Question

I need help in partial fraction!! With simple explaination would be nice!

Answer 1

`(x^4-7x^2+17x-10)/(x(x^2-3))`

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)


`(x^4-7x^2+17x-10)/(x(x^2-3))=x-(4x^2-17x+10)/(x(x^2-3))`

Now the second term can be decomposed into partial fractions.


`(4x^2-17x+10)/(x(x^2-3))=\frac{r_1}x+(r_2x+r_3)/(x^2-3)`

`(4x^2-17x+10)/(x(x^2-3))=(r_1(x^2-3)+x(r_2x+r_3))/(x(x^2-3))`

`4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)`

`4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1`

`\implies\begin{cases}r_1+r_2=4\\r_3=-17\\-3r_1=10\end{cases}\implies r_1=-\frac{10}3,r_2=\frac{22}3,r_3=-17=-\frac{51}3`

`\implies(4x^2-17x+10)/(x(x^2-3))=-(10)/(3x)+(22x-51)/(x^2-3)`

So


`(x^4-7x^2+17x-10)/(x(x^2-3))=x+(10)/(3x)-(22x-51)/(x^2-3)`