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# 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{align} 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{align}

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

Note by Arron Kau
3 years, 6 months ago

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