How do you determine the binomial factors of x^3+3x^2+3x+1?

1 Answer
May 28, 2017

(x+1)^3

Explanation:

Using the rational root theorem gives:

p = 1, q=1

All values of +-p/q are -1 and 1.

Now, plug in each value to the polynomial. By the remainder theorem, if the output is zero, the input must be a root.

(-1)^3+3(-1)^2+3(-1)+1=-1+3-3+1 = 0
therefore there is a root at x = -1

(1)^3+3(1)^2+3(1)+1 = 1+3+3+1=8
therefore there is NOT a root at x=1

Factors of the polynomial: (x - "root") = (x-(-1))=(x+1)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Now, divide the polynomial by (x+1):

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

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

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

= x^2+2x+1

You could do the same process with this new polynomial, or you could recognize it as a perfect square.

Either way, factoring this polynomial results in (x+1)^2

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So the final answer is (x+1)(x+1)^2 = (x+1)^3