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 and1 .
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 atx = -1
(1)^3+3(1)^2+3(1)+1 = 1+3+3+1=8
therefore there is NOT a root atx=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