Utilize your discriminant! 2

Since $x-b$ is a factor of $P(x)$ , we can say that $P(b) = 0$ .

$P(b) = ab^{3} + ab^{2} + ab + b = 0$

Since $b$ is non-zero, we can divide both sides by $b$ (and make our lives much easier as well):

$ab^{2} + ab + a + 1 = 0$

One could notice that this is a quadratic in terms of $b$ . For the conditions to be true (i.e. $x-b$ is a factor of $P(x)$ ), then the values of $a$ must have real solutions in $b$ . Therefore we use $B^{2} - 4AC ≥ 0$

$a^{2} - 4 \times a \times (a+1) ≥ 0$

$-3a^{2} - 4a ≥ 0$

$3a^{2} + 4a ≤ 0$

$a(3a+4) ≤ 0$

Critical values are $0$ and $-\frac {4} {3}$ .

It follows that $-\frac {4} {3} ≤ a ≤ 0$ . You could verify this by sketching a graph or by using other methods.