#1**+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

#2**+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