Question

The partial fraction decomposition of 9 x 11 6 x 2 23 x 21 can be written in the form of f ( x ) 2 x 3 g ( x ) 3 x 7 , where

Answer 1

Answer:

$f(x) = -1$

$g(x) =6$

$\frac{-1}{2x + 3} + \frac{6}{3x + 7}$

Step-by-step explanation:

The question is unreadable, however the real polynomial is:

The polynomial fraction is:

$P = \frac{9x + 11}{6x^2 + 23x + 21}$

And the decomposition is:

$\frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

The solution is as follows:

$P = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Substitute the expression for P

$\frac{9x + 11}{6x^2 + 23x + 21} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Expand the numerator of the polynomial

$\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}$

Take LCM

$\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)*(3x + 7) + g(x)*(2x + 3)}{(2x + 3)(x + 7)}$

Cancel out both denominators

$9x + 11} = f(x)*(3x + 7) + g(x)*(2x + 3)$

Represent f(x) as A and g(x) as B

$9x + 11} = A*(3x + 7) + B*(2x + 3)$

Open bracket

$9x + 11} = 3Ax + 7A + 2Bx + 3B$

$9x + 11} = 3Ax + 2Bx+ 7A + 3B$

$9x + 11} = (3A + 2B)x+ 7A + 3B$

By comparison:

$3A + 2B = 9$ ---- (1)

$7A + 3B =11$ ---- (2)

Make B the subject in (1)

$B = \frac{9 - 3A}{2}$

Substitute $B = \frac{9 - 3A}{2}$ in (2)

$7A + 3(\frac{9 - 3A}{2}) = 11$

Multiply through by 2

$2*7A +2* 3(\frac{9 - 3A}{2}) = 11*2$

$14A + 3(9 - 3A) = 22$

$14A + 27 - 9A = 22$

Collect Like Terms

$14A - 9A = 22-27$

$5A = -5$

$A = \frac{-5}{5}$

$A = -1$

Recall that:

$B = \frac{9 - 3A}{2}$

$B = \frac{9 - 3 * -1}{2}$

$B = \frac{9 + 3}{2}$

$B = \frac{12}{2}$

$B = 6$

A = -1 and B = 6

$3A + 2B = 9$ ---- (1)

$7A + 3B =11$ ---- (2)

So:

$f(x) = A$

$f(x) = -1$

$g(x) =B$

$g(x) =6$

And the decomposition is:

$\frac{-1}{2x + 3} + \frac{6}{3x + 7}$