Polynomials in an arbitrary field $F$ Let $p(t)$ be a polynomial of degree $n+1$ of an arbitrary field $F$. Suppose that $p(t)$ is a product of linear terms. Let $\lambda$ be one of its roots. So we can represent $p(t)$, up to a multiplicative constant as:
$$p(t) = (t-\lambda)\cdot (t-r_1)\cdot \;\dots \;\cdot(t-r_n)$$
Finally, let $d(t)$ be another polynomial such that:
$$
p(t) = (t-\lambda)d(t)
$$
Is it true that 
$$
d(t) = (t-r_1)\cdot \; \dots \cdot (t-r_n)
$$
?
The subtlety here is that we can't divide by $t-\lambda$ when $t=\lambda$. In the usual case where $F = \mathbb{R}$ or $F = \mathbb{C}$, the field is infinite, so the relation 
$$
d(t) = p(t)(t-\lambda)^{-1} \;\; t\neq\lambda
$$
gives us an infinite number of equations which uniquely determine $d(t)$. So my question here is concerned with the case of a finite field. 
 A: This is true. In fact, it’s equivalent to the Euclidean algorithm for long division. Let $h(t)$ be that product. Then $p(t) - p(t) = (t-\lambda)(d(t)-h(t))$. Since the degree is additive under product of nonzero polynomials, if $d(t) - h(t)$ is not identically $0$, it has negative degree! So $d(t) \equiv h(t)$.
Also, this holds for any arbitrary field. And we’re good to go.
A: 
The subtlety here is that we can't divide by $t−λ$ when $t=λ$.

That's not relevant. Polynomials are not functions and $t$ isn't equal to anything but $t$. $t$ is an indeterminate, not a variable.
A polynomial is equal to zero if and only if all of its coefficients are zero. This means that $F[t]$ is always an integral domain (in fact a PID). Being a domain, all non-zero elements, such as $t - \lambda$ are non-zero-divisors and the usual rule applies. That is if $ax = ay$ and $a$ is not a zero-divisor then $x = y$. Then $ax = ay$ and hence $d(t) = \prod (t - r_i)$.
If you take $a = t - \lambda$ and $x = d(t)$ and $y = \prod (t - r_i)$
