Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply the Remainder Theorem to find remainders.
• Use the Factor Theorem to identify factors of polynomials.
• Factorize polynomial expressions.
• Solve polynomial equations.
• Relate the roots of an equation to the factors of the polynomial.

Lesson summary

Polynomial functions are foundational in mathematics and have broad applications in various fields, from engineering and economics to computer science. Understanding the Remainder and Factor Theorems provides powerful tools for analyzing and manipulating these functions. For South African learners, mastering these concepts enhances problem-solving skills crucial for university studies and careers in STEM fields. Imagine using polynomial models to predict population growth in a specific region or designing a bridge with optimal load distribution – all rooted in the principles we'll explore this week.

Lesson notes

Polynomial Functions: A polynomial function is a function that can be written in the form: f(x) = a n x n + a n-1 x n-1 + ... + a 1 x + a 0 where n is a non-negative integer (the degree of the polynomial), and a n , a n-1 , ..., a 1 , a 0 are constants (coefficients).

Division Algorithm for Polynomials: When a polynomial f(x) is divided by a divisor d(x), we obtain a quotient q(x) and a remainder r(x), such that: f(x) = d(x) q(x) + r(x)* where the degree of r(x) is less than the degree of d(x). If d(x) is a linear divisor (i.e., of the form x - a), then the remainder r(x) will be a constant.

The Remainder Theorem: The Remainder Theorem states that if a polynomial f(x) is divided by (x - a), then the remainder is f(a).

Proof: From the division algorithm, we have: f(x) = (x - a) q(x) + r where r is the remainder (a constant). Substituting x = a into the equation, we get: f(a) = (a - a) q(a) + r f(a) = 0 q(a) + r f(a) = r Therefore, the remainder r is equal to f(a).

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

Solution: Using the Remainder Theorem, we need to find f(1). f(1) = (1) 3 - 2(1) 2 + 5(1) - 3 = 1 - 2 + 5 - 3 = 1 Therefore, the remainder is

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

Solution: Using the Remainder Theorem, we need to find f(-2). Note that we need to find f(a) where (x - a) = (x + 2) so a = -2. f(-2) = 2(-2) 4 + (-2) 3 - (-2) + 5 = 2(16) - 8 + 2 + 5 = 32 - 8 + 2 + 5 = 31 Therefore, the remainder is

3

1. The Factor Theorem: The Factor Theorem is a direct consequence of the Remainder Theorem. It states that (x - a) is a factor of f(x) if and only if f(a) =

0. Proof: If (x - a) is a factor of f(x), then f(x) = (x - a) q(x) for some polynomial q(x). Substituting x = a, we get: f(a) = (a - a) q(a) = 0 Conversely, if f(a) = 0, then from the Remainder Theorem, the remainder when f(x) is divided by (x - a) is

0. This means that f(x) is divisible by (x - a), and therefore (x - a) is a factor of f(x).

Example 3: Is (x - 2) a factor of f(x) = x 3 - 4x 2 + x + 6?

Solution: Using the Factor Theorem, we need to find f(2). f(2) = (2) 3 - 4(2) 2 + (2) + 6 = 8 - 16 + 2 + 6 = 0 Since f(2) = 0, (x - 2) is a factor of f(x).

Example 4: Is (x + 1) a factor of f(x) = x 4 + 3x 3 + x 2 - 3x - 2?

Solution: Using the Factor Theorem, we need to find f(-1). f(-1) = (-1) 4 + 3(-1) 3 + (-1) 2 - 3(-1) - 2 = 1 - 3 + 1 + 3 - 2 = 0 Since f(-1) = 0, (x + 1) is a factor of f(x). Using the Factor Theorem to Factorize Polynomials: The Factor Theorem can be used to factorize polynomials, especially cubic polynomials (degree 3).

Here's the general approach: Find a factor: Use trial and error, substituting different values for x into f(x) until you find a value a such that f(a) =

0. This means (x - a) is a factor. Look for integer values like -2, -1, 0, 1, 2 first.

Divide: Divide f(x) by (x - a) using polynomial long division or synthetic division to find the quotient q(x).

Factorize the quotient: Factorize the quotient q(x) (which will be a quadratic) using standard methods (factorization by grouping, quadratic formula).

Write the complete factorization: The complete factorization of f(x) is then (x - a) q(x), with q(x)* fully factorized.

Example 5: Factorize f(x) = x 3 - 6x 2 + 11x -

6. Solution: Find a factor: Let's try x = 1: f(1) = (1) 3 - 6(1) 2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 Therefore, (x - 1) is a factor.

Divide: Divide f(x) by (x - 1): ``` x^2 - 5x + 6 x - 1 | x^3 - 6x^2 + 11x - 6 (x^3 - x^2) ---------------- -5x^2 + 11x (-5x^2 + 5x) ---------------- 6x - 6 (6x - 6) ---------------- 0 ``` The quotient is q(x) = x 2 - 5x +

6. Factorize the quotient: x 2 - 5x + 6 = (x - 2)(x - 3)

Write the complete factorization: f(x) = (x - 1)(x - 2)(x - 3)

Solving Polynomial Equations: The Factor Theorem is also useful for solving polynomial equations of the form f(x) =

0. If you can factorize f(x), then the solutions (roots) of the equation are the values of x that make each factor equal to zero.

Example 6: Solve the equation x 3 - 6x 2 + 11x - 6 =

0. Solution: From Example 5, we know that x 3 - 6x 2 + 11x - 6 = (x - 1)(x - 2)(x - 3).

Therefore, the equation becomes: (x - 1)(x - 2)(x - 3) = 0 This equation is satisfied if x - 1 = 0 or x - 2 = 0 or x - 3 =

0. Therefore, the solutions are x = 1, x = 2, x =

3. Guided Practice (With Solutions)

Question 1: Find the remainder when f(x) = x 3 + 4x 2 - 7x + 2 is divided by (x + 3).

Solution: Using the Remainder Theorem, we need to find f(-3). f(-3) = (-3) 3 + 4(-3) 2 - 7(-3) + 2 = -27 + 36 + 21 + 2 = 32 Therefore, the remainder is

3

2. Commentary: This question directly applies the Remainder Theorem. The key is to substitute the correct value for 'x'. Remember to pay attention to the sign when d(x) = (x - a), so if we have (x+3), then a = -

3. Question 2: Is (x - 1) a factor of f(x) = 2x 4 - 3x 3 + x - 5?

Solution: Using the Factor Theorem, we need to find f(1).