The question has a linear factor in the denominator to the power of 3. I have only encountered 2 before. Have I started this question correctly? If I compare coefficients of x cubed the I find that A=2.

Printable View

- Sep 14th 2011, 08:07 AMStuck ManPartial fractions
The question has a linear factor in the denominator to the power of 3. I have only encountered 2 before. Have I started this question correctly? If I compare coefficients of x cubed the I find that A=2.

- Sep 14th 2011, 08:22 AMpiscoauRe: Partial fractions
the method used in your picture is not correct, when you take x=-3, the denominator will become 0.

So you better to use comparing the coefficient when you encounter this kind of problem. - Sep 14th 2011, 08:42 AMProve ItRe: Partial fractions
- Sep 14th 2011, 11:21 AMHallsofIvyRe: Partial fractions
No, what he is doing is perfectly valid. (Except, of course, for the missing $\displaystyle \frac{1}{x+1}$ term.) Yes, the original form has a denominator of x+ 3 but after multiplying through by the least common denominator, he has a polynomial that is equal to that for all x other than those values that make the denominator 0. And since a polynomial is

**continuous**, the value of the polynomial**at**3 gives the correct value for the rational function as well. - Sep 14th 2011, 08:44 PMProve ItRe: Partial fractions