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how do you solve for X when you have indecision

x^3 +3x^2+ 3x+1=0  

SydSu22  Nov 6, 2018
 #1
avatar+13718 
+2

You MIGHT recognize it as a perfect cube:  (x+1)^3        therefore x = -1

    Hope there is a better answer....

ElectricPavlov  Nov 7, 2018
 #4
avatar+67 
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Thank You! The concept of a perfect cube is very new to me so this actually clears up a lot of questions!

SydSu22  Nov 7, 2018
 #2
avatar+2793 
+2

this is sort of cheating but the way I do it is recognizing that the coefficients are a line of Pascal's Triangle 

and thus they are binary coefficients.

 

That means that

 

\(x^3 + 3x^2 + 3x + 1 = (x+1)^3\)

 

I doubt they expect you to know that so you'll have to use the rational root theorem.

Given the coefficients of \(x^3 \text{ and }1\) we know the only possible rational roots are \(\pm 1\)

 

\(f(1) = 1 + 3 + 3 + 1 = 6 \neq 0\\ f(-1) = -1 + 3 -3 + 1 = 0 \\ \text{so -1 is a root}\)

 

\(\text{so }x+1 \text{ is a factor we then divide }\\ \dfrac{x^3 + 3x^2 + 3x + 1}{x+1} = x^2 + 2x + 1\\ \text{The result has the same possible two roots so we check as before}\\ \left . x^2 + 2x + 1 \right |_{x=1} = 4 \neq 0\\ \left . x^2 + 2x + 1 \right |_{x=-1} = 0\\ \text{again }(x+1) \text{ is a factor we again we divide}\\ \dfrac{x^2+2x+1}{x+1}=x+1 \\ \text{and we see all 3 factors are }x+1 \text{ i.e.}\\ f(x) = (x+1)^3\)

 

And thus there is a single root, x=-1, that is repeated 3 times.

Rom  Nov 7, 2018
edited by Rom  Nov 7, 2018
 #3
avatar+67 
0

Thanks so much for the help! and you are right I have no idea what you did at first.

SydSu22  Nov 7, 2018

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