Question

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

Answer 1

`(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)`
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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`

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