Ninth Grade Mathematics: Remainder Theorem & Applications

Calculating the remainder when a high-degree polynomial is divided by a linear binomial can be tedious using long division. The Remainder Theorem provides a remarkable algebraic shortcut: it enables you to determine the exact remainder instantly by evaluating the polynomial at a single specific value, completely bypassing the division process.

Statement and Proof of the Remainder Theorem

The Remainder Theorem establishes a direct identity between polynomial evaluation and polynomial division:

The Remainder Theorem:
Let P(x) be any polynomial of degree n >= 1, and let 'a' be any real number.
If P(x) is divided by the linear divisor (x - a),
then the remainder R is equal to P(a).

Formal Proof

To prove this theorem, we invoke the fundamental Division Algorithm equation:

P(x) = (x - a) * Q(x) + R(x)

By the division algorithm, the degree of the remainder R(x) must be strictly less than the degree of the divisor (x - a). Because the divisor has degree 1, the remainder must have degree 0, which means R(x) is a constant real number, denoted simply by R:

P(x) = (x - a) * Q(x) + R

Since this identity holds true for every real value of x, we can substitute x = a into both sides of the equation:

P(a) = (a - a) * Q(a) + R
P(a) = 0 * Q(a) + R
P(a) = 0 + R
P(a) = R

This completes the proof: the remainder R is precisely equal to P(a).

Extended Forms of the Remainder Theorem

When the linear divisor is not in the simple monic form (x - a), the theorem generalizes seamlessly:

Divisor Form            Substitute x =           Remainder Formula
-------------------------------------------------------------------------
x - a                   x = a                    R = P(a)
x + a                   x = -a                   R = P(-a)
a*x - b                 x = b / a                R = P(b / a)
a*x + b                 x = -b / a               R = P(-b / a)

To find what value of x to substitute, simply set the linear divisor equal to zero and solve for x.

Example: Find the remainder when P(x) = 3*x^3 - 4*x^2 + 2*x - 5 is divided by 2*x - 1.

Step 1: Set divisor to zero:  2*x - 1 = 0  ===>  x = 1/2.
Step 2: Evaluate P(1/2):
        P(1/2) = 3*(1/2)^3 - 4*(1/2)^2 + 2*(1/2) - 5
               = 3*(1/8) - 4*(1/4) + 1 - 5
               = 3/8 - 1 + 1 - 5
               = 3/8 - 5 = 3/8 - 40/8 = -37/8.
The remainder is -37/8.

Determining Unknown Coefficients

A powerful application of the Remainder Theorem is finding unknown coefficients in polynomials when information about division remainders is provided.

Problem: When P(x) = x^3 + a*x^2 - 5*x + 6 is divided by x - 2, the remainder is 16.
         Find the value of 'a'.

Step 1: Apply the Remainder Theorem: R = P(2).
Step 2: P(2) = 2^3 + a*(2^2) - 5*(2) + 6
             = 8 + 4*a - 10 + 6
             = 4*a + 4.

Step 3: Set P(2) equal to the given remainder:
        4*a + 4 = 16
        4*a = 12
        a = 3.

Chapter Exercises

Exercise 1: Using the Remainder Theorem, find the remainder when the polynomial P(x) = 2x^4 - 5x^3 + 2*x^2 - x + 7 is divided by x - 2. Verify your answer by comparing with polynomial long division.

Exercise 2: Find the remainder when P(x) = x^3 - 6x^2 + 11x - 6 is divided by x + 3.

Exercise 3: The polynomials P(x) = ax^3 + 3x^2 - 13 and Q(x) = 2x^3 - 5x + a leave the same remainder when divided by x - 2. Determine the numerical value of a.

Solutions and Explanations

Solution 1: According to the Remainder Theorem, the remainder when P(x) is divided by x - 2 is equal to P(2). Evaluate P(2): P(2) = 2(2^4) - 5(2^3) + 2(2^2) - 2 + 7 = 2(16) - 5(8) + 2(4) - 2 + 7 = 32 - 40 + 8 - 2 + 7 = 5. By long division, dividing 2x^4 - 5x^3 + 2x^2 - x + 7 by x - 2 yields the quotient 2x^3 - x^2 - 1 and leaves a final constant remainder of 5. Both methods produce the exact same remainder of 5, validating the theorem.

Solution 2: For divisor x + 3, set x + 3 = 0, giving x = -3. By the Remainder Theorem, the remainder is P(-3). Evaluating: P(-3) = (-3)^3 - 6((-3)^2) + 11(-3) - 6 = -27 - 6*(9) - 33 - 6 = -27 - 54 - 33 - 6 = -120. Thus, dividing P(x) by x + 3 yields a remainder of -120.

Solution 3: When P(x) is divided by x - 2, the remainder is R1 = P(2). Evaluating P(2): R1 = a(2^3) + 3(2^2) - 13 = 8a + 3(4) - 13 = 8a + 12 - 13 = 8a - 1. When Q(x) is divided by x - 2, the remainder is R2 = Q(2). Evaluating Q(2): R2 = 2(2^3) - 5(2) + a = 2(8) - 10 + a = 16 - 10 + a = 6 + a. Since the problem states that both remainders are equal, set R1 = R2: 8a - 1 = 6 + a. Subtracting a from both sides yields 7a - 1 = 6. Adding 1 gives 7a = 7, so a = 1.