Remainder theorem:
If the polynomial $P(x)$ is divided by $(x-a)$, then the remainder is $P(a)$.
Factor theorem:
If $P(a)=0$, then $(x-a)$ is a factor of $P(x)$
I am wondering if they are equivalent – if both imply the other, or one is a consequence of the other.
My intuition is that they are equivalent, but I am not sure how to prove. It seems clear that the remainder theorem implies the factor theorem (by definition of factor). I am not sure about the other direction.
Am I on the right track? Thank you.
Best Answer
Yes, both are equivalent to the Division algorithm with linear divisor $\,x\!-\!a$. Below, if rings are unfamiliar, let $R[x] = $ polynomials whose coefficients are complex (or real or rational). As for integers, $\, f\bmod g :=\,$ remainder left in $f\div g = \,$ polynomial (long) division with remainder.
Theorem $\ $ TFAE for a polynomial $\,f\in R[x],\,$ and $\,a\in R\,$ a commutative ring.
$(0)\ \ \ f = (x\!-\!a)q + r\ $ for some $\,q\in R[x],\ r\in R\ \ \ $ [Monic Linear Division Algorithm]
$(1)\ \ \ f\bmod x\!-\!a = f(a)\ \ \ \ \ \ \ $ [Remainder Theorem]
$(2)\ \ \ f(a) = 0\,\Rightarrow\, x\!-\!a\mid f\ \ \ $ [Factor Theorem]
Proof $\ (0\Rightarrow 1)\ \ \ f = (x\!-\!a)q + r\,\overset{\large x\,=\,a}\Longrightarrow\, r=f(a)\,\Rightarrow\,f\bmod x\!-\!a = r = f(a) $
$(1\Rightarrow 2)\ \ \ f\bmod x\!-\!a = f(a) = 0\,\Rightarrow\, x\!-\!a\mid f$
$(2\Rightarrow 0)\ \ \ g := f-f(a)\,$ has $\,g(a) = 0\ $ so $\ g = f-f(a) = (x\!-\!a)q$
Remark $ $(Euclidean) polynomial division with (smaller) remainder is always achievable for divisors that are monic (lead coef $\color{#c00}{= 1}$ or a $\rm\color{#c00}{unit}$ = invertible), such as $\,\color{#c00}1x-a\,$ above, see here.