SOLUTION: Help with factoring I need to factor and then solve for x (x+1)(2x+1)^3 -3(2x+1)^2 (x+1)=0 I know that factored it is: 2(x+1)(2x+1)(2x^2+2x-1) And solved for x it is: x=-1

Algebra ->  Expressions-with-variables -> SOLUTION: Help with factoring I need to factor and then solve for x (x+1)(2x+1)^3 -3(2x+1)^2 (x+1)=0 I know that factored it is: 2(x+1)(2x+1)(2x^2+2x-1) And solved for x it is: x=-1      Log On


   

Question 775977: Help with factoring I need to factor and then solve for x
(x+1)(2x+1)^3 -3(2x+1)^2 (x+1)=0
I know that factored it is:
2(x+1)(2x+1)(2x^2+2x-1)
And solved for x it is:
x=-1, x= -.5, x=(-sqrt(3)-1)/2)(about -1.366), x= (sqrt(3)-1)/2)(about .366)
But no matter how I work it I am unable to end up where I am supposed to be, please mkae steps as clear as possible, thank you.
Found 2 solutions by MathLover1, stanbon:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

%28x%2B1%29%282x%2B1%29%5E3+-3%282x%2B1%29%5E2+%28x%2B1%29=0..factor out %28x%2B1%29%282x%2B1%29%5E2
%28x%2B1%29%282x%2B1%29%5E2%28%282x%2B1%29+-3%29=0
%28x%2B1%29%282x%2B1%29%5E2%282x%2B1+-3%29=0
%28x%2B1%29%282x%2B1%29%5E2%282x-2%29=0
%28x%2B1%29%282x%2B1%29%5E2%2A2%28x-1%29=0
2%28x%2B1%29%282x%2B1%29%282x%2B1%29%28x-1%29=0..use zero product rule to find solutions
if %28x%2B1%29=0 => x=-1
if %282x%2B1%29=0=> 2x=-1=>x=-1%2F2...this solution doubles
if %28x-1%29=0=> x=1
so, here you have three possible real solutions



Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Help with factoring I need to factor and then solve for x
(x+1)(2x+1)^3 -3(2x+1)^2 (x+1)=0
Factored it is:
(x+1)(2x+1)^2[2x+1-3] = 0
2(x+1)(2x+1)^2*(x-1) = 0
And solved for x it is:
x= -1, x= +1 , x = -1/2 with multiplicity two
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Cheers,
Stan H.
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