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.
• Completely factorize polynomial expressions.
• Solve polynomial equations.
• Apply these theorems to solve problems.

Lesson summary

Polynomial functions are fundamental to understanding relationships in various fields, from engineering and economics to computer science. The Remainder and Factor Theorems provide powerful shortcuts for analyzing these functions, particularly when finding roots (x-intercepts) and factors. These theorems save us from tedious long division and empower us to solve complex polynomial equations more efficiently. In a South African context, understanding polynomial functions can be applied to modelling population growth, resource allocation, and even financial trends. For example, businesses can use polynomial models to predict sales based on marketing expenditure.

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 called coefficients (a n ≠ 0). x is the variable.

Examples of polynomial functions include: Linear function: f(x) = 2x + 3 (degree 1)

Quadratic function: f(x) = x 2 - 5x + 6 (degree 2)

Cubic function: f(x) = x 3 + 2x 2 - x - 2 (degree 3) 2.2 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 it as: f(x) = (x - a)q(x) + R Where: q(x)* is the quotient (another polynomial). R is the remainder (a constant since we're 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 - 4x 2 + 5x - 2 is divided by (x - 2). Using the Remainder Theorem, we need to find f(2): f(2) = (2) 3 - 4(2) 2 + 5(2) - 2 f(2) = 8 - 16 + 10 - 2 f(2) = 0 Therefore, the remainder is

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

Note: (x + 1) is the same as (x - (-1)) so a = -

1. Using the Remainder Theorem, we need to find f(-1): f(-1) = 2(-1) 4 - 3(-1) 3 + (-1) - 5 f(-1) = 2(1) - 3(-1) - 1 - 5 f(-1) = 2 + 3 - 1 - 5 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: (x - a) is a factor of the polynomial f(x) if and only if f(a) =

0. In other words, if substituting x = a into the polynomial makes the polynomial equal to zero, then (x - a) divides evenly into f(x). This means there's no remainder.

Example 3: Show that (x - 1) is a factor of f(x) = x 3 - 6x 2 + 11x -

6. We need to find f(1): f(1) = (1) 3 - 6(1) 2 + 11(1) - 6 f(1) = 1 - 6 + 11 - 6 f(1) = 0 Since f(1) = 0, (x - 1) is a factor of f(x).

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

8. We need to find f(-2): f(-2) = (-2) 3 + 8 f(-2) = -8 + 8 f(-2) = 0 Since f(-2) = 0, (x + 2) is a factor of f(x). 2.4 Using the Factor Theorem to Factorize Polynomials The Factor Theorem is particularly useful for factorizing polynomials, especially cubic and higher-degree polynomials.

Here's the process: Find a factor using trial and error: Substitute different values for x (small integers like 0, 1, -1, 2, -2, etc.) into f(x) until you find a value a such that f(a) =

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

Divide the polynomial by the factor: Use long division or synthetic division to divide f(x) by (x - a). This gives you the quotient q(x).

Factorize the quotient: Factorize the quotient q(x). If it's a quadratic, you can use standard quadratic factorization techniques. Write the polynomial as a product of its factors: f(x) = (x - a)q(x). Combine all the factors you found.

Example 5: Factorize f(x) = x 3 - 4x 2 + x + 6 completely.

Trial and error: f(0) = 6 (not 0) f(1) = 1 - 4 + 1 + 6 = 4 (not 0) f(-1) = -1 - 4 - 1 + 6 = 0 Aha! So (x + 1)* is a factor. Divide by (x + 1): We'll use synthetic division here. ``` -1 | 1 -4 1 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 polynomial as a product of its factors: f(x) = (x + 1)(x - 2)(x - 3) Therefore, f(x) = (x + 1)(x - 2)(x - 3) is the complete factorization. 2.5 Solving Polynomial Equations using the Factor Theorem To solve a polynomial equation f(x) = 0 using the Factor Theorem, follow these steps: Factorize the polynomial: Use the Factor Theorem (as described above) to factorize f(x) completely.

Set each factor equal to zero: If f(x) = (x - a)(x - b)(x - c) = 0, then either (x - a) = 0, (x - b) = 0, or (x - c) =

0. Solve for x: Solve each of the resulting equations to find the roots (solutions) of the polynomial equation.

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

0. We already factorized this polynomial in Example 5: f(x) = (x + 1)(x - 2)(x - 3).

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

0. This means: x + 1 = 0 => x = -1 x - 2 = 0 => x = 2 x - 3 = 0 => x = 3 Therefore, the solutions to the equation x 3 - 4x 2 + x + 6 = 0 are x = -1, x = 2, x =

3. Guided Practice (With Solutions)

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

Solution: Using the Remainder Theorem, we need to find f(3): f(3) = (3) 4 - 2(3) 3 + (3) 2 - 5(3) + 4 f(3) = 81 - 54 + 9 - 15 + 4 f(3) = 25 Therefore, the remainder is

2

5. Commentary: This is a straightforward application of the Remainder Theorem. Substitute the value of a (in this case, 3) into the polynomial and evaluate. Pay close attention to the order of operations (exponents first).

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