SHS Mathematics //Tricks, Questions & Answers

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Find the value of x³ + y³ – 12xy + 64, when x + y = – 4. Solution: Given, x + y = -4 Or x + y + 4 = 0….(i) Let the given polynomial be: p(x) = x³ + y³ – 12xy + 64 = x³ + y³ + 4³ – 3(4xy) {since 64 = 4³} We know that is a + b + c = 0 then a³ + b³ + c³ = 3abc That means if x + y + 4 = 0, x³ + y³ + 4³ = 3(x)(y)(4) {from (i)} So, p(x) = 3(x)(y)(4) – 3(4xy) = 12xy – 12xy = 0 7. Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2. [Hint: Factorise x² – 3x + 2] Solution: Let p(x) = 2x⁴ – 5x³ + 2x² – x + 2 Let us factorise x² – 3x + 2. x² – 3x + 2 = x² – 2x – x + 2 = x(x – 2) -1(x – 2) = (x – 1)(x – 2) Hence, 1 and 2 are the zeroes of x² – 3x + 2. Now, substitute x = 1 and x = 2 in p(x). p(1) = 2(1)⁴ – 5(1)³ + 2(1)² -1 + 2 = 2 – 5 + 2 – 1 + 2 = 6 – 6 = 0 p(2) = 2(2)⁴ – 5(2)³ + 2(2)² – 2 + 2 = 2(16) – 5(8) + 2(4) = 32 – 40 + 8 = 40 – 40 = 0 Therefore, p(x) = 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2.

Thus, x³ – 3x² – 9x – 5 = (x + 1)(x² – 4x – 5) Consider quotient, x² – 4x – 5 = x² – 5x + x – 5 = x(x – 5) + 1(x – 5) = (x + 1)(x – 5) Therefore, p(x) = x³ – 3x² – 9x – 5 = (x + 1)(x + 1)(x – 5) = (x + 1)2(x – 5).

5. Factorize the polynomial p(x) = x³ – 3x² – 9x – 5. Solution: Given, p(x) = x³ – 3x² – 9x – 5 By trial and error method, substitute x = -1 in p(x). p(-1) = (-1)³ – 3(-1)² – 9(-1) – 5 = -1 – 3 + 9 – 5 = -9 + 9 = 0 Since p(-1) = 0 and by the factor theorem we can say x + 1 is a factor of p(x). By dividing p(x) by x + 1 we can get the remaining factors.

3. If x + y = 12 and xy = 27, find the value of x3 + y3. Solution: Given, x + y = 12 xy = 27 x3 + y3 = (x + y) (x2 – xy + y2) = (x + y) [(x + y)2 – 3xy] = 12 × (122 – 3 × 27) = 12 × (144 – 81) = 12 × 63 = 756 4. Factorise: 12x2 – 7x + 1 Solution: Let the given polynomial be: p(x) = 12x2 – 7x + 1 = 12x2 – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1) = (3x – 1)(4x – 1) Hence, the factors of 12x2 – 7x + 1 are (3x – 1) and (4x – 1).

Polynomials Questions with Answers 1. For the polynomial (x3 + 2x + 1)/5 – (7/2)x2 – x6, write (i) the degree of the polynomial (ii) the coefficient of x3 (iii) the coefficient of x6 (iv) the constant term Solution: Given polynomial is: (x3 + 2x + 1)/5 – (7/2)x2 – x6 Or (1/5)x3 + (2/5)x + (1/5) – (7/2)x2 – x6 (i) Degree of the polynomial = 6 {since the highest power of variable x is 6} (ii) Coefficient of x3 = (1/5) (iii) Coefficient of x6 = -1 (iv) Constant term = 1/5 2. Find the value of a, if x – a is a factor of x3 – ax2 + 2x + a – 1. Solution: Let p(x) = x3 – ax2 + 2x + a – 1 Given that x – a is a factor of p(x). ⇒ p(a) = 0 i.e., a3 – a(a)2 + 2a + a – 1 = 0 a3 – a3 + 2a + a – 1 = 0 3a – 1 = 0 3a = 1 a = 1/3 Therefore, a = 1/3.

V. Polynomial Graphs: x-Intercepts: The x-intercepts (or zeros) of a polynomial function are the values of x for which f(x) = 0. They can be found by solving the equation f(x) = 0. y-Intercept: The y-intercept is the point where the polynomial intersects the y-axis. It can be found by evaluating f(0). Turning Points: Turning points or local extrema are the high or low points on the graph of a polynomial. They occur where the polynomial changes direction. End Behavior: The end behavior of a polynomial describes how the polynomial behaves as x approaches positive or negative infinity. It is determined by the degree and leading coefficient of the polynomial. VI. Factoring Polynomials: Greatest Common Factor (GCF): Factor out the greatest common factor from the polynomial expression. Factoring by Grouping: Group the terms of a polynomial and look for common factors that can be factored out. Factoring Trinomials: Factor trinomials of the form ax^2 + bx + c using various factoring techniques such as the ac-method or by findingperfect square trinomials or difference of squares. VII. Solving Polynomial Equations: Zero Product Property: If the product of two factors is zero, then at least one of the factors must be zero. Use this property to solve polynomial equations by setting the polynomial equal to zero and factoring. Quadratic Formula: Use the quadratic formula to solve quadratic equations of the form ax^2 + bx + c = 0, where a, b, and c are real numbers. VIII. Applications of Polynomial Functions: Modeling Data: Polynomial functions can be used to model real-world data sets, such as population growth, economic trends, or physical phenomena. Optimization Problems: Polynomial functions can be used to solve optimization problems by finding the maximum or minimum values of a quantity. Geometry: Polynomial functions can be used to solve problems involving areas, volumes, and geometric shapes. Engineering and Physics: Polynomial functions are widely used in engineering and physics to describe various physical phenomena, such as motion, electrical circuits, and fluid dynamics. Conclusion: Polynomial functions are essential mathematical concepts that provide a foundation for higher-level mathematics and have numerous real-world applications. This lesson note has covered key definitions, properties, operations, graphing, factoring, solving equations, and applications of polynomial functions. By providing clear explanations, examples, and opportunities for practice, students can develop a solid understanding and proficiency in working with polynomial functions.

Introduction: Polynomial functions are fundamental mathematical concepts taught in high school mathematics. They play a vital role in various areas of mathematics, science, engineering, and economics. This teaching note aims to provide a detailed guide for teaching polynomial functions to high school students. It includes key definitions, properties, and examples to enhance understanding and application. I. Definition of Polynomial Functions: A polynomial function is a function of the form f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0, where n is a non-negative integer, a_n, a_{n-1}, ..., a_1, a_0 are real numbers, and a_n ≠ 0. II. Degree and Leading Coefficient: Degree of a Polynomial: The highest exponent of the variable in a polynomial determines its degree. For example, in the polynomial f(x) = 3x^4 - 2x^2 + 5, the degree is 4. Leading Coefficient: The coefficient of the term with the highest exponent is called the leading coefficient. In the previous example, the leading coefficient is 3. III. Types of Polynomials: Constant Polynomial: A polynomial of degree zero, where all terms are constant. Example: f(x) = 5. Linear Polynomial: A polynomial of degree one. Example: f(x) = 3x + 2. Quadratic Polynomial: A polynomial of degree two. Example: f(x) = x^2 - 4x + 3. Cubic Polynomial: A polynomial of degree three. Example: f(x) = x^3 + 2x^2 - x - 2. IV. Operations with Polynomials: Addition and Subtraction: Polynomials can be added or subtracted by combining like terms. Example: (2x^2 + 3x + 4) + (x^2 - 2x - 1) = 3x^2 + x + 3. Multiplication: To multiply polynomials, distribute each term of one polynomial by each term of the other polynomial and combine like terms. Example: (2x + 3)(x - 4) = 2x^2 - 5x - 12. Division: Polynomial division involves dividing one polynomial by another. Synthetic division and long division methods can be used to perform polynomial division.