# Cubic factoring question

I'm trying to figure out how a colleague factored an expression. I don't get how:

$$a^3+a^2b-(b+1)=(a-1)[a^2+a(b+1)+(b+1)]$$

Multiplying the result I see it's true, but not sure how he got there..is this a general rule for cubic expressions of this form?

Thanks,

P

• This question is quite broad. The statement is true, but we can't find how exactly your colleague obtained it - you can ask them, of course. However, if we want to exhibit some particular properties of the left-hand side expression, then we can follow some ideas. For example, it's easy to see that $a=1$ is a root of the polynomial, hence it's logical to factor $(a-1)$ out. – TZakrevskiy May 18 '14 at 14:35
• The "$\Longrightarrow$" should be "=" really. – user88595 May 18 '14 at 14:37
• math.stackexchange.com/questions/800320/cubic-trig-equation – lab bhattacharjee May 18 '14 at 14:39
• If we look at the cubic, it is clear that it is $0$ at $a=1$. So $a-1$ divides the cubic. Now do the division as usual. – André Nicolas May 18 '14 at 14:49

I'd do it like this, since it is linear in $b$:$$a^3+a^2b-(b+1)=(a^2-1)b+(a^3-1) = (a-1)\left ((a+1)b+a^2+a+1\right) =$$ if you want the inner bracket expressed as a polynomial in $a$ $$(a-1)\left(a^2+(b+1)a+(b+1)\right)$$
Note that though the common factor here is obvious, in a more complex case $p(a)b+q(a)$ it would be possible to use the Euclidean algorithm to find any common factors between $p(a)$ and $q(a)$.
Alternatively note that $a=1$ is a root of the original expression and use polynomial division to extract the factor $a-1$.