Remainder-Factor Theorem

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

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

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

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


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


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

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

*Note: *There are no restrictions on the constant c 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.


Completely factorize f(x)=6x323x26x+8 f(x) = 6x^3 - 23x^2 - 6x+8.

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

f(1)=6236+8=15f(2)=489212+8=48f(4)=38436824+8=0f(12)=345343+8=0f(12)=34534+3+8=412f(23)=179+1029+4+8=0 \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 x4,x12, x-4, x- \frac {1}{2}, and x+23x+ \frac {2}{3} are factors of f(x) f(x), implying

f(x)=A(x)(x4)(x12)(x+23). 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) A(x) has degree 0, so is a constant we denote by A A. Substituting x=0 x=0, we obtain A=6 A=6. Therefore,

f(x)=6(x4)(x12)(x+23)=(x4)(2x1)(3x+2). 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) g(x) is a polynomial that leaves a remainder of 1 when divided by x1 x-1 and leaves a remainder of 4 when divided by x+2 x+2. What is the remainder when g(x) g(x) is divided by (x1)(x+2) (x-1)(x+2)?

We have g(x)=(x1)(x+2)q(x)+r(x) g(x) = (x-1)(x+2)q(x) + r(x). Since (x1)(x+2) (x-1)(x+2) has degree 2, the remainder r(x) r(x) has degree at most 1, hence r(x)=Ax+B r(x)=Ax+B for some constants A A and B B. By the remainder factor theorem, we have g(1)=1 g(1)=1 and g(2)=4 g(-2)=4. Substituting x=1 x=1 and 2 -2, we obtain

1=g(1)=(11)(1+2)q(1)+r(1)=A(1)+B, 1 = g(1) = (1-1)(1+2)q(1) + r(1) = A (1) + B, 4=g(2)=(21)(2+2)q(2)+r(2)=A(2)+B. 4 = g(-2) = (-2-1)(-2+2)q(-2)+r(-2) = A(-2)+B.

This implies 1=A+B,4=2A+B 1 = A + B, 4 = -2A + B, which has solution A=1,B=2 A=-1, B=2. Thus, the remainder when g(x) g(x) is divided by (x1)(x+2) (x-1)(x+2) is x+2 -x+2. _\square


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

Solution: Consider the cubic polynomial j(x)=h(x)x+1 j(x)= h(x) - x +1. Then j(1)=1,j(2)=0,j(3)=0 j(1) = 1, j(2) = 0, j(3) = 0 and j(4)=0. j(4) = 0. By the Remainder-Factor Theorem, we have j(x)=A(x)(x2)(x3)(x4) j(x)=A(x) (x-2)(x-3)(x-4), where A(x) A(x) is a polynomial. Since j(x) j(x) is a cubic, it follows that A(x) A(x) has degree 0, hence is a constant which we denote by A A. Substituting x=1 x=1, we obtain

1=j(1)=A(12)(13)(14)A=16. 1 = j(1) = A(1-2)(1-3)(1-4) \Rightarrow A = -\frac {1}{6}.

Thus, h(x)=j(x)+x1=16(x2)(x3)(x4)+x1. h(x) = j(x)+x-1 = -\frac {1}{6} (x-2)(x-3)(x-4)+x-1. Hence,

h(5)=16(52)(53)(54)+51=3. 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 n. _\square


Suppose k(x) k(x) is a polynomial with integer coefficients and c c is an integer root of k(x) k(x). Show that k(x)xc \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 k(c)=0.

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

k(x)=knxn+kn1xn1++k1x+k0, k(x) = k_n x^n + k_{n-1} x^{n-1} + \ldots + k_1 x + k_0,

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

0=k(c)=kncn+kn1cn1++k1c+k0. 0 = k(c) = k_n c^n + k_{n-1} c^{n-1} +\ldots + k_1 c + k_0.

Using the fact that xici=(xc)(xi1+xi2c++xci2+ci1) x^i - c^i=(x-c)(x^{i-1} + x^{i-2}c + \ldots + xc^{i-2} + c^{i-1}), we obtain

k(x)=k(x)0=k(x)k(c)=kn(xncn)+kn1(xn1cn1)++k1(xc)+k0(11)=(xc)Q(x).\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) Q(x) is integer combinations of polynomials with integer coefficients, it follows that Q(x) Q(x) is also a polynomial with integer coefficients. _\square

Note by Arron Kau
7 years, 5 months ago

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