What's the easiest way to factor this cubic term?

Question

A while ago a student has contacted me to explain some exercises in factoring to him. I have since lost contact but one of the exercises I still have and I cannot figure out how you are supposed to do it without looking for zeroes.

$$a^3 + 6a^2b + 11ab^2 + 6b^3$$

$11$ being $5 + 6$ seems kind of suspicious but splitting up the terms in the middle to pull out any of the three linear factors via grouping feels too complicated to me. Is there an easier way?

By the way, the solution is \begin{align} a^3 + 6a^2b + 11ab^2 + 6b^3 = (a + b)(a + 2b)(a + 3b) \end{align}

Answer 1

To combine the first two comments:

The polynomial is homogeneous in degree $3$ so if $x=a/b$ then you are trying to factor $$b^3(x^3+6x^2+11x+6)$$

For the bracketed expression, you might either

• spot that the odd and even coefficients sum to the same value with $1+11=6+6$ so $(x+1)$ is a factor, i.e. $(a+b)$ is a factor of the original expression, or
• hope there are rational factors, which by the rational root theorem would have to be of the form $(x\pm1),(x\pm2),(x\pm3),(x\pm6)$, and testing in this case shows three of these work