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 completely.
• Solve polynomial equations.
• Apply theorems to solve related problems.

Lesson summary

Polynomial functions are foundational in mathematics, serving as building blocks for more advanced topics like calculus and differential equations. Understanding the Remainder and Factor Theorems allows us to efficiently analyze and manipulate these functions, leading to solutions for various real-world problems. In South Africa, understanding polynomials can be applied in diverse fields, such as: Engineering: Designing structures like bridges or buildings often involves polynomial equations to model stress and strain.

Finance: Calculating compound interest or projecting investment growth relies on polynomial models.

Lesson notes

Polynomial Functions: A polynomial function is an expression 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) and a n , a n-1 , ..., a 1 , a 0 are constants (coefficients).

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

Why this works: Consider dividing f(x) by (x - a).

We can write: f(x) = (x - a)q(x) + r where q(x) is the quotient and r is the remainder (a constant since we're dividing by a linear term). Now, if we substitute x = a, we get: f(a) = (a - a)q(a) + r => f(a) = 0*q(a) + r => f(a) = r How to use it: To find the remainder when f(x) is divided by (x - a), simply evaluate f(a).

Example 1: Find the remainder when f(x) = x 3 - 2x 2 + x - 5 is divided by (x - 2). Here, a = 2. f(2) = (2) 3 - 2(2) 2 + (2) - 5 = 8 - 8 + 2 - 5 = -3 Therefore, the remainder is -

3. Example 2: Find the remainder when f(x) = 2x 4 + x 3 - 5x + 3 is divided by (x + 1). Here, a = -1 (because x + 1 = x - (-1)). f(-1) = 2(-1) 4 + (-1) 3 - 5(-1) + 3 = 2 - 1 + 5 + 3 = 9 Therefore, the remainder is

9. The Factor Theorem: The Factor Theorem is a special case of the Remainder Theorem. It states that (x - a) is a factor of the polynomial f(x) if and only if f(a) =

0. Why this works: From the Remainder Theorem, we know f(x) = (x - a)q(x) + f(a). If f(a) = 0, then f(x) = (x - a)q(x), which means (x - a) divides f(x) exactly, i.e., (x-a) is a factor of f(x).

How to use it: To check if (x - a) is a factor of f(x), evaluate f(a). If f(a) = 0, then (x - a) is a factor. If f(a) ≠ 0, then (x - a) is not a factor.

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

6. Here, a = 3. f(3) = (3) 3 - 4(3) 2 + (3) + 6 = 27 - 36 + 3 + 6 = 0 Since f(3) = 0, (x - 3) is a factor of f(x).

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

8. Here, a = -2 (because x + 2 = x - (-2)). f(-2) = (-2) 3 + 5(-2) 2 + 2(-2) - 8 = -8 + 20 - 4 - 8 = 0 Since f(-2) = 0, (x + 2) is a factor of f(x). Using Remainder and Factor Theorems for Complete Factorization: Once you've found one factor using the Factor Theorem, you can use polynomial division (long division or synthetic division) to find the remaining factor(s). This is essential for completely factorizing the polynomial.

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

Find a factor using the Factor Theorem: Let's try factors of 6 (±1, ±2, ±3, ±6). f(-1) = (-1) 3 + 6(-1) 2 + 11(-1) + 6 = -1 + 6 - 11 + 6 =

0. So, (x + 1) is a factor. Divide f(x) by (x + 1): We will use synthetic division here, as it's more concise. ``` -1 | 1 6 11 6 | -1 -5 -6 ---------------- 1 5 6 0 ``` The quotient is 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 (f(x) = 0). By finding the factors, we can find the roots (solutions) of the equation.

Example 6: Solve the equation x 3 - 4x 2 + x + 6 =

0. From Example 3, we know (x - 3) is a factor of x 3 - 4x 2 + x +

6. Using polynomial division (either long or synthetic), we find that x 3 - 4x 2 + x + 6 = (x - 3)(x 2 - x - 2).

Factoring the quadratic: x 2 - x - 2 = (x - 2)(x + 1).

Therefore, x 3 - 4x 2 + x + 6 = (x - 3)(x - 2)(x + 1) =

0. The solutions are x = 3, x = 2, and x = -

1. Guided Practice (With Solutions)

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

Solution: Apply the Remainder Theorem: a = 1 f(1) = (1) 4 - 3(1) 3 + 2(1) 2 - (1) + 1 = 1 - 3 + 2 - 1 + 1 = 0 Therefore, the remainder is

0. This also means (x-1) is a factor.

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

Solution: Apply the Factor Theorem: a = -3 g(-3) = (-3) 3 + 5(-3) 2 + 2(-3) - 24 = -27 + 45 - 6 - 12 = 0 Since g(-3) = 0, (x + 3) is a factor of g(x).

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

Solution: Find a factor: Try factors of 4 (±1, ±2, ±4). h(1) = (1) 3 - (1) 2 - 4(1) + 4 = 1 - 1 - 4 + 4 =

0. So, (x - 1) is a factor.

Divide h(x) by (x - 1): Using synthetic division: ``` 1 | 1 -1 -4 4 | 1 0 -4 ---------------- 1 0 -4 0 ``` The quotient is x 2 -

4. Factorize the quotient: x 2 - 4 = (x - 2)(x + 2) (Difference of squares)

Write the complete factorization: h(x) = (x - 1)(x - 2)(x + 2)

Question 4: Solve the equation 2x 3 - 5x 2 + x + 2 =

0. Solution: Find a factor: Try factors of 2/2 (±1, ±2, ±1/2). f(1) = 2(1) 3 - 5(1) 2 + 1 + 2 = 2 - 5 + 1 + 2 =

0. So, (x - 1) is a factor. Divide 2x 3 - 5x 2 + x + 2 by (x - 1): Using synthetic division: ``` 1 | 2 -5 1 2 | 2 -3 -2 ---------------- 2 -3 -2 0 ``` The quotient is 2x 2 - 3x - 2.