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# algebra

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

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

Feb 6, 2021

#1
+168
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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 $$\boxed{-1}$$

Feb 6, 2021
edited by textot  Feb 6, 2021