Remainder-Factor Theorem

Remainder Theorem: For a polynomial $f(x)$, the remainder of $f(x)$ when dividing by $x-c$ is $f(c)$.

Proof: Dividing $f(x)$ by $x-c$, we obtain

$f(x)=(x-c)q(x)+r(x),$

where $r(x)$ is the remainder. Since $x-c$ has degree 1, it follows that the remainder $r(x)$ has degree 0, thus is a constant. Let $r(x)=R$. Substituting $x=c$, we obtain $f(c)=(c-c)q(c)+R=R$. Thus, $R=f(c)$ as claimed. $_\square$

Factor Theorem: Let $f(x)$ be a polynomial such that $f(c) =0$ for some constant $c$. Then $x-c$ is a factor of $f(x)$. Conversely, if $x-c$ is a factor of $f(x)$, then $f(c)=0$.

Proof:

(Forward direction) Since $f(c)=0$, by the Remainder Theorem, we have $f(x) = (x-c)h(x) + R = (x-c)h(x)$. Hence, $x-c$ is a factor of $f(x)$.

(Backward direction) If $x-c$ is a factor of $f(x)$, then (by definition) the remainder when $f(x)$ is divided by $x-c$ would be 0. By the Remainder Theorem, this is equal to $f(c)$. Hence, $f(c)=0$. $_\square$

*Note: *There are no restrictions on the constant $c$. It could be a real number, a complex number, or even a matrix!

The Remainder-Factor theorem is often used to help factorize polynomials without the use of long division. When combined with the Rational Roots theorem, this gives us a powerful factorization tool.

Technique

Completely factorize $f(x) = 6x^3 - 23x^2 - 6x+8$.

Solution: From the rational root theorem, we try numbers of the form $\frac {a}{b}$, where $a$ divides 8 and $b$ divides 6. By the Remainder-Factor Theorem, we just need to calculate the values for which $f \left( \frac {a}{b} \right)=0$. Through trial and error, we obtain

$\begin{aligned} f(1) & = 6-23 - 6 + 8 = -15 \\ f(2) & = 48 - 92 - 12 + 8 = -48 \\ f(4) & = 384 - 368 - 24 + 8 = 0 \\ f\left(\frac {1}{2}\right) & = \frac {3}{4} - 5 \frac {3}{4} - 3 + 8 = 0 \\ f \left(-\frac {1}{2} \right) & = -\frac {3}{4} - 5\frac {3}{4} + 3 + 8 = 4 \frac {1}{2} \\ f \left(-\frac {2}{3} \right) & = - 1 \frac {7}{9} + 10 \frac {2}{9} + 4 + 8 = 0\\ \end{aligned}$

This shows $x-4, x- \frac {1}{2},$ and $x+ \frac {2}{3}$ are factors of $f(x)$, implying

$f(x) = A(x) (x-4)\left(x- \frac {1}{2}\right) \left(x+\frac {2}{3}\right).$

Since the left hand side has degree 3, it follows that $A(x)$ has degree 0, so is a constant we denote by $A$. Substituting $x=0$, we obtain $A=6$. Therefore,

$f(x) = 6 (x-4)\left(x- \frac {1}{2}\right) \left(x+\frac {2}{3}\right) = (x-4)(2x-1)(3x+2). \, _\square$

Application and Extensions

$g(x)$ is a polynomial that leaves a remainder of 1 when divided by $x-1$ and leaves a remainder of 4 when divided by $x+2$. What is the remainder when $g(x)$ is divided by $(x-1)(x+2)$?

We have $g(x) = (x-1)(x+2)q(x) + r(x)$. Since $(x-1)(x+2)$ has degree 2, the remainder $r(x)$ has degree at most 1, hence $r(x)=Ax+B$ for some constants $A$ and $B$. By the remainder factor theorem, we have $g(1)=1$ and $g(-2)=4$. Substituting $x=1$ and $-2$, we obtain

$1 = g(1) = (1-1)(1+2)q(1) + r(1) = A (1) + B,$ $4 = g(-2) = (-2-1)(-2+2)q(-2)+r(-2) = A(-2)+B.$

This implies $1 = A + B, 4 = -2A + B$, which has solution $A=-1, B=2$. Thus, the remainder when $g(x)$ is divided by $(x-1)(x+2)$ is $-x+2$. $_\square$

In an attempt to discover a formula for the Fibonacci numbers, Alex finds a cubic polynomial $h(x)$ such that $h(1)=1$, $h(2)=1$, $h(3)=2$ and $h(4)=3$. What is the value of $h(5)$?

Solution: Consider the cubic polynomial $j(x)= h(x) - x +1$. Then $j(1) = 1, j(2) = 0, j(3) = 0$ and $j(4) = 0.$ By the Remainder-Factor Theorem, we have $j(x)=A(x) (x-2)(x-3)(x-4)$, where $A(x)$ is a polynomial. Since $j(x)$ is a cubic, it follows that $A(x)$ has degree 0, hence is a constant which we denote by $A$. Substituting $x=1$, we obtain

$1 = j(1) = A(1-2)(1-3)(1-4) \Rightarrow A = -\frac {1}{6}.$

Thus, $h(x) = j(x)+x-1 = -\frac {1}{6} (x-2)(x-3)(x-4)+x-1.$ Hence,

$h(5) = -\frac {1}{6} (5-2)(5-3)(5-4) + 5 - 1 = 3.$

Note: The closed form of the Fibonacci sequence is an exponential function. This cannot be approximated using a polynomial function for large values of $n$. $_\square$

Suppose $k(x)$ is a polynomial with integer coefficients and $c$ is an integer root of $k(x)$. Show that $\frac {k(x)}{x-c}$ is also a polynomial with integer coefficients.

At first glance, it seems impossible to work directly from the Factor Theorem's result that $k(c)=0$.

Solution: Let the degree of $k(x)$ be $n$, and let

$k(x) = k_n x^n + k_{n-1} x^{n-1} + \ldots + k_1 x + k_0,$

where $k_i$ are integers. By the remainder-factor theorem,

$0 = k(c) = k_n c^n + k_{n-1} c^{n-1} +\ldots + k_1 c + k_0.$

Using the fact that $x^i - c^i=(x-c)(x^{i-1} + x^{i-2}c + \ldots + xc^{i-2} + c^{i-1})$, we obtain

$\begin{aligned} k(x) & = k(x)- 0 = k(x)-k(c) \\ &= k_n (x^n-c^n) + k_{n-1} ( x^{n-1} - c^{n-1}) + \ldots + k_1 (x-c) + k_0 (1-1) \\ & = (x-c) Q(x). \end{aligned}$

Since $Q(x)$ is integer combinations of polynomials with integer coefficients, it follows that $Q(x)$ is also a polynomial with integer coefficients. $_\square$