I have this partial fraction: $${3x+7}\over{(x-4)^2+25}$$ As far as I can tell, I do not think this can be decomposed. Is that a correct assumption?

Sorry for the very short question, there isn't much work I could show, I think.

Thank you!

• Yes, in $\mathbb{R}$ is this fraction irreducible. – user376343 Nov 8 '18 at 23:29

We have: $$\dfrac{3x+7}{(x-4)^2+25}=\dfrac{3x+7}{x^2-8x+16+25}=\dfrac{3x+7}{x^2-8x+41}.$$ Since discriminant of $$x^2-8x+41$$ is negative denominator can not be decomposed into product of two (real) linear expresions. Therfore you can not simplify original expression meaningfully.
$$\frac{3x+7}{x^2-8x+41} = \frac{3x-12}{x^2-8x+41} + \frac{19}{x^2-8x+41} = \left( \frac{3}{2} \right) \left( \frac{2x-8}{x^2-8x+41} \right) + \left( \frac{19}{(x-4)^2+25} \right)$$
For the 19 part, we would expect to use a substitution $$x-4 = 5 u$$