$\displaystyle (9x^2) / [(x -1)^2 * (x + 2)] $
It's part of a set of partial fractions I have to do.. I can't seem to solve this?
Substitute values of x? do you know the method of partial fractions?
multiply through by the denominator of the left side, we get:
$\displaystyle 9x^2 = A(x - 1)(x + 2) + B(x + 2) + C(x - 1)^2$
Can you take it from here? You can multiply out and equate coefficients, or you can choose convenient values of x that wipes out two variables at a time
Thank you!
One last question...
$\displaystyle
\frac {17x^2+23x+12}{(3x+4)(x^2+4)} = \frac{A}{3x+4} + \frac{B}{x^2+4}
$
By cover-up rule, I got
$\displaystyle
A = \frac {17(-4/3)^2+23(-4/3)+12}{(-4/3)^2+4}
= 2
$
But since$\displaystyle (x^2 +4) $ isn't linear, does it mean I follow the steps you mentioned earlier (multiply through by denominator, etc)?
what is this "cover-up" rule you are referring to?
yes, but you have to change the numerator. The numerator must always be one degree less than the denominator. so for the part where the denominator is $\displaystyle x^2 + 4$ you must write:
But since$\displaystyle (x^2 +4) $ isn't linear, does it mean I follow the steps you mentioned earlier (multiply through by denominator, etc)?
$\displaystyle \frac {Bx + C}{x^2 + 4}$