Remainder Factor Theorem

The remainder factor theorem is actually two theorems that relate the roots of a polynomial with its linear factors. The theorem is often used to help factorize polynomials without the use of long division. Especially when combined with the rational root theorem, this gives us a powerful tool to factor polynomials.

Remainder Theorem:

For a polynomial \( f(x)\), the remainder of \( f(x)\) upon division by \( x-c\) is \( f(c)\).

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 \( f(c)=0\).

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

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

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

\( g(x)\) 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 \( (x-1)(x+2)?\)

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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, and 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

\[\begin{align} 1 &= g(1) = (1-1)(1+2)q(1) + r(1) = A (1) + B\\ 4 &= g(-2) = (-2-1)(-2+2)q(-2)+r(-2) = A(-2)+B. \end{align}\]

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 \( h(5)?\)

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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 and thus is a constant which we denote by \( A\). Substituting \( x=1\), we obtain

\[ 1 = j(1) = A(1-2)(1-3)(1-4) \implies A = -\frac {1}{6}.\]

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

\[ h(5) = -\frac {1}{6} (5-2)(5-3)(5-4) + 5 - 1 = 3. \ _\square\]

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

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.

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At first glance, it seems impossible to work directly from the factor theorem's result that \( k(c)=0\).

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

\[ k(x) = k_n x^n + k_{n-1} x^{n-1} + \cdots + k_1 x + k_0,\]

where \( k_i\) are integers. Then by the remainder factor theorem,

\[ 0 = k(c) = k_n c^n + k_{n-1} c^{n-1} +\cdots + k_1 c + k_0.\]

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

\[\begin{align} k(x) & = k(x)- 0 = k(x)-k(c) \\ &= k_n (x^n-c^n) + k_{n-1} \big( x^{n-1} - c^{n-1}\big) + \cdots + k_1 (x-c) + k_0 (1-1) \\ & = (x-c) Q(x). \end{align}\]

Since \( Q(x)\) is integer combinations of polynomials with integer coefficients, it follows that \( \frac {k(x)}{x-c}=Q(x)\) is also a polynomial with integer coefficients. \( _\square \)

Remainder:

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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 and thus is a constant. Let \( r(x)=R\). Then substituting \( x=c\), we obtain \( f(c)=(c-c)q(c)+R=R\). Thus, the remainder is \(f(c)\), as claimed. \( _\square \)

Factor:

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(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 \( f(x)\). \( _\square \)

(Backward direction)
If \( x-c\) is a factor of \( f(x)\), then (by definition) the remainder of \( f(x)\) upon division 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!