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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!

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

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    $\begingroup$ Tom, the given fraction has a sum of two squares in denominator, thus cannot be factored. $\endgroup$ – user376343 Nov 8 '18 at 23:32
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    $\begingroup$ @user376343 Did i miss the point of the question or that is just a remark? $\endgroup$ – Thom Nov 8 '18 at 23:35
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    $\begingroup$ It is just a remark. It is good to see things in their compact form (moreover if this form is given). $\endgroup$ – user376343 Nov 8 '18 at 23:39
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In case the purpose is to find an antiderivative

$$\frac{3x+7}{x^2-8x+41} = \frac{3x-12}{x^2-8x+41} + \frac{19}{x^2-8x+41} = \left( \frac{3}{2} \right) \left( \frac{2x-8}{x^2-8x+41} \right) + \left( \frac{19}{(x-4)^2+25} \right) $$

For the 19 part, we would expect to use a substitution $$ x-4 = 5 u $$

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