Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply the Remainder Theorem to find the remainder.
• Use the Factor Theorem to identify factors of polynomials.
• Factorize cubic and higher-degree polynomials.
• Solve polynomial equations to find their roots.

Lesson summary

Polynomial functions are fundamental building blocks in mathematics. They appear in various fields, from engineering and physics to economics and computer graphics. Understanding the Remainder and Factor Theorems equips you with powerful tools for analyzing and solving problems involving polynomials. In South Africa, these skills are crucial for careers in sectors like engineering (designing infrastructure like bridges, which involves polynomial modeling), finance (modeling investment growth), and data analysis (predicting trends using polynomial regression).

Lesson notes

2.1 Polynomial Functions: A Quick Review A polynomial function is a function of 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) a n , a n-1 , ..., a 1 , a 0 are constants (coefficients) and a n ≠ 0. x is the variable

Examples: f(x) = 3x 2 - 2x + 1 (Quadratic polynomial, degree 2) f(x) = x 3 + 5x - 7 (Cubic polynomial, degree 3) f(x) = 4x 5 - x 2 + 9 (Polynomial of degree 5) 2.2 The Remainder Theorem The Remainder Theorem states that if a polynomial f(x) is divided by x - a, then the remainder is f(a). Why does this work? When we divide f(x) by x - a, we can write: f(x) = (x - a)q(x) + r where: q(x)* is the quotient (another polynomial) r is the remainder (a constant since we are dividing by a linear expression) Now, substitute x = a into the equation: 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 + x - 5 is divided by x -

2. Using the Remainder Theorem, the remainder is f(2): f(2) = (2) 3 - 2(2) 2 + (2) - 5 f(2) = 8 - 8 + 2 - 5 f(2) = -3 Therefore, the remainder is -

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

1. Note that x + 1 can be written as x - (-1), so a = -

1. Using the Remainder Theorem, the remainder is f(-1): f(-1) = 2(-1) 4 + 3(-1) 3 - (-1) + 1 f(-1) = 2(1) + 3(-1) + 1 + 1 f(-1) = 2 - 3 + 1 + 1 f(-1) = 1 Therefore, the remainder is 1. 2.3 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. In other words, if the remainder when f(x) is divided by x - a is zero, then x - a is a factor of f(x). Why does this work? From the Remainder Theorem, we know f(x) = (x - a)q(x) + r. If f(a) = 0, then r =

0. Therefore, f(x) = (x - a)q(x), which means x - a divides f(x) exactly, making it a factor.

Example 1: Is x - 1 a factor of f(x) = x 3 - 3x 2 + 2x? Using the Factor Theorem, we need to check if f(1) = 0. f(1) = (1) 3 - 3(1) 2 + 2(1) f(1) = 1 - 3 + 2 f(1) = 0 Since f(1) = 0, x - 1 is a factor of f(x).

Example 2: Is x + 2 a factor of f(x) = x 4 + x 3 - 4x 2 - 4x? Note that x + 2 can be written as x - (-2), so a = -

2. Using the Factor Theorem, we need to check if f(-2) = 0. f(-2) = (-2) 4 + (-2) 3 - 4(-2) 2 - 4(-2) f(-2) = 16 - 8 - 16 + 8 f(-2) = 0 Since f(-2) = 0, x + 2 is a factor of f(x). 2.4 Using the Remainder and Factor Theorems to Factorize Polynomials The Factor Theorem is extremely useful for factorizing cubic and higher-degree polynomials.

The process involves: Finding a factor: Use trial and error to find a value a such that f(a) =

0. This means x - a is a factor.

Polynomial Division: Divide f(x) by x - a to find the quotient q(x). You can use long division or synthetic division.

Factorizing the Quotient: Factorize the quotient q(x), which will be a polynomial of a lower degree than f(x).

Writing the Complete Factorization: Write f(x) as the product of all its factors.

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

6. Finding a factor: Let's try x = 1: f(1) = (1) 3 - 6(1) 2 + 11(1) - 6 = 1 - 6 + 11 - 6 =

0. Since f(1) = 0, x - 1 is a factor.

Polynomial Division: Divide x 3 - 6x 2 + 11x - 6 by x - 1 using long division (or synthetic division, if you're familiar with it): ``` 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 x 2 - 5x +

6. Factorizing the Quotient: Factorize x 2 - 5x + 6: (x - 2)(x - 3)

Writing the Complete Factorization: Therefore, f(x) = (x - 1)(x - 2)(x - 3). Guided Practice (With Solutions)

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

2. Solution: Using the Remainder Theorem, we need to find f(-2). f(-2) = (-2) 3 + 4(-2) 2 - 3(-2) + 2 f(-2) = -8 + 16 + 6 + 2 f(-2) = 16 The remainder is

1

6. Commentary: This question directly applies the Remainder Theorem. Be careful with the signs.

Question 2: Is x - 3 a factor of f(x) = x 4 - 5x 3 + 6x 2 + 4x - 24?

Solution: Using the Factor Theorem, we need to check if f(3) = 0. f(3) = (3) 4 - 5(3) 3 + 6(3) 2 + 4(3) - 24 f(3) = 81 - 135 + 54 + 12 - 24 f(3) = 0 Since f(3) = 0, x - 3 is a factor of f(x).

Commentary: This directly tests understanding of the Factor Theorem.

Question 3: Factorize f(x) = x 3 - x 2 - 4x + 4 completely.

Solution: Find a factor: Try x = 1: f(1) = (1) 3 - (1) 2 - 4(1) + 4 = 1 - 1 - 4 + 4 =

0. So, x - 1 is a factor.

Polynomial Division: Divide x 3 - x 2 - 4x + 4 by x - 1: ``` x^2 - 4 x - 1 | x^3 - x^2 - 4x + 4 (x^3 - x^2) ----------------- 4x + 4 (-4x + 4) ----------------- 0 ``` The quotient is x 2 -

4. Factorize the Quotient: x 2 - 4 = (x - 2)(x + 2) (difference of squares)

Complete Factorization: f(x) = (x - 1)(x - 2)(x + 2).

Commentary: This combines both theorems to perform a full factorization.