Question

partial fraction decomposition 5x^2-9x+19 over (x-4)(x^2+5) can someone help?

Answer 1

I can :)
You have a linear factor (x-4) and an irreducible quadratic (x^2+5)
So you should look for something in the form:
\[\frac{5x^2-9x+19}{(x-4)(x^2+5)}=\frac{A}{(x-4)}+\frac{Bx+C}{(x^2+5)}\]
Get a common denominator:
\[\frac{A(x^2+5)+(x-4)(Bx+C)}{(x-4)(x^2+5)}\]
From here set the numerators equal.
\[5x^2-9x+19=Ax^2+5A+Bx^2+Cx-4Bx-4C\]
From here you know that:
A+B=5; 5A-4C=19; C-4B=-9
Solving you get that:
A=3,B=2, C=-1
So you get:\[\frac{3}{(x-4)}+\frac{2x-1}{(x^2+5)}\]

Answer 2

Sorry it took so long to type up.