.... part, thanks! – 2017-02-11