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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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