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 polynomials and solve polynomial equations.
• Find unknown coefficients in polynomials.

Lesson summary

Polynomial functions are a cornerstone of advanced mathematics, providing the foundation for understanding complex relationships in various fields. The Remainder and Factor Theorems are powerful tools that simplify working with these functions, particularly when dealing with finding roots and factoring. In South Africa, understanding these concepts is crucial for students pursuing careers in engineering, finance, data science, and various other STEM fields. These theorems allow us to efficiently analyse and manipulate equations that model real-world scenarios, from predicting economic growth to designing efficient structures.

Lesson notes

Polynomial Functions: A Quick Recap 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 the coefficients a n , a n-1 , ..., a 1 , a 0 are constants. The Remainder Theorem The Remainder Theorem states: When a polynomial f(x) is divided by (x - a), the remainder is f(a). Why does this work? When we divide f(x) by (x - a), we can express the division as: f(x) = (x - a) * q(x) + r where q(x) is the quotient and r is the remainder (which is a constant since we are dividing by a linear expression). If we substitute 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 + x - 5 is divided by (x - 3).

Solution: According to the Remainder Theorem, the remainder is f(3). f(3) = (3) 3 - 2(3) 2 + (3) - 5 f(3) = 27 - 18 + 3 - 5 f(3) = 7 Therefore, the remainder is

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

Solution: Since the divisor is (x + 1), we can rewrite it as (x - (-1)).

Therefore, a = -1. f(-1) = 2(-1) 4 + (-1) 3 - 5(-1) 2 + 3(-1) - 2 f(-1) = 2(1) - 1 - 5(1) - 3 - 2 f(-1) = 2 - 1 - 5 - 3 - 2 f(-1) = -9 Therefore, the remainder is -

9. The Factor Theorem The Factor Theorem states: (x - a) is a factor of a polynomial f(x) if and only if f(a) =

0. Why does this work? The Factor Theorem is a direct consequence of the Remainder Theorem. If f(a) = 0, then the remainder when f(x) is divided by (x - a) is

0. This means that (x - a) divides f(x) exactly, making it a factor of f(x).

Example 3: Determine whether (x - 2) is a factor of f(x) = x 3 - 4x 2 + 5x -

2. Solution: We need to check if f(2) = 0. f(2) = (2) 3 - 4(2) 2 + 5(2) - 2 f(2) = 8 - 16 + 10 - 2 f(2) = 0 Since f(2) = 0, (x - 2) is a factor of f(x).

Example 4: Determine whether (x + 3) is a factor of f(x) = x 3 + 2x 2 - 5x -

6. Solution: We need to check if f(-3) = 0. f(-3) = (-3) 3 + 2(-3) 2 - 5(-3) - 6 f(-3) = -27 + 18 + 15 - 6 f(-3) = 0 Since f(-3) = 0, (x + 3) is a factor of f(x). Using the Factor Theorem to Factorize Polynomials If we know one factor of a polynomial, we can use the Factor Theorem in conjunction with long division (or synthetic division – if applicable) to find the other factors.

Example 5: Factorize f(x) = x 3 - 6x 2 + 11x - 6 completely, given that (x - 1) is a factor.

Solution: Since (x - 1) is a factor, we can 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 ``` Therefore, f(x) = (x - 1)(x 2 - 5x + 6). Now, we can factorize the quadratic expression: x 2 - 5x + 6 = (x - 2)(x - 3) So, f(x) = (x - 1)(x - 2)(x - 3). Solving Polynomial Equations using the Factor Theorem Once we have factored a polynomial, we can easily find its roots (the solutions to the equation f(x) = 0). The roots 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, (x - 1)(x - 2)(x - 3) =

0. This means x - 1 = 0 or x - 2 = 0 or x - 3 =

0. So, x = 1, x = 2, or x =

3. Finding Unknown Coefficients The Remainder and Factor Theorems can be used to find unknown coefficients in a polynomial.

Example 7: When f(x) = x 3 + ax 2 - x + b is divided by (x - 2), the remainder is

5. When f(x) is divided by (x + 1), the remainder is -

4. Find the values of a and b.

Solution: Using the Remainder Theorem: f(2) = 5 => (2) 3 + a(2) 2 - (2) + b = 5 => 8 + 4a - 2 + b = 5 => 4a + b = -1 (Equation 1) f(-1) = -4 => (-1) 3 + a(-1) 2 - (-1) + b = -4 => -1 + a + 1 + b = -4 => a + b = -4 (Equation 2) Now we have a system of two linear equations: 4a + b = -1 a + b = -4 Subtracting Equation 2 from Equation 1: (4a + b) - (a + b) = -1 - (-4) 3a = 3 a = 1 Substituting a = 1 into Equation 2: 1 + b = -4 b = -5 Therefore, a = 1 and b = -

5. Guided Practice (With Solutions)

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

Solution: Using the Remainder Theorem, the remainder is f(2): f(2) = (2) 4 - 3(2) 2 + 2(2) - 1 f(2) = 16 - 12 + 4 - 1 f(2) = 7 The remainder is

7. Commentary: This question directly applies the Remainder Theorem. Identifying the value of 'a' (in (x-a)) is crucial.

Question 2: Is (x + 2) a factor of f(x) = x 3 + 5x 2 + 2x - 8?

Solution: Using the Factor Theorem, we need to check if f(-2) = 0: f(-2) = (-2) 3 + 5(-2) 2 + 2(-2) - 8 f(-2) = -8 + 20 - 4 - 8 f(-2) = 0 Since f(-2) = 0, (x + 2) is a factor of f(x).

Commentary: This question assesses understanding of the Factor Theorem and the substitution of negative values.

Question 3: Given that (x - 1) is a factor of f(x) = x 3 + x 2 + kx - 3, find the value of k.