Trick to factor: $n^3 + 6n^2 + 9n + 4$

Question

Is there any way to tell if

$$n^3 + 6n^2 + 9n + 4$$

has a factored (condensed) form, and, if so, what that factored form is? If the answer is yes to both, may one explain the more general principle?

Answer 1

Notice that, for $f(n)=an^3+bn^2+cn^1+dn^0$, we have $$f(-1)=a-b+c-d=(a+c)-(b+d)$$

In other words, if the sum of the even co-efficients is equal to the sum of the odd co-efficients, $(x+1)$ can be taken out. Here, this holds

Then we have:

$$(n+1)(pn^2+qn+r)=n^3+6n^2+9n+4$$ Expanding out the LHS, we get: $$pn^3+(p+q)n^2+(q+r)n+r=n^3+6n^2+9n+4$$ Can you see that $p=1, q=5, r=4$ by matching terms?

So we have:

$$n^3+6n^2+9n+4=(n+1)(n^2+5n+4)$$

Factorise the second bracket and you're done.

Answer 2

$$ n^3+6n^2+9n+4 = (n^3+1)+(6n^2+6n)+(3n+3)$$ $$ = (n+1)(n^2-n+1)+6n(n+1)+3(n+1)=$$

$$ = (n+1)(n^2-n+1+6n+3)$$ $$ = (n+1)(n^2+5n+4) =$$ $$= (n+1)^2(n+4)$$

Answer 3

The general "principle" is called Polynomial factorization. We have $$ n^3+6n^2+9n+4=(n+1)^2(n+4). $$ There are many interesting algorithms. In this particular example, it is enough to do the rational root test. It immediately gives that a rational root must be of the form $\pm 1,\pm 4$, and then you just try. For a more demanding example, then, try to factorize $$ n^9 + 3n^7 + 3n^6 + 3n^5 + 6n^4 + 4n^3 + 3n^2 + 3n + 1. $$ The result is very short and nice.

Answer 4

Let's factor it!

$$\begin{align*}n^3+6n^2+9n+4 & =n^3+6n^2+8n+n+4\\ & =n(n^2+6n+8)+n+4\\ & =n(n+4)(n+2)+n+4\\ & =(n+4)(n^2+2n+1)\\ & =(n+1)^2(n+4)\end{align*}$$

Answer 5

$$ 1+9 = 10 $$ $$ 6+4 = 10 $$ Either $1$ or $-1$ is a root. I'm betting on $-1$

Answer 6

By inspection, $1+9=6+4$ so that $-1$ is a root. Now by long division,

$$\frac{n^3+6n^2+9n+4}{n+1}=n^2+5n+4$$ and you can use the quadratic formula.

Or by inspection, $-1$ is again a root and

$$\frac{n^2+5n+1}{n+1}=n+4.$$