Factoring a 5 term polynomial

Question

I am struggling to factor $n^4 + 4n^3 + 8n^2 + 8n +4$.

I have tried grouping the terms a couple of times, but got nowhere. What am I missing?

Answer 1

The first thing I would try are degree one factors, which by the Rational Roots Theorem must have the form $n + d$ where $d$ is an integer divisor of $4$.

Once these possibilities are exhausted, the only factors left to check are quadratic ones:

$$ n^4 + 4n^3 + 8n^2 + 8n + 4 = (n^2 + an + b)(n^2 + cn + d) $$

where $a,b,c,d$ are integers with the restrictions implied by collecting like powers of $n$. For example, $bd = 4$. There are not many of these to check.

This particular polynomial yields to a trick for finding square-free factors. One takes the derivative of the polynomial $4n^3 + 12n^2 + 16n + 8$, and computes the greatest common divisor of the derivative with the original:

$$ 4n^3 + 12n^2 + 16n + 8 = 4(n^3 + 3n^2 + 4n + 2) = 4(n+1)(n^2 + 2n + 2) $$

One way or the other, one finds the original polynomial factors as $(n^2 + 2n + 2)^2$.

Answer 2

This is an approach based on pattern recognition.

Recall the Pascal's triangle $$1\\ 1 \quad 1\\ 1 \quad 2 \quad 1\\ 1 \quad 3 \quad 3 \quad 1\\ 1 \quad 4 \quad 6 \quad 4 \quad 1$$ The coefficients of the polynomial at hand are $$1 \quad 4 \quad 8 \quad 8 \quad 4$$ They look "close" to $5^{th}$ row of above triangle. This suggest us to rewrite our polynomial as a sum $(n+1)^4$ plus some small pieces:

$$n^4 + 4n^3 + 8n^2 + 8n + 4 = (n+1)^4 + 2n^2 + 4n + 3$$

The coefficient of the small piece $$2\quad 4 \quad 3$$ now looks like twice the $3^{rd}$ row of above triangle. This suggest us to rewrite the small piece as a sum of $2(n+1)^2$ plus an even smaller piece.

$$2n^2 + 4n + 3 = 2(n+1)^2 + 1$$

Combine these, we get something that one can complete the square. Complete the square and we are basically done!

$$n^4 + 4n^3 + 8n^2 + 8n + 4 = (n+1)^4 + 2(n+1)^2 + 1\\ = ((n+1)^2 + 1)^2 = (n^2 + 2n + 2)^2 $$