Answer:
[tex]f(x) = -1[/tex]
[tex]g(x) =6[/tex]
[tex]\frac{-1}{2x + 3} + \frac{6}{3x + 7}[/tex]
Step-by-step explanation:
The question is unreadable, however the real polynomial is:
The polynomial fraction is:
[tex]P = \frac{9x + 11}{6x^2 + 23x + 21}[/tex]
And the decomposition is:
[tex]\frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}[/tex]
The solution is as follows:
[tex]P = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}[/tex]
Substitute the expression for P
[tex]\frac{9x + 11}{6x^2 + 23x + 21} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}[/tex]
Expand the numerator of the polynomial
[tex]\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)}{2x + 3} + \frac{g(x)}{3x + 7}[/tex]
Take LCM
[tex]\frac{9x + 11}{(2x + 3)(3x + 7)} = \frac{f(x)*(3x + 7) + g(x)*(2x + 3)}{(2x + 3)(x + 7)}[/tex]
Cancel out both denominators
[tex]9x + 11} = f(x)*(3x + 7) + g(x)*(2x + 3)[/tex]
Represent f(x) as A and g(x) as B
[tex]9x + 11} = A*(3x + 7) + B*(2x + 3)[/tex]
Open bracket
[tex]9x + 11} = 3Ax + 7A + 2Bx + 3B[/tex]
[tex]9x + 11} = 3Ax + 2Bx+ 7A + 3B[/tex]
[tex]9x + 11} = (3A + 2B)x+ 7A + 3B[/tex]
By comparison:
[tex]3A + 2B = 9[/tex] ---- (1)
[tex]7A + 3B =11[/tex] ---- (2)
Make B the subject in (1)
[tex]B = \frac{9 - 3A}{2}[/tex]
Substitute [tex]B = \frac{9 - 3A}{2}[/tex] in (2)
[tex]7A + 3(\frac{9 - 3A}{2}) = 11[/tex]
Multiply through by 2
[tex]2*7A +2* 3(\frac{9 - 3A}{2}) = 11*2[/tex]
[tex]14A + 3(9 - 3A) = 22[/tex]
[tex]14A + 27 - 9A = 22[/tex]
Collect Like Terms
[tex]14A - 9A = 22-27[/tex]
[tex]5A = -5[/tex]
[tex]A = \frac{-5}{5}[/tex]
[tex]A = -1[/tex]
Recall that:
[tex]B = \frac{9 - 3A}{2}[/tex]
[tex]B = \frac{9 - 3 * -1}{2}[/tex]
[tex]B = \frac{9 + 3}{2}[/tex]
[tex]B = \frac{12}{2}[/tex]
[tex]B = 6[/tex]
A = -1 and B = 6
[tex]3A + 2B = 9[/tex] ---- (1)
[tex]7A + 3B =11[/tex] ---- (2)
So:
[tex]f(x) = A[/tex]
[tex]f(x) = -1[/tex]
[tex]g(x) =B[/tex]
[tex]g(x) =6[/tex]
And the decomposition is:
[tex]\frac{-1}{2x + 3} + \frac{6}{3x + 7}[/tex]
