The Factor Theorem states that a linear divisor $ax+b$ is a factor of a polynomial $P(x)$ if and only if $P(\frac{-b}{a})=0$.
As is the case with numbers, the division of a polynomial by any of its factors always produces a remainder of $0$ (often phrased as “leaving no remainder”).
Thus, the Factor Theorem is just a natural consequence of the Remainder Theorem being applied with the fact that division by factors produces no remainder. Once we know how to apply Remainder Theorem, we just have to infer that the result of the substitution, $P(\frac{-b}{a})$, will be $0$.
Example 1
Determine if the linear divisor $x+3$ is a factor of the polynomial $x^2+5x+6$.
Solution As with Remainder Theorem, we find the $x$ value that must be substituted into the polynomial: $$ \begin{equation}\begin{aligned} x+3&=0\\ \therefore x&=-3\\ \end{aligned}\end{equation} $$
Then we substitute into the polynomial to see if the result is $0$: $$ \begin{equation}\begin{aligned} P(-3)&=(-3)^2+5(-3)+6\\ &=9-15+6\\ &=0\\ \end{aligned}\end{equation} $$
According to the Factor Theorem, the substitution $P(-3)$ producing $0$ means that the linear divisor is a factor of $P(x)=x^2+5x+6$.
Example 2
Show that the polynomial $3x+1$ is a factor of $3x^2-2x-1$. First we find the $x$ value to make the divisor $0$: $$ \begin{equation}\begin{aligned} 3x+1&=0\\ 3x&=-1\\ \therefore x&=\frac{-1}{3}\\ \end{aligned}\end{equation} $$
Then we substitute: $$ \begin{equation}\begin{aligned} P(\frac{-1}{3})&=3(\frac{-1}{3})^2-2(\frac{-1}{3})-1\\ &=3(\frac{1}{9})-2(\frac{-1}{3})-1\\ &=\frac{1}{3}+\frac{2}{3}-1\\ &=0\\ \end{aligned}\end{equation} $$
Because the substitution produces a result of $0$, by the Factor Theorem the linear divisor $3x+1$ is a factor of the polynomial $3x^2-2x-1$.
Finding unknown coefficients
We can use the Factor Theorem in order to evaluate the coefficients in a polynomial, given a divisor. It should be noted that for a polynomial to be divisible (can be divided) by divisor, it is by extension divisible by the factors of that divisor.
Consider the number $24$. It is divisible by $8$. It is also, by extension, divisible by every factor of $8$ ($1$, $2$, $4$, and $8$ itself).
Example 1
If $px^3+qx^2-19x+30$ is divisible by $x^2-5x+6$, determine the values of the constants $p$ and $q$.
Solution First, we acknowledge that the Factor Theorem only applies to linear divisors (in the form $ax+b$) so we cannot use the divisor $x^2-5x+6$ directly. However, because the polynomial is divisible by both the divisor and its factors, we can apply the Factor Theorem to the linear factors of $x^2-5x+6$. Hence, we must first factorize the divisor: $$ \begin{equation}\begin{aligned} &x^2-5x+6\\ &=x^2-3x-2x+6\\ &=(x^2-3x)-(2x-6)\\ &=x(x-3)-2(x-3)\\ &=(x-3)(x-2)\\ \end{aligned}\end{equation} $$
We can safely say that since $px^3+qx^2-19x+30$ is divisible by $x^2-5x+6$, it is also divisible by $x-3$ AND $x-2$ (the linear factors of $x^2-5x+6$). Using the first linear factor, $x-3$, we apply the Factor Theorem: $$ \begin{equation}\begin{aligned} x-3&=0\\ \therefore x&=3\\ P(3)&=p(3)^3+q(3)^2-19(3)+30\\ &=27p+9q-57+30\\ 0&=27p+9q-27\\ \end{aligned}\end{equation} $$
Note that last line where there is a $0$. This is because the Factor Theorem mandates that a linear factor must produce a result of $P(\frac{-b}{a})=0$. An equation can be formed: $$ \begin{equation}\begin{aligned} 27p+9q&=27\\ \therefore 3p+q=3\\ \end{aligned}\end{equation} $$
We then apply the same to the other linear factor, $x-2$: $$ \begin{equation}\begin{aligned} x-2&=0\\ \therefore x&=2\\ P(2)&=p(2)^3+q(2)^2-19(2)+30\\ &=8p+4q-38+30\\ 0&=8p+4q-8\\ \end{aligned}\end{equation} $$
This can be simplified into: $$ \begin{equation}\begin{aligned} 8p+4q&=8\\ \therefore 2p+q&=2\\ \end{aligned}\end{equation} $$
We can then solve these two equations simultaneously: $$ \begin{equation}\begin{aligned} 3p+q&=3\\ 2p+q&=2\\ \therefore p&=1\\ \end{aligned}\end{equation} $$
Substituting $p$ as $1$, we get: $$ \begin{equation}\begin{aligned} 3(1)+q&=3\\ \therefore q&=3-3\\ &=0\\ \end{aligned}\end{equation} $$
Therefore, we found $p$ to be $1$ and $q$ to be $0$, using the Factor Theorem and some properties of factors.
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