Algebra of. polynomials2

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1 polynomials2 Algebra of Polynomial functions are used to model many physical situations. One such example, the effect of gravity on falling objects, was investigated experimentally by Galileo Galilei ( ), who measured the effects of gravity on falling objects by rolling a ball down a groove on an inclined ramp. Galileo determined that, under gravity (in the absence of air resistance), all objects would fall at the same rate and that the distance, d, travelled by the rolling ball was proportional to the square of the time, t (i.e. d t 2 ). Furthermore, in the absence of air resistance, it was found later that the height, h, of a falling object above ground level at time t could be modelled by the quadratic function h(t) = at 2 + bt + c, where h(t) represents the object s height at time t and a, b and c are real coefficients representing key features of the object s motion:. a represents the effect of gravity. b represents the object s initial speed. c represents the object s initial position. The quadratic function h(t) = at 2 + bt + c is an example of a polynomial function.

2 MathsWorld Mathematical Methods Units 1 & What is a polynomial? Function notation Functions are usually denoted by lower case letters, such as f, g and h, and indicate the dependence of one quantity on another. If p is used to denote a function, then the equation y = p(x), read as y equals p of x, indicates that y is a function of x. The variable x is the independent variable and the variable y is the dependent variable of the function p. We assume throughout this chapter that x is any real number unless otherwise stated. Tip A facility with function notation, such as p(x), is important for developing a good mathematical understanding of algebra and calculus and for doing mathematics effectively with a CAS. Definition of polynomial functions Polynomials are formed from the addition of successive power functions of the form ax n, where n is a positive integer or zero and a is a constant. For example, if we add two power functions together, such as 2x and 3x 2, we obtain 3x 2 + 2x. For the polynomial function given by p(x) = 2x 2 + 5x + 7, the terms are 2x 2, 5x and 7, with 2x 2 called the leading term and 7 the constant term. By convention, polynomial functions are usually written in descending order of powers so that the leading term contains the largest power and is written first. The power of the independent variable in the leading term tells us the degree of a polynomial function. Thus, p(x) = 2x 2 + 5x + 7 is a polynomial function of degree 2. In each term, the number component is known as the coefficient of that term. Thus, for p(x) = 2x 2 + 5x + 7, the coefficients of each term are 2, 5 and 7 respectively, with 2 being the leading coefficient. Linear functions are polynomial functions of degree 1, as the highest power of the independent variable (i.e. x) is 1. Quadratic functions are polynomial functions of degree 2, as the highest power of the independent variable (i.e. x) is 2. Polynomial functions can also be of a degree greater than 2. For example:. cubic polynomials (degree 3): e.g. p(x) = 2x 3 3x quartic polynomials (degree 4): e.g. p(x) = x x + 6x 5 3 Terms p(x) = 2x 2 + 5x + 7 Leading term Constant term Coefficients p(x) = 2x 2 + 5x + 7 Leading coefficient 36

3 2 Algebra of polynomials Example 1 For each polynomial below, state its degree and give the coefficient of x: a p(x) = x 3 6x + 6 b q(x) = 3x 5 4x a The degree of p(x) is 3 as x 3 is the term containing the highest power of x. The coefficient of x is 6. b The degree of q(x) is 5 as x 5 is the term containing the highest power of x. The coefficient of x is Polynomial functions can be expressed in a variety of different forms. For example, the function p(x) = (x 2) 2 can be shown to be identical to x 2 4x + 4, and the function q(x) = (x 2)(x + 1) 3x can be shown to be identical to x 2 4x 2. So p(x) and q(x) are polynomial functions. How can we tell if a function is a polynomial function? 1 For example, p(x) = x + 4 and q(x) = x + 4 are not polynomial functions. Why not? x 2 1 In the case of p(x), it is not a polynomial function because ---- = x 2, which is a term with x 2 a negative power. Alternatively, p(x) is not a polynomial function because it contains a term with the independent variable x in the denominator. In the case of q(x), it is not a polynomial function because x = x Polynomial functions can contain only powers of x that are either positive integers or zero. Definition of a polynomial function A polynomial function, p(x), is a function of an independent variable, x, that can be expressed in the form p(x) = a n x n + a n 1 x n 1 + a n 2 x n a 1 x + a 0, where the coefficients a n, a n 1, a n 2,, a 1 and a 0 are real numbers and n is either a positive integer or zero. The degree of p(x) is n for a n 0 and is given by the highest power of x. The coefficient of the highest power of p(x) is the leading coefficient and the term that is independent of x is the constant term. Identical polynomials Two polynomials are identical if they:. are of the same degree. have the same number of terms. have corresponding terms with identical coefficients. 37

4 MathsWorld Mathematical Methods Units 1 & 2 Example 2 Find the values of a, b, c and d if the polynomial functions p(x) = 2x 4 + 8x 3 x 9 and q(x) = ax 4 2bx 3 + (c 3)x 3d are identical. By equating the coefficients of x 4, a = 2. The other coefficients can be found by equating coefficients and solving three linear equations in turn: 2b = 8 c 3 = 1 3d = 9 b = 4 c = 2 d = 3 Hence, p(x) and q(x) are identical polynomial functions if a = 2, b = 4, c = 2 and d = 3. Tip When using a CAS it is important to be able to recognise equivalent expressions and equations. The outputs may be expressed in puzzling forms and so moving between forms and identifying equivalence becomes a very important skill when doing and learning mathematics with a CAS. The TI-Nspire can be used to verify whether two polynomial functions are identical, as shown in the screenshot opposite. If two expressions are equivalent, the calculator output will be true. Addition and subtraction of polynomials Polynomial functions can be added and subtracted by collecting like terms and simplifying. Example 3 If p(x) = x 4 3x and q(x) = 2x 4 6x 2 + x 3, find: a p(x) + q(x) and state its degree b q(x) p(x) and state its degree a Substitute for p(x) and q(x) and then add or subtract like terms to simplify. p(x) + q(x) = (x 4 3x 2 + 4) + (2x 4 6x 2 + x 3) = 3x 4 9x 2 + x + 1 p(x) + q(x) is a polynomial function of degree 4. b Substitute for p(x) and q(x) and then add or subtract like terms to simplify. q(x) p(x) = (2x 4 6x 2 + x 3) (x 4 3x 2 + 4) = x 4 3x 2 + x 7 q(x) p(x) is a polynomial function of degree 4. 38

5 TI-Nspire 10.1 ClassPad Algebra of polynomials Tip With a CAS, polynomial functions can be defined or stored and then operated on. The screenshot opposite shows how each of p(x) and q(x) in example 3 can be defined or stored as a function and then q(x) p(x) found. 2.1 exercise State the degree of each polynomial function listed below. a p(x) = x 2 9x 80 b p(x) = 5x 3 6x 2 + c f(x) = 9x 2 + 8x 5 2 d h(t) = 0.09t t t Let p(x) = 2x 3 + 5x 7. a State the coefficient of x 3. b State the coefficient of x. c State the degree of q(x) where q(x) = x 4 + p(x). 3 Let f(x) = 2x 2 4x + 5 and g(x) = x 3 5x. Find: a i f(x) + g(x) ii f(x) g(x) b State the degree of f(x) g(x). 4 Let f(x) = 3x 3 4x and g(x) = x 3 + 4x 2. Find: a i f(x) + 2g(x) ii f(x) g(x) b c State the degree of f(x) + 2g(x). Does f(x) + f(x) = 2f(x)? Give supporting evidence for your answer. 5 Let f(x) = x 3 + 2x + 1 and g(x) = x 3 2x. a b c Find f(x) + g(x). State the degree of f(x) + g(x). Find f(x). 6 Let f(t) = 2t 3 5t 2 + t + 4 and h(t) = at 3 + bt 2 + ct + d, where a, b, c and d are real numbers. Under what conditions are f(t) and h(t) identical? 7 Let p(x) = 4x 3 + 7x 2 5x + 8 and q(x) = ax 3 7bx 2 + (c 1)x 2d, where a, b, c and d are real numbers. State the values of a, b, c and d so that p(x) and q(x) are identical. 8 Let f(x) = 3x 4 + 5x 2 + 3x 4 and g(x) = 3x 4 + (m n)x 2 + (m + n)x 4, where m and n are real numbers. Find the values of m and n so that f(x) and g(x) are identical. 9 Let f(x) = x 3 + (2a 2 a 1)x 2 2x and g(x) = x 3 + (a 2 a + 8)x 2 2x, where a is a real number. Find the values of a so that f(x) and g(x) are identical. 2 39

6 MathsWorld Mathematical Methods Units 1 & 2 10 Let p(x) = 2x 3 3x 2 + (3c 3)x 4d and q(x) = (2 a)x 3 b --x + 6x + 8, where a, b, c 2 2 and d are real numbers. State the values of a, b, c and d so that p(x) and q(x) are identical. 11 Let f(x) = 9. Is f(x) a polynomial? Give reasons. Substitution Substitution into a function p(x) means replacing x by a number or expression and evaluating or simplifying the result. For example, if p(x) = x 2, then p( 3) is ( 3) 2, which simplifies to 9. That is, p( 3) denotes the function s value when x = 3. We are mainly concerned with substituting numerical values into polynomial functions (i.e. numerical substitution). Tip It is expected that numerical substitution for simple cases including polynomial functions can be done without resorting to technology. A CAS should be used for substitutions that are either difficult to calculate or for multiple substitutions. Example 4 Given that p(x) = x 3 6x + 2, find: a p(2) b p( 1) a p(2) = (2) 3 6(2) + 2 b p( 1) = ( 1) 3 6( 1) + 2 = = = 2 = 7 Example 5 The quadratic function h(t) = 5t for t 0 describes the height, h metres, above ground level of a falling tennis ball after t seconds. a How far above ground level is the tennis ball after it has fallen for 3 seconds? b What does h(1.5) represent? a The height of the tennis ball above ground level after falling for 3 seconds is given by h(3). h(3) = 5(3) = 55 The tennis ball is 55 metres above ground level after falling for 3 seconds. b h(1.5) represents the height of the tennis ball above ground level after falling for 1.5 seconds. 40

7 TI-Nspire 10.1 ClassPad Algebra of polynomials Tip Numerical substitutions can be performed with a CAS by defining or storing the function and calculating a value directly as shown. 2.1 Multiple substitution and tables of values Using a CAS, substitutions with awkward numerical values and multiple substitutions, such as tables of values, can be made. Example 6 Lucas, Joe, Kai and Sam have been given the following problem to solve in their mathematics class. A square sheet of paper measures 24 cm by 24 cm. A square of side length x cm is cut out of each corner and the sides are then turned up to form an open rectangular box. Find the maximum volume of this open rectangular box. After some discussion on how to make such a box with maximum volume, Lucas thought 2 cm squares should be cut from the corners. Joe thought 3 cm squares, Kai thought 4 cm squares and Sam thought 5 cm squares should be cut from each corner. Which of these four suggestions produces an open rectangular box with the larger volume? Let V cm 3 represent the volume of the open rectangular box and x cm represent the side length of the square cut from each corner. We first find an expression for V in terms of x. Then, we substitute x = 2, 3, 4 and 5 into V to determine which of the four suggestions produces a box with the larger volume. continued 41

8 TI-Nspire 1.4, 2.2 ClassPad 1.4, 2.2 MathsWorld Mathematical Methods Units 1 & 2 The figures below help to construct an expression for the volume of this open rectangular box in terms of x. 24 cm x cm (24 2x) cm 24 cm (24 2x) cm x cm The volume of a rectangular box is given by V = lwh. For a height of x cm and a base length and width of (24 2x) cm respectively, then V(x) = x(24 2x) 2. For multiple substitutions, a table of values is useful. The table of values below displays each of Lucas, Joe, Kai and Sam s suggestions and indicates that, of these four, the open rectangular box of volume 1024 cm 3 gives the larger volume. This occurs when squares of side length 4 cm are cut out of each corner, which is Kai s suggestion. Substituting different values can also be performed at the home screen as shown. Tip An alternative to the Function Table is to use the Spreadsheet feature of a CAS. Insert the required values of the independent variable in column A. Then, just as in a standard computer-based spreadsheet, insert Al (24 2Al) 2 in cell B1 and fill down, as shown in the screenshots. 42

9 2 Algebra of polynomials exercise If p(x) = x 2 5x + 4, find: a p(1) b p( 2) 13 If f(x) = 2x 3 x 2 7x + 6, find: a f(0) b f(1) c f( 2) 14 If f(x) = x 2 1 2x + 6, find f continued If f(x) = x 3 x 2, find f The number of diagonals, d, in an n-sided regular polygon is given by the function 1 3 d(n) = --n n 2 a Find d(4). b What does d(4) represent? c Construct a table of values to show the number of diagonals for the following regular polygons. i pentagon ii hexagon iii octagon iv decagon v dodecagon (12 sides) d Does an equilateral triangle have any diagonals? Explain your answer briefly. 17 The height, h metres, above ground level of a falling stone after t seconds, t 0, is given by h(t) = 5t a Find the stone s initial height above ground level. b How far above ground level is the stone after it has fallen 2 seconds? 18 A paint manufacturer knows that the daily production cost, C dollars, of making x litres of paint is modelled by C(x) = 0.001x x + 5. a State the degree of C(x). b Find the daily production cost if the paint manufacturer makes: i 100 L of paint. ii 400 L of paint. 19 The area, A square metres, of a rectangular yard of width x metres is given by A(x) = 20x x 2. a Find the area of the rectangular yard if its width is: i 2 metres. ii 10 metres. b If the area of the rectangular yard is 64 square metres, use a table of values to find its dimensions. c Given that the maximum area of the yard occurs for an integer value of x, use a table of values to find the dimensions of the yard of maximum area. 43

10 MathsWorld Mathematical Methods Units 1 & 2 20 Everyone at a party shook hands with each other as a way of exchanging greetings. x The number of handshakes, H, exchanged between x people is given by H = -- ( x 1). 2 a Construct a table of values for the number of handshakes exchanged between 5, 10, 50 and 100 people, respectively. b If 136 handshakes were exchanged, use a table of values to determine how many people at the party shook hands. 21 A farmer has decided to construct a rectangular paddock with Existing fence 80 metres of fencing. One side of the rectangular paddock has an existing fence that does not require replacement. x a If the width of the paddock is x metres and its area is A square metres, find an expression for A(x). b Given that the maximum area occurs for an integer value of x, use a table of values to determine the maximum area that can be formed for this rectangular paddock. c What relationship exists between the length and the width of the paddock when it is of maximum area? Algebraic substitution The previous section discussed how numerical substitution is performed. Algebraic expressions can also be substituted for the independent variable in polynomial functions. Example 7 Consider p(x) = x 3 6x + 2. a i Find p( t). ii State the degree of p( t) as a polynomial function in t. b i Find p(t 2 ). ii State the degree of p(t 2 ) as a polynomial function in t. a i p( t) = ( t) 3 6( t) + 2 (substituting t for x) = t 3 + 6t + 2 ii p( t) is a polynomial function in t of degree 3. b i p(t 2 ) = (t 2 ) 3 6(t 2 ) + 2 (substituting t 2 for x) = t 6 6t ii p(t 2 ) is a polynomial function in t of degree 6. 44

11 TI-Nspire 10.1 ClassPad 10.1 TI-Nspire 10.1 ClassPad Algebra of polynomials Example 8 If f(x) = x + 2, find f (a 2 ) 2f (a) + 2 in terms of a. 2.1 f(a 2 ) 2f(a) + 2 = (a 2 + 2) 2(a + 2) + 2 = a a = a 2 2a Tip With a CAS, algebraic substitutions can be performed in the same way as numerical substitutions. The solution to example 8 is shown in the screenshot opposite. Functions of more than one variable In many applications, a function may involve more than one variable. For example, the volume, V, of a cylinder is given by V = πr 2 h where r is the radius of its circular base and h is its height. In function terms, we can write: V(r, h) = πr 2 h Thus, V(2, 1) would represent the volume of a cylinder of base radius 2 and height 1 and V(r, 3) would represent the volume of a cylinder with base radius r (a variable) and height 3 (a constant). Tip A CAS allows functions of two or more variables to be defined. In the case above, here is what we find: exercise 2.1 continued 22 Consider f(x) = x 2 + 4x 5. a i Find f( 2t). ii State the degree of f( 2t) as a polynomial function in t. b i Find f(t 3 ). ii State the degree of f(t 3 ) as a polynomial function in t. 45

12 MathsWorld Mathematical Methods Units 1 & 2 23 Consider g(x) = x 3 x 2. a i Find g(t 2 ). ii State the degree of g(t 2 ) as a polynomial function in t. b i Find g(t 8 ). ii State the degree of g(t 8 ) as a polynomial function in t. 24 Consider f(x) = 2x 2 + 4x 1. a Find f( t 3 ). b State the degree of f( t 3 ) as a polynomial function in t. 25 If f(x) = x 4, find f(a 2 ) f(a) in terms of a. 26 If f(x) = x + 2, find f(a 2 ) 2f(a) + 3 in terms of a. 27 The volume, V cm 3, of a cone with base radius r cm and height h cm is given by 1 V(r, h) = --πr. 3 h a Find the volume of a cone with height 5 cm and base radius 2 cm. b What would happen to the volume of a cone if its base radius were doubled? c What would happen to the volume of a cone if its height were trebled? 28 Let f(x) be a polynomial function of degree n. If t m is substituted for x in f(x), where m is a positive integer, what will be the degree of the new polynomial function in t? 29 A farmer has decided to construct a rectangular yard with p metres of fencing. One side of the rectangular yard is an Existing fence existing fence that does not require replacement. x a If the width of the yard is x metres and its area is A square metres, find an expression for A(x). p b The farmer knows that the maximum area of the yard occurs when x = p Find A -- and hence state the maximum area of the yard in terms of p. 4 c What relationship exists between the length and width of the yard for this maximum area? 30 Let f(x) = ax 2 + bx + c, where a, b and c are constants and a 0. a Find f b a b If k 0 and f(k) = f(2k), find k. 46

13 Finding polynomial coefficients instantly SAC analysis task 2 Algebra of polynomials Analysis task 1 finding polynomial coefficients instantly SAC 2.1 The aim of this analysis task is to consider a different way of determining the coefficients of a polynomial function of any degree. Part 1: Polynomials with single-digit non-negative coefficients a For each of the following, evaluate f(10), where the coefficients of f(x) in each case are single-digit non-negative coefficients. i f(x) = 7x + 5 ii f(x) = 2x + 9 iii f(x) = x 2 + 4x + 3 iv f(x) = 7x 2 + 3x + 7 v f(x) = 2x 3 + 3x 2 + 6x + 1 b From question a, suggest a rule relating f(10) to the coefficients of f(x). c Use the rule developed in question b to determine g(x) in each case below. i g(10) = 86, where g(x) is a linear function. ii g(10) = 312, where g(x) is a quadratic function. iii g(10) = 5869, where g(x) is a cubic function. Part 2: Polynomials with two-digit non-negative coefficients d For each of the following, evaluate f(100), where the coefficients of f(x) in each case are two-digit non-negative coefficients. i f(x) = 19x + 45 ii f(x) = 65x x + 42 iii f(x) = 23x x x + 88 e From question d, suggest a rule relating f(100) to the coefficients of f(x). f Use the rule developed in question e to determine g(x) in each case. i g(100) = , where g(x) is a quadratic function. ii g(100) = , where g(x) is a cubic function. g Given that h(10) = 247, where h(x) is a linear or quadratic function, what might h(x) be? h Consider h(x) = (x + 3)(x + 2) + (2x + 1)(x + 3) + (x + 2)(x + 5). Given that h(x) can be expressed in the form h(x) = ax 2 + bx + c, determine the values of a, b and c by using a carefully selected substitution

14 Polygonal numbers SAC analysis task MathsWorld Mathematical Methods Units 1 & 2 Part 3: More general cases i From the results obtained in parts 1 and 2, describe how you would find the coefficients of f(x) when the coefficients are non-negative and have k digits. So far only non-negative coefficients with k digits have been considered. What about coefficients with k digits that are negative or positive integers or zero? j Evaluate each of the following expressions where f(x) in each case is a polynomial consisting of single-digit integer coefficients. i f(100) + 909, where f(x) = 3x + 4. ii f(100) , where f(x) = 2x 2 3x + 5. iii f(100) , where f(x) = x 3 6x 2 + 3x 4. k From question j, construct a general rule that would enable someone to calculate the single-digit integer coefficients of a polynomial of any degree. Extension Construct a general rule that would enable someone to calculate the coefficients of a polynomial of any degree for the general case of integer coefficients with k digits. Analysis task 2 polygonal numbers 2.2 SAC A polygonal number is a number that can be arranged as a regular polygon. The diagrams below illustrate the first four triangular, square, pentagonal and hexagonal numbers, respectively. The first number in each case is 1 (i.e. we begin with one dot). Then additional dots are used to create the polygonal shape. A triangular number, for example, is a number that can be arranged in the shape of an equilateral triangle, as shown below. So the first four triangular numbers illustrated below are 1, 3, 6 and 10. a List the first four: i square numbers ii pentagonal numbers iii hexagonal numbers 48

15 2 Algebra of polynomials The rth polygonal number of an n-sided polygon, P, is given by the formula r P(n, r) = -- [( n 2)r ( n 4) ], where, for example, r = 4 and n = 3 correspond to 2 the fourth triangular number. b Find a formula for the rth triangular number, P(3, r), in terms of r. c Construct a table of values that includes the 10th, 30th, 70th and 200th triangular numbers. d If the rth triangular number, P(3, r), is 300, find the value of r. e State the formula for the rth pentagonal number, P(5, r), in terms of r. f If the rth pentagonal number, P(5, r), is 3725, find the value of r. g State the formula for the rth hexagonal number, P(6, r), in terms of r. h If the rth hexagonal number, P(6, r), is 2016, find the value of r. i Find a formula for the sum of two consecutive triangular numbers. What conclusion can you make? j Are there any numbers less than 500 (and greater than 1) which are both triangular and: i square? ii pentagonal?

16 MathsWorld Mathematical Methods Units 1 & Expansion Two or more polynomial functions can be multiplied together. For example, given f(x) and g(x), we can find f(x) g(x). The process of multiplying is often called expanding. When expanding, we generally want to express a polynomial in standard form, that is, p(x) = a n x n + a n 1 x n 1 + a n 2 x n a 1 x + a 0. Distributive law To expand a polynomial function, the distributive law is applied. The distributive law states that a(b + c) = ab + ac In this, the factor a is distributed to both terms b and c in the factor (b + c). Example 1 Express 4(x 1) 2(x + 3) in the form ax + b. 4(x 1) 2(x + 3) = 4x 4 2x 6 (expanding by the use of the distributive law) = 2x 10 This is in the form ax + b, with a = 2 and b = 10. Example 2 Express (x + 3)(x 5) in the form ax 2 + bx + c. (x + 3)(x 5) = x(x + 3) 5(x + 3) (using the distributive law) = x 2 + 3x 5x 15 (expanding both brackets) = x 2 2x 15 This is in the form ax 2 + bx + c, with a = 1, b = 2 and c = 15. Example 3 Express (x 2)(x 2 + 2x + 3) in the form ax 3 + bx 2 + cx + d. (x 2)(x 2 + 2x + 3) = x(x 2 + 2x + 3) 2(x 2 + 2x + 3) (using the distributive law) = x 3 + 2x 2 + 3x 2x 2 4x 6 = x 3 x 6 This is in the required form with a = 1, b = 0, c = 1 and d = 6. 50

17 2 Algebra of polynomials Perfect squares A special type of expansion is the perfect square expansion. This can be shown as: (x + a) 2 = (x + a)(x + a) = x(x + a) + a(x + a) = x 2 + ax + ax + a 2 = x 2 + 2ax + a 2 Similarly, (x a) 2 = x 2 2ax + a Perfect squares In expanded form, perfect squares consist of three terms:. (x + a) 2 = x 2 + 2ax + a 2. (x a) 2 = x 2 2ax + a 2 The perfect square expansion (x + a) 2 = x 2 + 2ax + a 2 can be represented in a diagram as shown. x a The square of side length (x + a) units with area (x + a) 2 square units has been split into two differentsized squares of area x 2 and a 2 square units respectively, and two rectangles each of area ax square units. This diagram helps show that: x x 2 ax (x + a) 2 = x 2 + ax + ax + a 2 = x 2 + 2ax + a 2 We can see from the general formula that two of the three terms in any perfect square expansion are positive square terms. The other term is twice the product of the original two terms. a ax a 2 Example 4 Express (2x + 3) 2 in the form ax 2 + bx + c. The general result can be used as follows. (2x + 3) 2 = (2x) 2 + 2(3)(2x) = 4x x + 9 (note that (2x) 2 = 4x 2 and not 2x 2 ) This is in the required form with a = 4, b = 12 and c = 9. Alternatively, the distributive law can be used: (2x + 3) 2 = (2x + 3)(2x + 3) = 2x(2x + 3) + 3(2x + 3) = 4x 2 + 6x + 6x + 9 = 4x x

18 MathsWorld Mathematical Methods Units 1 & 2 Example 5 Which of the following expressions are perfect squares for any real value of x? Give a reason in each case. a x 2 + 7x + 49 b 16x x 9 c 4x x a For x 2 + 7x + 49 to be a perfect square for any real value of x, the middle term would need to be ±2(x)(7) = ±14x. As the middle term is 7x, the expression is not a perfect square. b 16x x 9 is not a perfect square as 9 is not the square of a real number. (The first and last terms need to be positive square terms.) c 4x x is a perfect square. First express it in descending powers of x: 4x 2 20x As 4x 2 = (2x) 2 and 25 = 5 2, the middle term needs to be ±2(2x)(5) = ±20x. As it is 20x, we have: 4x 2 20x + 25 = (2x 5) 2. Example 6 If 9x 2 + px + 4 is a perfect square and p > 0, find the value of p. First note that 9x 2 = (3x) 2 and 4 = 2 2. The middle term will be ±2(3x)(2) = ±12x. By equating coefficients, p = ±12. But p > 0, so p = 12. Difference of two squares Expand (x + a)(x a). Thus: (x + a)(x a) = x(x a) + a(x a) = x 2 ax + ax a 2 = x 2 a 2 Thus the product of the two factors is the difference of two squares. It is usual to write this expansion in the reverse order since it is most useful in factorisation (see next section). Difference of two squares x 2 a 2 = (x + a)(x a) Example 7 Express (3x + 4)(3x 4) in the form ax 2 + bx + c. The general result can be used as follows (3x + 4)(3x 4) = (3x) (using the difference of two squares) = 9x 2 16 (note that (3x) 2 = 9x 2 and not 3x 2 ) This is in the required form with a = 9, b = 0 and c = continued

19 2 Algebra of polynomials Alternatively, the distributive law can be used: (3x + 4)(3x 4) = 3x(3x 4) + 4(3x 4) = 9x 2 12x + 12x 16 = 9x exercise Express each of the following in the form ax + b. a 2(x 1) b 3(3 x) c 4(x + 2) d 2(x + 3) 3(x + 1) e 5(x 2) + 2(x + 5) f 3(x 2) 2(x + 7) 2 Express each of the following in the form ax 2 + bx + c. a x(x 2) b 2x(x + 1) c 3x(x 4) d (x + 3)(x + 2) e (x + 4)(x + 1) f (x 3)(x 2) g (2x + 1)(x 2) h (x 3)(5 2x) i (x + 4)(4x 1) j (x + 2)(x 4) + (x 3)(x + 1) k (2x 1)(x + 4) (x + 1)(3x + 1) 3 Express each of the following in the form ax 2 + bx + c. a (x + 2) 2 b (x 5) 2 c (x 7) 2 d (2x 1) 2 e (5x + 3) 2 f (2 3x) 2 g ( 2x 1) 2 h (px + 3) 2 i (mx + n) 2 j 1 x k 6 1 x -- 2 l (1 x) Earlier in the chapter, it was stated that the function q(x) = (x 2)(x + 1) 3x is identical to x 2 4x 2. Verify that this statement is correct. 5 Which of the following expressions are perfect squares for all real values of x? Give a reason for each answer. a x 2 + 5x + 25 b 9x 2 6x 1 c x + 9x 2 6 If px 2 + 4x + 1 is a perfect square for any real value of x, find the value of p. 7 If 25x 2 + px + 4 is a perfect square for any real value of x and p < 0, find the value of p. 8 If 9x x + p is a perfect square for any real value of x, find the value of p. 9 Express each of the following in the form ax 2 + bx + c. a (x + 4)(x 4) b (x 8)(x + 8) c (2x 3)(2x + 3) d ( 3x 5) ( 3x + 5) e (6x 7)(6x + 7) f (mx 3)(mx + 3) 10 Express each of the following in the form ax 3 + bx 2 + cx + d. a x(x 1)(x + 1) b 2x(x 1) 2 c x(x + 3) 2 d (x 2 + x 1)(x + 1) e (x 1)(2x 2 + 3x 5) f (3 4x)(x 2 + 5) 53

20 MathsWorld Mathematical Methods Units 1 & 2 Perfect cubes A perfect cube is of the form (x + a) 3. It can be expanded as follows: (x + a) 3 = (x + a)(x + a)(x + a) = (x + a)(x + a) 2 = (x + a)(x 2 + 2ax + a 2 ) = x(x 2 + 2ax + a 2 ) + a(x 2 + 2ax + a 2 ) = (x 3 + 2ax 2 + a 2 x) + (ax 2 + 2a 2 x + a 3 ) = x 3 + 3ax 2 + 3a 2 x + a 3 Similarly, (x a) 3 = x 3 3ax 2 + 3a 2 x a 3. Perfect cubes. (x + a) 3 = x 3 + 3ax 2 + 3a 2 x + a 3. (x a) 3 = x 3 3ax 2 + 3a 2 x a 3 Example 8 Express (x 2) 3 in the form ax 3 + bx 2 + cx + d. The general result for a perfect cube can be used as follows. (x 2) 3 = x 3 3(2)x 2 + 3(2) 2 x (2) 3 (replacing a by 2 in (x a) 3 ) = x 3 6x 2 +12x 8 This is in the required form with a = 1, b = 6, c = 12 and d = 8. (Alternatively, the distributive law could be applied directly.) Sums and differences of two cubes Consider the expansion of (x + a)(x 2 ax + a 2 ): (x + a)(x 2 ax + a 2 ) = x(x 2 ax + a 2 ) + a(x 2 ax + a 2 ) = (x 3 ax 2 + a 2 x) + (ax 2 a 2 x + a 3 ) = x 3 ax 2 + ax 2 + a 2 x a 2 x + a 3 = x 3 + a 3 Thus, the product of the two factors is the sum of two cubes. Similarly, the following product gives us the difference of two cubes. (x a)(x 2 + ax + a 2 ) = x 3 a 3. It is usual to write these expansions in the reverse order since they are most useful in factorisation (see next section). 54

21 TI-Nspire 1.9 ClassPad Algebra of polynomials Sum/difference of two cubes. x 3 + a 3 = (x + a)(x 2 ax + a 2 ). x 3 a 3 = (x a)(x 2 + ax + a 2 ) 2.2 Example 9 Express (x + 3)(x 2 3x + 9) in the form ax 3 + bx 2 + cx + d. Comparing (x + 3)(x 2 3x + 9) to (x + a)(x 2 ax + a 2 ) we see that a = 3. Hence: (x + 3)(x 2 3x + 9) = x = x This is in the required form with a = 1, b = c = 0 and d = 27. (Alternatively, the distributive law could be used directly.) exercise Express each of the following in the form ax 3 + bx 2 + cx + d. a (x 1) 3 b (x + 3) 3 c (1 x) 3 d (2x 1) 3 e (2x + 3) 3 f (4x 5) 3 12 Express each of the following in the form ax 3 + bx 2 + cx + d. a (x + 1)(x 2 x + 1) b (x 2)(x 2 + 2x + 4) c (x + 4)(x 2 4x + 16) d 3(x 3)(x 2 + 3x + 9) continued Tip A CAS can be used to expand expressions, as shown in the screenshot opposite. However, it is important that by-hand skills are developed for the range of examples met in this chapter so far. 55

22 Proving algebraic formulas with geometry SAC analysis task MathsWorld Mathematical Methods Units 1 & 2 Analysis task 3 SAC proving algebraic formulas with geometry 2.3 In this analysis task, diagrams are constructed and used to represent formulas for expanding expressions, such as perfect squares and the difference of two squares. For each diagram, explain your reasoning carefully, label your completed diagram accurately and use the area formulas for a square and a rectangle to verify the correctness of your diagram. The figure opposite illustrates a geometric representation of the perfect square formula (x + a) 2 = x 2 + 2ax + a 2. a Construct a diagram to illustrate that x a x x 2 ax a ax a 2 (x + 3)(x + 2) = x 2 + 5x + 6 for all positive values of x. b Construct a diagram to determine whether or not x x + 81 is a perfect square for all positive values of x. c Use the diagram of two squares of side length x units and 3 units, respectively, as shown below, to demonstrate that (x 3) 2 = x 2 6x + 9 for all positive values of x. x 3 d Hence, construct a similar diagram to demonstrate the general result that (x a) 2 = x 2 2ax + a 2 for all positive values of x. e Construct a diagram to demonstrate that (x 3)(x + 3) = x 2 9 for all positive values of x. f Hence, construct a diagram to demonstrate that (x a)(x + a) = x 2 a 2 for all positive values of x. g Construct a diagram to demonstrate that (x + a) 2 (x a) 2 = 4ax for all positive values of x. 56

23 2 Algebra of polynomials 2.3 Factorisation 2.3 Polynomial functions can be expressed in many different forms. For example:. f(x) = (x + 3)(x 2)(2x 3) (1). g(x) = 2x 3 x 2 15x + 18 (2). h(x) = (x 1)(2x 2 + x 14) + 4 (3) Are any of these functions equivalent for all real values of x, although expressed differently? One way to find this out might be to expand (1) and (3) and then compare the results. For example, here is what we get if we expand (3). (x 1)(2x 2 + x 14) + 4 = x(2x 2 + x 14) 1(2x 2 + x 14) + 4 = 2x 3 + x 2 14x 2x 2 x = 2x 3 + x 2 2x 2 14x x + 18 = 2x 3 x 2 15x + 18 By expanding (3), we see that (2) and (3) are equivalent for all real values of x. How could the form of (2) be changed to determine whether it is equivalent to (1)? This requires the opposite algebraic process of expanding, which is factorising. When factorising, we aim to write an algebraic expression as a product of its linear factors where possible. An example of this is f(x) = (x + 3)(x 2)(2x 3), which is a product of three linear factors. Similarly, in arithmetic, when we express 6 as a product of its prime factors, we write 6 as 2 3. Hence, 2 and 3 are factors of 6. Polynomials can be factorised using a variety of techniques. Highest common factor An important factorisation technique is to take out the highest common factor (HCF) where possible. This is best thought of as applying the distributive law in reverse. Example 1 Factorise 6x x. The HCF of 6x 2 and 12x is 6x. So: 6x x = 6x(x + 2) (taking out 6x as the HCF) Example 2 Factorise 5mt 2 20m 2 t. The HCF of 5mt 2 and 20m 2 t is 5mt. So: 5mt 2 20m 2 t = 5mt(t 4m) (taking out 5mt as the HCF) 57

24 MathsWorld Mathematical Methods Units 1 & 2 Factorising using the difference of two squares In section 2.2, we saw that the expansion of (x + a)(x a) is x 2 a 2. As factorisation is the reverse of expansion, we can express x 2 a 2 as a product of two linear factors: (x + a)(x a). This is known as the difference of two squares factorisation. Example 3 Factorise 5y y 2 45 = 5(y 2 9) (removing 5 as the HCF) = 5(y + 3)(y 3) (difference of two squares) Example 4 Factorise 25x 2 (m x) 2. 25x 2 (m x) 2 = (5x) 2 (m x) 2 = [5x + (m x)][5x (m x)] (difference of two squares) = (4x + m)(6x m) Warning Factorising using perfect squares In section 2.2 we saw that (x + a) 2 = x 2 + 2ax + a 2 and (x a) 2 = x 2 2ax + a 2. As factorisation is the reverse of expansion, we should check a quadratic expression to see if it is a perfect square. For example, 9x 2 12x + 4 is a perfect square as 9x 2 12x + 4 = (3x 2) 2. Example 5 Factorise y y y y + 25 = y 2 + 2(5)(y) = (y + 5) 2 Brackets, brackets, brackets! It is essential that brackets are used in situations like example 4, particularly when there are multiple negative signs. For example, if brackets are not kept in the second line, then it is easy to make the sign error (5x + m x)(5x m x) and hence produce the incorrect result (4x + m)(4x m). 58

25 2 Algebra of polynomials Example 6 Factorise 7x 2 28x x 2 28x + 28 = 7(x 2 4x + 4) = 7(x 2 2(2)(x) ) = 7(x 2) 2 (taking out 7 as the HCF) 2.3 exercise Factorise the following by finding the HCF. a 3p 3q b x c 6 12w d y y 2 e 2x 2 + 4x f 8y 2 6y g x 2 y 7xy h 3x 2 9x i xy + x 2 y 2 j a 2 b 2 + a 3 b 3 k xy + x 2 y 2 + x 3 y 3 l 8mn + 4m 3 n 10mn 4 2 Factorise the following using the difference of two squares. a x 2 16 b 36 t 2 c 4y 2 64 d x 2 y 2 16 e 5ax 2 45a f 20x 3 5xy 2 g 16y 2 (4 y) 2 h 3(x + 1) 2 27 i (x 4) 2 (x + 1) 2 3 Factorise the following by showing that they are perfect squares. a x 2 + 6x + 9 b m 2 8m + 16 c 5r 2 10r + 5 d 3x 2 24x + 48 e 18x xy + 8y 2 f 3x xy + 75y 2 4 Factorise the following. a 9s 2 25t 2 b 12 12b + 3b 2 c 4t 2 12t + 9 d (x y) 2 9 e (x y) 2 (x + 2y) 2 f (x 1) 2 + 4(x 1) + 4 Factorising quadratic expressions Up to this stage, we have found factors of quadratic expressions by first removing any common factors, and then checking whether the difference of two squares applies or whether the expression is a perfect square. When these methods do not apply, we can use our knowledge of expansion to find linear factors involving integers when they exist. Warning It doesn t always work Factorising quadratic expressions as the product of linear factors by the following methods only applies when the linear factors involve integers. In general, quadratic expressions may have linear factors including surds, or may not have linear factors with real numbers at all. 59

26 MathsWorld Mathematical Methods Units 1 & 2 From our earlier work on expansion, we know that (x + m)(x + n) = x(x + n) + m(x + n) = x 2 + nx + mx + mn = x 2 + (m + n)x + mn Note that the constant term is the product of two numbers and the coefficient of x is the sum of these numbers. So to factorise a quadratic expression, we look at the constant term, find its possible factors, and choose the factors so that their sum is the coefficient of x. Example 7 Factorise: a x 2 + 5x + 6 b x 2 7x + 6 c x 2 5x 6 a The constant term 6 has integer factors 6 and 1, 6 and 1, 3 and 2 or 3 and 2. Which combination has a sum of 5, the coefficient of x? Clearly, = 5, so the factorisation is: x 2 + 5x + 6 = (x + 3)(x + 2) b From part a, we want a pair of factors with a sum of 7. Clearly, 6 + ( 1) = 7, so the factorisation is: x 2 7x + 6 = (x 6)(x 1) c The integer factors of 6 are 6 and 1, 6 and 1, 3 and 2 or 3 and 2. As = 5, the factorisation is: x 2 5x 6 = (x 6)(x + 1) Tip We can find all the sums by writing the factors of the constant term vertically. In example 7 part c, this would be: We now pick out the factors with the sum we want: 6 and 1 in this case Factors Sums Example 8 Factorise: a x 2 + 6x 16 b x 2 8x 20 Using the above tip, find the factors of the constant term and their sums. a The constant term is 16. b The constant term is Factors Sums Factors Sums The coefficient of x is 6, so: The coefficient of x is 8, so: x 2 + 6x 16 = (x + 8)(x 2) x 2 8x 20 = (x 10)(x + 2) 60

27 2 Algebra of polynomials For quadratic expressions where the coefficient of x is not 1, the first step is to see whether there is an HCF that can be removed at the start of the factorisation process. 2.3 Example 9 Factorise 2x 2 + 6x x 2 + 6x + 4 = 2(x 2 + 3x + 2) = 2(x + 2)(x + 1) (taking out 2 as the HCF) The next example shows what we can try when all else fails. In this course, only simple cases where the coefficient of x 2 is a small positive integer are considered. Example 10 Factorise: a 2x 2 x 6 b 3x 2 8x + 4 a As the coefficient of x 2 is 2 (and the HCF is 1), we need to modify the previous method. The factors of 2x 2 are 2x and x. The constant term 6 has integer factors 6 and 1, 6 and 1, 3 and 2 or 3 and 2. Consider all the ways to form a pair of linear factors and find the x term in each case when we expand their product. Possible factorisation Line (5) gives the x term we require, so: 2x 2 x 6 = (2x + 3)(x 2) Resulting x term (1) (2x + 6)(x 1) 6x 2x = 4x (2) (2x 1)(x + 6) x + 12x = 11x (3) (2x 6)(x + 1) 6x + 2x = 4x (4) (2x + 1)(x 6) x 12x = 11x (5) (2x + 3)(x 2) 3x 4x = x (6) (2x 2)(x + 3) 2x + 6x = 4x (7) (2x 3)(x + 2) 3x + 4x = x (8) (2x + 2)(x 3) 2x 6x = 4x Note: We can simplify the working by observing that the term in x is found by crossmultiplying the numbers in the brackets. So set out the working as follows, multiplying along the arrows and adding: 2x x Coefficient of x : The one we want is circled. So 2x 2 x 6 = (2x + 3)(x 2) as before. continued 61

28 MathsWorld Mathematical Methods Units 1 & 2 b The factors of 3x 2 are 3x and x. The constant term +4 has integer factors 4 and 1, 4 and 1, 2 and 2 or 2 and 2. Use the cross-product method shown in part a: 3x x Coefficient of x: So 3x 2 8x + 4 = (3x 2)(x 2). Tip We can often reduce the amount of work by some simple observations. In example 10 part a, since the HCF was 1, then neither of the linear factors could possibly have an HCF > 1. So combinations (1), (3), (6) and (8) could all be rejected, as one of the linear factors had an HCF of 2. In the cross-product method, we could have simply tried: 2x x In example 10 part b, the constant term was positive but the term in x was negative. So only negative factors could work and we could have simply tried: 3x x exercise Factorise the following. a x 2 + 7x + 6 b r 2 + 8r + 15 c x x + 18 d p 2 5p + 6 e y 2 12y + 27 f x 2 + 2x 15 g x 2 + x 56 h x 2 x 6 i x 2 2x 48 j 2x 2 7x + 6 k 3m 2 5m 8 l 4t 2 + 9t + 2 m 3x 2 15x + 18 n 2m 2 18m + 28 o 4t 2 36t Simplify the following. x a 2 4 x b 2 1 x c 2 7x d ( x 2) 2 x 2 + 3x + 2 x 2 9 Further factorisation techniques In section 2.2, it was noted that:. (x + a) 3 = x 3 + 3ax 2 + 3a 2 x + a 3. (x a) 3 = x 3 3ax 2 + 3a 2 x a 3 These can be used to factorise in special cases. x 2 x 2 continued 5x x

29 2 Algebra of polynomials Example 11 Express x 3 + 9x x + 27 in the form (x + a) 3. Try to write the expression in the form that corresponds to a perfect cube: x 3 + 9x x + 27 = x 3 + 3(3)(x 2 ) + 3(3 2 )(x) + (3) 3 = (x + 3) In section 2.2, we found the following results for the sum and difference of two cubes:. x 3 + a 3 = (x + a)(x 2 ax + a 2 ). x 3 a 3 = (x a)(x 2 + ax + a 2 ) Example 12 Factorise the following: a x b 2x 3 16 a x = x = (x + 3)(x 2 3x + 9) (using the sum of two cubes with a = 3) Can the quadratic factor be factorised as the product of linear factors involving integers? From the coefficients of the quadratic, we require two integers whose sum is 3 and product is 9. As no pair of integers satisfy this, x 2 3x + 9 has no such linear factors. b This is a difference of two cubes in disguise. 2x 3 16 = 2(x 3 8) (taking out 2 as the HCF) = 2(x 2)(x 2 + 2x + 4) (using the difference of two cubes with a = 2) Can the quadratic factor be factorised as the product of linear factors involving integers? From the coefficients of the quadratic, we require two integers whose sum is 2 and product is 4. As no pair of integers satisfy this, x 2 + 2x + 4 has no such linear factors. exercise Express the following as perfect cubes. a x 3 6x x 8 b x 3 + 3x 2 y + 3xy 2 + y 3 c x x x + 64 d x 3 6x 2 y + 12xy 2 8y 3 8 Factorise the following. a x 3 1 b x c x 3 64 d 8x 3 1 e x 3 8y 3 f 27x g 8a b 3 h x i 8x 3 27y 3 continued 63

30 TI-Nspire 1.9 ClassPad 1.9 Exploring the coefficients of polynomials SAC analysis task MathsWorld Mathematical Methods Units 1 & 2 9 Factorise the following. a 2x b x 4 8x c x x d (x + 1) 3 (x 2) 3 e (x + 1) 3 (2x 1) 3 f 27 (1 x) 3 Tip A CAS can be used to factorise expressions, as shown in the screenshot opposite. However, it is important that by-hand skills are developed for the range of examples met in this chapter so far. Analysis task 4 SAC exploring the coefficients of polynomials 2.4 This analysis task investigates the relationships between the coefficients of quadratic polynomials in factorised and expanded forms. a Expand (x + a)(x + 2) for 5 a 5, where a is an integer, and display your results in a table. b Use the results from question a to describe briefly how the coefficient of x is formed in each expansion. c Use the results from question b to describe briefly how the term independent of x is formed in each expansion. d Based on the above results, state a general expansion for (x + a)(x + b). e Use the results obtained above to write down the following quadratics in expanded form without using the distributive law. i (x + 4)(x + 3) ii (x 4)(x 3) iii (x 8)(x + 2) iv (x + 3) 2 v (x + 5)(x 4) vi x(x + 12) vii (x 7)(x + 7) f Given that x 2 + mx + n = (x + a)(x + b), where a, b, m and n are integers, express: i m in terms of a and b ii n in terms of a and b g Given that x 3 + mx 2 + nx + p = (x + a)(x + b)(x + c), where a, b, c, m, n and p are integers, express: i m in terms of a, b and c ii n in terms of a, b and c iii p in terms of a, b and c 64

31 2 Algebra of polynomials 2.4 Remainder and factor theorems Division of polynomials The process of the operation of division in arithmetic involves the:. divisor, which is the number that divides another number quotient. dividend, which is the number being divided. quotient, which is the result of the division dividend. remainder, which is the number that remains. divisor For example, the calculation of using long division is shown opposite with the quantities above identified remainder Thus, when the dividend 167 is divided by the divisor 6, the quotient is 27 and remainder is 5. This can be expressed in two equivalent ways: = or 167 = The division process Dividend Remainder = Quotient Divisor Divisor or Dividend = Divisor Quotient + Remainder Example 1 Divide x 2 + 5x + 3 by x + 1, stating the answer in each of the equivalent ways above. (1)(4) x + 4 x + 1 ) x 2 + 5x + 3 (2) x 2 + x (3) 4x + 3 (5) 4x + 4 (6) 1 x Thus, 2 + 5x = x or equivalently x + 1 x + 1 x 2 (1) ---- = x x (2) x(x + 1) = x 2 + x (3) (x 2 + 5x) (x 2 + x) = 4x; bring down the 3 to give 4x + 3 (4) 4x = 4 x (5) 4(x + 1) = 4x + 4 (6) (4x + 3) (4x + 4) = 1 Dividend Divisor Quotient Remainder This statement is true for all values of x. As the remainder is not zero, x + 1 is not a factor of x 2 + 5x + 3. Note that unlike division of numbers, the remainder could be a negative number. 65

32 TI-Nspire 1.9 ClassPad 1.9 MathsWorld Mathematical Methods Units 1 & 2 Example 2 Divide x + 2 into 2x 3 x 2 3x + 4, and state the quotient q(x) and the remainder r. The division is shown below. 2x 2 5x + 7 x + 2 ) 2x 3 x 2 3x + 4 2x 3 + 4x 2 5x 2 3x 5x 2 10x 7x + 4 7x Thus, q(x) = 2x 2 5x + 7 and r = 10. We can summarise the result as: 2x 3 x 2 3x + 4 = 2x x , or x + 2 x + 2 2x 3 x 2 3x + 4 = (x + 2)(2x 2 5x + 7) 10 Example 3 Let p(x) = 2x 2 + 9x 2. Divide p(x) by 2x 1, and express the answer in the form p(x) = (2x 1)q(x) + r. The division is shown below. x + 5 2x 1 ) 2x 2 + 9x 2 2x 2 x 10x 2 10x 5 3 Thus, q(x) = x + 5 and r = 3. So: 2x 2 + 9x 2 = (2x 1)(x + 5) + 3 Tip A CAS can perform the division process using the propfrac command. The screenshot opposite shows the result for example 1. 66

33 2 Algebra of polynomials exercise For each of the following, divide the first polynomial by the second, stating the quotient q(x) and the remainder r in each case. a x 2 5x + 7, x 2 b x 2 + 9x + 10, x + 1 c x 3 + 2x 2 x + 4, x 1 d x 3 3x 2 + 2x 7, x + 2 e x 3 5x 2 + 3x 2, x 3 f x 3 2x + 8, x + 1 g 2x 3 3x 2 + 2x 1, x 2 h x 3 + x 2 2x + 2, x For each of the following, divide the first polynomial by the second, stating the quotient q(x) and the remainder r in each case. a 3x 2 7x + 6, 3x 1 b 2x 2 + x 8, 2x + 3 c 2x 3 5x 2 + 6x 1, 2x 1 3 For each polynomial pair, p(x) and d(x), listed below, express p(x) in the form d(x) q(x) + r, where d(x) is the divisor, q(x) is the quotient and r is the remainder. a p(x) = x 2 4x + 6, d(x) = x + 4 b p(x) = x 3 + 8, d(x) = x + 2 c p(x) = x 3 6x x 6, d(x) = x 3 px ( ) 4 For each polynomial pair, p(x) and d(x), listed below, express dx ( ) in the form r q(x) , where d(x) is the divisor, q(x) is the quotient and r is the remainder. dx ( ) a p(x) = x 2 x + 3, d(x) = x + 2 b p(x) = x 3 + 3x 2 x 5, d(x) = x 1 c p(x) = x 3 6x x 6, d(x) = x 2 5 Consider p(x) = x 4 2x 3 + x 2 3x + 5 and d(x) = x 2. Find the quotient q(x) and remainder r when p(x) is divided by d(x). 6 Consider p(x) = x 4 + x 2 6 and d(x) = x + 1. Find the quotient q(x) and remainder r when p(x) is divided by d(x). 7 If a polynomial p(x) of degree n is divided by x a, what is the degree of the resulting quotient q(x)? Remainder theorem In this section, we introduce the remainder theorem and illustrate how a remainder can be calculated without the need for division. 2.4 Example 4 a If p(x) = x 2 + 5x + 3: i find p( 1). ii compare the answer with the long division result in example 1. b If p(x) = 2x 2 + 9x 2: i find p ii compare the result with the long division result in example 3. continued 67

34 MathsWorld Mathematical Methods Units 1 & 2 a i p( 1) = ( 1) 2 + 5( 1) + 3 = = 1 ii In example 1, we found that the remainder when x 2 + 5x + 3 was divided by x + 1 was 1. Thus, p( 1) is equal to the remainder in that example. b i p 1 = = = 3 ii In example 3, we found that the remainder when 2x 2 + 9x 2 was divided by 2x 1 was 3. Thus, p 1 is equal to the remainder in that example The results in example 4 suggest a simple way of finding remainders from division of a polynomial by a linear expression. If p(x) is divided by x a, then the remainder r is given by r = p(a). For instance, in example 1, we previously found that: p(x) = x 2 + 5x + 3 = (x + 1)(x + 4) 1. It is easy to see why p( 1) = 1 gives the remainder. In general, if we divide a polynomial p(x) by x a, then p(x) = (x a)q(x) + r, where q(x) is the quotient and r is the remainder. Substituting x = a gives: p(a) = (a a)q(a) + r = 0 q(a) + r = r Remainder theorem If p(x) is divided by x a, then the remainder r is given by r = p(a). Note that a can be a fraction. Example 5 Without dividing, find the remainder when x 3 4x 2 + 3x 2 is divided by x + 2. Let p(x) = x 3 4x 2 + 3x 2. The remainder r is given by p( 2). p( 2) = ( 2) 3 4( 2) 2 + 3( 2) 2 = = 32 Hence, p( 2) = 32 and so the remainder is