If $m$ is a real number such that $m^2+1 = 3m$, find the value of the expression below.

\[\large \frac{2m^5-5m^4+2m^3-8m^2}{m^2+1}\]

Guest Feb 6, 2021

#1**+1 **

Try reducing the big expression as much as possible.

First, notice that the denominator can be simplified to 3m, since m^2+1=3m:

\(2m^5-5m^4+2m^3-8m^2\over3m\)

Divide the top and bottom by m:

\(2m^4-5m^3+2m^2-8m\over3\)

notice that the m^4 term and the m^2 term can be factored like this:

\(2m^2(m^2+1) -5m^3-8m\over3\)

The value in the parenthesis can be replaced with 3m:

\(\frac{2m^2(3m)-5m^3-8m} {3} \\ =\frac{6m^3-5m^3-8m}{3}\\=\frac{m^3-8m}{3}\)

Add then subtract 9m from the numerator:

\(\frac{m^3-8m+9m-9m}{3}\\ = \frac{m^3+m-9m}{3}\\ = \frac{m(m^2+1)-9m}{3}\\ = \frac{m(3m)-9m}{3}\\ = m^2-3m\)

Rearranging the equation, \(m^2-3m=-1\), so the answer to this question is \(\boxed{-1}\)

textot Feb 6, 2021