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

##### 1 Answer

#### 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:

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

Now, divide the polynomial by

#(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

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

So the final answer is