alg2/geo hw

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

    Hope there is a better answer....

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.