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

(m^5 + 2m^3 - 9m + 27)/(m^2 + 1)

Guest Jul 22, 2021

#1**+1 **

Here you go guest:

if m^{2} + 1 = 3m

we can simplify like this:

(m^{5}+2m^{3}-9m+27)/3m

= (m^{3}(m^{2}+2)-9m+27)/3m

= (m^{3}(3m+1)-9m+27)/3m

= (3m^{4}+(m^{3}-9m)+27)/3m

= (3m^{4}+m(m^{2}-9)+27)/3m

= (3m^{4}+m(3m+10)+27)/3m

= (3m^{4}+3m^{2}+10m+27)/3m

= (3m^{2}(m^{2}+1)+10m+27)/3m

= (3m^{2}(3m)+10m+27)/3m

= (9m^{3}+10m+27)/3m

= (m(9m^{2}+10)+27)/3m

= (m(27m+1)+27)/3m

= (27m^{2}+m+27)/3m

= (27(m^{2}+1)+m)/3m

= (27(3m)+m)/3m

= (81m+m)/3m

=(82m)/(3m)

__= 82/3__

__= 27 1/3__

__P.S. I probably made a mistake somewhere, but this is the general idea of how to do it!__

SWest Jul 23, 2021