Question

(2r+1)^6-16(2r+1)^4=0

Answer 1

\[(2r+1)^{4}[(2r+1)^{2}-(4)^{2}]=0\]
\[(2r+1)^{4}(2r+1-4)(2r+1+4)=0\]
\[(2r+1)^{4}(2r-3)(2r+5)=0\]

Answer 2

either (2r+1)^4=0 which gives r=-(1/2) with multiplicity 4
or (2r-3)=0 which gives r=3/2
or (2r+5)=0 which gives r=-(5/2)

Answer 3

ok, then go back and check solutions in the equation

Answer 4

Hello, why is that nitz's job? Maybe you should check the solutions, given that you didn't have to solve the problem!

\[(2r+1)^6-16(2r+1)^4 = (2(-1/2)+1)^6 - 16(2(-1/2)+1)^4 = 0^6-16(0)^4 = 0\]
\[(2(3/2)+1)^6 -16(2(3/2)+1)^4 = 4^6-16(4)^4 = 4^6-4*4*4^4 = 0\]
\[(2(-5/2)+1)^6 - 16(2(-5/2)+1)^4 = (-4)^6 - 16(-4)^4 = 0\]