CfE Higher Mathematics Course Materials Topic 4: Polynomials and quadratics

SCHOLAR Study Guide CfE Higher Mathematics Course Materials Topic 4: Polynomials and quadratics Authored by: Margaret Ferguson Reviewed by: Jillian Hornby Previously authored by: Jane S Paterson Dorothy A Watson Heriot-Watt University Edinburgh EH14 4AS, United Kingdom.

First published 2014 by Heriot-Watt University. This edition published in 2016 by Heriot-Watt University SCHOLAR. Copyright 2016 SCHOLAR Forum. Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educational purposes within their establishment providing that no profit accrues at any stage, Any other use of the materials is governed by the general copyright statement that follows. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, without written permission from the publisher. Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the information contained in this study guide. Distributed by the SCHOLAR Forum. SCHOLAR Study Guide Course Materials Topic 4: CfE Higher Mathematics 1. CfE Higher Mathematics Course Code: C747 76

Acknowledgements Thanks are due to the members of Heriot-Watt University's SCHOLAR team who planned and created these materials, and to the many colleagues who reviewed the content. We would like to acknowledge the assistance of the education authorities, colleges, teachers and students who contributed to the SCHOLAR programme and who evaluated these materials. Grateful acknowledgement is made for permission to use the following material in the SCHOLAR programme: The Scottish Qualifications Authority for permission to use Past Papers assessments. The Scottish Government for financial support. The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA) curriculum. All brand names, product names, logos and related devices are used for identification purposes only and are trademarks, registered trademarks or service marks of their respective holders.

1 Topic 1 Polynomials and quadratics Contents 4.1 Looking back at quadratics from National 5.................... 3 4.1.1 Factorising: Difference of two squares................. 7 4.1.2 Factorising: Trinomials.......................... 8 4.1.3 Completing the square.......................... 10 4.1.4 Features of quadratics.......................... 15 4.1.5 Determining equations of quadratics.................. 21 4.1.6 Sketching quadratic functions...................... 26 4.1.7 Quadratic equations: Solving graphically................ 34 4.1.8 Quadratic equations: Solving by factorising............... 36 4.1.9 Quadratic equations: Solving using the quadratic formula....... 37 4.1.10 Determining and interpreting the nature of the roots using the discriminant................................ 38 4.2 Polynomials..................................... 44 4.2.1 Synthetic division............................. 46 4.2.2 Division by (x - a)............................. 48 4.2.3 Factor theorem.............................. 52 4.2.4 Factorising polynomials.......................... 53 4.3 Finding coefficients and solving polynomials................... 55 4.3.1 Finding coefficients of polynomials................... 55 4.3.2 Solving polynomials............................ 59 4.4 Determining functions from graphs......................... 61 4.5 Completing the square............................... 68 4.6 Solving quadratic equations and inequations................... 72 4.6.1 Solving quadratic equations....................... 72 4.6.2 Solving quadratic inequations...................... 75 4.7 The discriminant................................... 79 4.7.1 Nature of roots............................... 79 4.7.2 Points of intersection of lines and parabolas.............. 82 4.8 Points of intersection of a line and a curve or two curves............. 86 4.9 Learning points................................... 90 4.10 End of topic test................................... 92

2 TOPIC 1. POLYNOMIALS AND QUADRATICS Learning objectives By the end of this topic, you should be able to: identify when an expression is a polynomial; factorise polynomials; solve polynomial equations; apply factor theorem to find unknown coefficients; determine functions from graphs; use the method of completing the square to determine the turning point of a quadratic; solve quadratic equations and inequations; determine, interpret and use: the discriminant; the conditions for tangency; find the coordinates of the points of intersection of: a line and a curve; two curves.

TOPIC 1. POLYNOMIALS AND QUADRATICS 3 1.1 Looking back at quadratics from National 5 Summary Factorising When factorising always ask yourself three questions: 1. Is there a simple common factor? 2. Is it a difference of two squares? 3. Is it a trinomial? and remember you could have a simple common factor and a difference of two squares or a simple common factor and a trinomial. Completing the square Completing the square has a method which requires practice. You will need this to be able to find the turning point of a parabola. Always make the expression take the form ax 2 + bx + c. Write the expression in the form (x + k) 2 k 2 + c OR [(x + k) 2 k 2 ]+c. Tidy up your answer by calculating k 2 and expanding any square brackets []. Graphs of quadratics Features of quadratic functions The shape of the graph of a quadratic will be smiley if the x 2 term is positive e.g. y = 5x 2 looks like The shape of the graph of a quadratic will be sad if the x 2 term is negative e.g. y = x 2 looks like If the shape is smiley the nature will be a minimum. If the shape is sad the nature will be a maximum. The graph is symmetrical and the equation of the axis of symmetry will take the form x = a. The zeros or roots of a quadratic are the point(s) where the graph crosses the x-axis and will take the form (p,0). The y-intercept is the point where the graph crosses the y-axis and will take the form (0,c) where c can be identified from y = ax 2 + bx + c.

4 TOPIC 1. POLYNOMIALS AND QUADRATICS Determining equations of quadratic functions from graphs If the equation of the quadratic takes the form y = kx 2 : find the coordinates of a point on the graph (Note: you cannot use the origin); substitute the values of x and y into y = kx 2 ; calculate the value of k; state the equation of the function. If the equation of the quadratic takes the form y = ± (x a) 2 + b: find the coordinates of the turning point from the graph; replace a with the x-coordinate; replace b with the y-coordinate; state the equation of the function Sketching the graph of a quadratic function Find the coordinates of the y-intercept If the equation takes the form ax 2 + bx + c = 0 then (0,c). If the equation takes the form y = (x m)(x n) then substitute x = 0 into the function to calculate the y-coordinate (0, m n) If the equation takes the form y = (x a) 2 + b then substitute x = 0 into the function to calculate the y-coordinate. Find the coordinates of the zeros or roots If the equation takes the form y = (x m)(x n) then the roots are (m,0) and (n,0). If the equation takes the form y = ax 2 + bx + c you must factorise the expression first. Find the equation of the axis of symmetry Find the value in the middle of the zeros by inspection or calculating the average of the roots m + n 2. State the equation in the form x = m + n 2. If the equation takes the form y = (x a) 2 + b then the equation of the axis of symmetry is x = a. Find the coordinates of the turning point The x-coordinate is the value in the middle of the roots. Substitute for x into the equation of the function to determine the y-coordinate. If the function is in the form y = (x a) 2 + b then the turning point is (a,b).

TOPIC 1. POLYNOMIALS AND QUADRATICS 5 Sketching the graph Identify the shape of the function. Identify the nature of the turning point. Draw a set of axes. Plot the points for the roots, turning point and y-intercept. (You may not always know the coordinates of the roots.) Bearing in mind the shape, sketch the graph. Write the coordinates beside the roots, turning point and y-intercept on your graph. Label your graph with its equation e.g. y = x 2 +2x 3 Solving a quadratic equation graphically The zeros or roots of a quadratic are the point(s) where the graph crosses the x-axis and will take the form (p,0) and (q,0). The solution is x = p and x = q. Solving a quadratic equation by factorising A quadratic equation of the form y = ax 2 + bx + c must be factorised to take the form y = (x p)(x q). The roots are (p,0) and (q,0). To calculate the solution pull the brackets apart: x p = 0 and x q = 0. The solution is x = p and x = q. Solving a quadratic equation using the quadratic formula From a quadratic equation of the form y = ax 2 + bx + c you must identify the values of a, b and c. Substitute a, b and c into the quadratic formula. x = b± b 2 4ac 2a Remember to split the expression into: x = b+ b 2 4ac 2a x = b b 2 4ac 2a and

6 TOPIC 1. POLYNOMIALS AND QUADRATICS Identifying and interpreting the nature of the roots of a quadratic using the discriminant From a quadratic equation of the form y = ax 2 + bx + c you must identify the values of a, b and c. The discriminant of a quadratic is b 2 4ac. If b 2 4ac > 0 (positive), the roots are real and distinct. There are two solutions. If b 2 4ac = 0, the roots are real and equal. There is one solution. There are two really but they are both the same. If b 2 4ac < 0 (negative), there are no real roots. There are no solutions. If b 2 4ac is positive and a square number, the roots are real and rational. If b 2 4ac is positive and not a square number, the roots are real and irrational.

TOPIC 1. POLYNOMIALS AND QUADRATICS 7 1.1.1 Factorising: Difference of two squares Factorising: Difference of two squares The following example shows how to factorise the difference of two squares. Go online Examples 1. Problem: Factorise 49x 2 16 We have 49x 2 16 = 7 2 x 2 4 2 which is a difference of two squares 49x 2 16 = (7x 4)(7x +4) 2. Problem: Factorise 50a 2 18 We have a simple common factor of 2 giving 50a 2 18 = 2(25a 2 9) 25a 2 9 is a difference of two squares so 2(25a 2 9) = 2(5 2 a 2 3 2 ) hence 50a 2 18 = 2(5a 3)(5a +3) Factorising a difference of two squares practice Q1: Factorise e 2 81 Go online Q2: Factorise 16g 2 49h 2 Q3: Factorise 27y 2 12z 2

8 TOPIC 1. POLYNOMIALS AND QUADRATICS 1.1.2 Factorising: Trinomials A trinomial expression has three terms. Quadratic expressions are normally called trinomials. Here are examples of some trinomials, x 2 +2x +3 3a 2 +4a +1 4k 2 2k 2 2m 2 + 7m 15 Factorising a trinomial Go online The following example shows you how to factorise a trinomial expression. The factorised form of x2 + 11x + 30 is (x + a)(x + b) where a b = a + b = +11. +30 and 1. Begin by finding the factors of +30. 2. Find the pair which add to make +11... This gives us the values a = +6 and b = +5. So we can write... x2 + 11x + 30 = (x + 6)(x + 5). Multiply out the brackets to check. Examples 1. Problem: Factorise x 2 x 12. We want two numbers which add to make -1 and multiply to make -12.

TOPIC 1. POLYNOMIALS AND QUADRATICS 9 1 12 (sum = 11); 2 6 (sum = 4); 3 4 (sum = 1)... we nearly have it; the sign is wrong so change the signs... 3 4 (sum = -1). The numbers are 3 and -4. So we have x 2 x 12 = (x 4)(x +3). 2. Problem: Factorise 2y 2 +5y +2. As we now have 2y 2 finding the factorised form requires more thought. We want two terms which multiply to make 2y 2. 2y y =2y 2 so our brackets start (2y )(y ) Now we want two numbers which multiply to make +2, i.e. 2 1 or 1 2. In this example we need to check that the sum of the products of the inner and outer terms gives us the middle term 5y. Multiplying the inner terms gives 2y and multiplying the outer terms gives 2y (sum 4y) but the middle term we want is 5y so... Multiplying the inner terms gives 1y and multiplying the outer terms 4y (sum 5y). This gives us the middle term we want. So we have 2y 2 +5y +2=(2y +1)(y +2). 3. Problem: Factorise 14g 2 20g +6. We should always check for a simple common factor first. This question has a common factor of 2 giving 2(7g 2 10g +3). Next we want two terms to make 7g 2 7g g =7g 2 so our brackets start (7g )(g ) Now we want two numbers which multiply to make +3 i.e. 3 1or1 3. In this example we need to check that the sum of the products of the inner and outer terms gives the middle term 10g. (7g +3)(g +1) makes the product of the inner terms 3g and the outer terms 7g (sum 10g) we nearly have it but we wanted 10g... the sign is wrong but we know that 3 1 also makes 3... (7g 3)(g 1) = 7g 2 10g +3 Hence 14g 2 20g +6=2 ( 7g 2 10g +3 ) =2(7g 3)(g 1)

10 TOPIC 1. POLYNOMIALS AND QUADRATICS Factorising a trinomial practice Go online Q4: Factorise x 2 +9x +20. (Hint: To check whether your answer is correct, multiply out the brackets and you should get the original trinomial.) Q5: Factorise x 2 +6x 16. Q6: Factorise 3g 2 +4g +1. Q7: Factorise 2j 2 +2j 12. 1.1.3 Completing the square Some trinomials are also called quadratic expressions or quadratics. Here are some quadratics, x 2 +6x +5 y 2 3 n 2 +7n The general form of a quadratic expression is ax 2 + bx + c, where a cannot be equal to 0. The quadratic x 2 2x +4does not factorise but it is possible to rearrange it into the form (x 1) 2 +3. This form is the result of the process called completing the square. Key point After completing the square, a quadratic expression has the form (x + k) 2 + m An example will help to demonstrate the method. Examples 1. Problem: Express x 2 +6x 1 in the form (x + k) 2 + m by completing the square.

TOPIC 1. POLYNOMIALS AND QUADRATICS 11 Step 1 x 2 +6x =(x +3) 2 3 2 First, concentrate on the x 2 and x terms. Express these in the form (x + k) 2 k 2 To find k we halve +6. Here k = 3. Step 2 (x +3) 2 3 2 =(x +3) 2 9 Simplify the square term on the end. If you were to expand (x +3)(x +3)you would get x 2 +6x +9Since we only want the x 2 +6x all we have to do is -9. Step 3 x 2 +6x 1 =(x +3) 2 9 1 Now we have to include the -1 from the original expression. Step 4 x 2 +6x 1=(x +3) 2 10 Tidy up the expression. x 2 +6x 1 expressed in the form (x + k) 2 + m is (x +3) 2 10 where k = 3 and m = 10. 2. Problem: Express x 2 10x +2in the form (x + k) 2 + m by completing the square. Step 1 x 2 10x =(x 5) 2 5 2 First, concentrate on the x 2 and x terms. Express these in the form (x + k) 2 k 2. To find k we halve -10. Here k = 5. Step 2 (x 5) 2 5 2 =(x 5) 2 25 Simplify the square term on the end. If you were to expand (x 5)(x 5) you would get x 2 10x +25 Since we only want the x 2 10x all we have to do is -25. Step 3 x 2 10x+2 = (x 5) 2 25+2Now we have to include the +2 from the original expression. Step 4 x 2 10x+2 = (x 5) 2 23 Tidy up the expression. x 2 10x +2expressed in the form (x + k) 2 + m is (x 5) 2 23 where k = m = 23. 5 and

12 TOPIC 1. POLYNOMIALS AND QUADRATICS If, however, the coefficient of x 2 is negative then it is necessary to take the negative out as a common factor of the first two terms before following the steps above. It is usual for these types of questions to be presented slightly differently as shown below. Examples 1. Problem: Express 1 2x x 2 in the form by completing the square. m (x + k) 2 Step 1 x 2 2x +1 Change the order of the terms. To do this swap the first and last terms. Step 2 x 2 2x = [x 2 +2x] Take out the negative as a common factor of the first two terms. Step 3 [x 2 +2x] = [(x +1) 2 1 2 ] Concentrate on the x 2 and x terms. Express] the first 2 terms in the form [(x + k) 2 k 2 To do this we half +2. Simplify the square term on the end. Step 4 [(x +1)2 1 2 ]= [(x +1) 2 1] If you were to expand (x +1)(x +1)you would get x 2 +2x +1Since we only want the x 2 +2x all we have to do is +1. Step 5 x2 2x+1 = [(x +1) 2 1]+1Now we have to include the +1 from the original expression. Step 6 [(x +1)2 1]+1 = (x +1) 2 +1+1 Expand the square brackets. Step 7 (x +1) 2 +1+1 = (x +1) 2 +2Tidy up the expression. Step 8 (x +1)2 +2=2 (x +1) 2 Swap the terms around. 1 2x x 2 expressed in the form m (x + k) 2 is 2 (x +1) 2 where k = 1and m = 2.

TOPIC 1. POLYNOMIALS AND QUADRATICS 13 2. Problem: Express 7+8x x 2 in the form m (x + k) 2 by completing the square. 7+8x x 2 = x 2 +8x +7 = [ x 2 8x ] +7 [ = (x 4) 2 4 2] +7 [ ] = (x 4) 2 16 +7 = (x 4) 2 +16+7 = (x 4) 2 +23 = 23 (x 4) 2 This strategy for completing the square is worth remembering. Look at it now and the examples which follow in the interactivity. Completing the square The following activity shows a method for completing the square. Step 1 Swap the first and last terms around if the x 2 is negative. Step 2 If the x 2 is negative take the - out as a common factor of the first 2 terms. Step 3 Write the expression in the form ±[(x + k) 2 k 2 ]+c Step 4 Simplify the k 2 term. Step 5 Expand the square bracket if there is one. Step 6 Tidy up to leave the expression in the form ±(x + k) 2 + m. Step 7 If you wish you can change (x + k) 2 + m to m (x + k) 2. Go online Examples 1. Problem: Express x 2 8x +5in the form (x k) 2 + m by completing the square.

14 TOPIC 1. POLYNOMIALS AND QUADRATICS 2. Problem: Express 1+6x x 2 in the form m (x k) 2 by completing the square. Completing the square exercise Go online These questions are for practice only. Q8: Express x 2 4x 7 in the form (x + k) 2 + m. Q9: Express x 2 8x +7in the form (x + k) 2 + m. Q10: Express x 2 +6x 7 in the form (x + k) 2 + m. Q11: Express x 2 12x +2in the form (x + k) 2 + m. Q12: Express 14 4x x 2 in the form m (x + k) 2. Q13: Express 6 24x x 2 in the form m (x + k) 2. Q14: Express 3+4x x 2 in the form m (x + k) 2.

TOPIC 1. POLYNOMIALS AND QUADRATICS 15 1.1.4 Features of quadratics A quadratic function has a graph called a parabola. There are two basic shapes of parabola: A good way to describe the shape of this function is a smiley face. 1. The function has a minimum turning point because if you run your finger over the shape of the graph it decreases to a low point then turns and increases. This is the nature of the function. A parabola is symmetrical and has a line of symmetry. The equation of the line of symmetry takes the form x = a. The line of symmetry passes through the turning point so the coordinates of the turning point are (a,...). The line of symmetry is normally called the axis of symmetry. A good way to describe the shape of this function is a sad face. 2. The nature of this function is a maximum because if you run your finger over the shape of the graph it increases to a high point then turns and decreases. The y-intercept is the place where the graph crosses the y axis. The coordinates of the y-intercept are (0, c). The graph may cross the x axis in two places, one place or not at all. If the graph crosses the x axis these point(s) are called the zeros, roots or x-intercepts. The coordinates of the zeros or roots are (p,0)and (q, 0).

16 TOPIC 1. POLYNOMIALS AND QUADRATICS Key point The shape of the graph of a quadratic will be smiley if the x 2 term is positive e.g. y = 5x 2 looks like, The shape of the graph of a quadratic will be sad if the x 2 term is negative e.g. y = x 2 looks like, Example Problem: A quadratic function has the equation y = 2x 2 +5x 1. State the shape of the graph and its nature. The x 2 term is negative because its coefficient is -2 so it is a sad face and looks like this, The nature of the function is a maximum because it increases to a high point, turns and decreases.

TOPIC 1. POLYNOMIALS AND QUADRATICS 17 Examples 1. Problem: Identify the coordinates of the zeros and the y-intercept of the quadratic y = x 2 +6x + 8. The zeros or roots are the points where the graph crosses the x axis so we get (2,0) and (4,0). The y-intercept is the point where the graph crosses the y axis so we get (0,8). Notice the +8 on the end of the equation y = x 2 + 6x + 8, this also helps to identify the y-intercept. 2. Problem: Identify the equation of the axis of symmetry and the coordinates of the turning point of the quadratic y = x 2 +6x 8 y 2 4 x -8 The axis of symmetry goes vertically half way between the zeros. The zeros are at 2 and 4, 3 lies half way between 2 and 4. An easy way to calculate it is to find the average of the roots 2+4 2 = 6 2 = 3. So the equation of the axis of symmetry is x = 3. We now know that the maximum turning point is (3,...).

18 TOPIC 1. POLYNOMIALS AND QUADRATICS If we substitute x = 3 into y = x 2 +6x 8 we get, y = - (3) 2 + 6 3 8 = - 9 + 18 8 = 1 The coordinates of the maximum turning point are (3, 1). Identifying the features of a quadratic function exercise Go online Identifying the shape of a quadratic function Q15: Identify the shape of the quadratic y = 3x 2. a) b) Q16: Identify the shape of the quadratic y = 2x 2. a) b) Identifying the nature of a quadratic function Q17: Identify the nature of the quadratic y = 3x 2. a) Maximum b) Minimum Q18: Identify the nature of the quadratic y = 2x 2. a) Maximum b) Minimum

TOPIC 1. POLYNOMIALS AND QUADRATICS 19 Identifying the equation of the line of symmetry of a quadratic Q19: Identify the equation of the quadratic y = x 2 6x +8. Identifying the coordinates of the turning point of a quadratic function Q20: Identify the coordinates of the turning point of y = x 2 +8x +15.

20 TOPIC 1. POLYNOMIALS AND QUADRATICS Identifying the zeros of a quadratic function Q21: Identify the coordinates of the zeros or roots of the turning point of y = x 2 x 6. Identifying the y-intercept of a quadratic function Q22: Identify the coordinates of the y-intercept of the turning point of y = x 2 2x 3.

TOPIC 1. POLYNOMIALS AND QUADRATICS 21 1.1.5 Determining equations of quadratics Quadratic functions of the form y = kx 2 This is a graph of the quadratic function y = kx 2. The curve is called a parabola. Ifwe change the value of k notice how the shape of the quadratic changes. If we set k = 1 we will achieve the following this quadratic. Notice that the point (1,k) lies on this curve, in this case the point (1,1). Go online If we set k = 2 we will achieve this quadratic. Notice that the point (1,k) lies on this curve, in this case the point (1,-2). Key point If the equation of a quadratic function is of the form y = kx 2 then: the turning point will be at the origin, (0,0); the graph will pass through the point (1,k).

22 TOPIC 1. POLYNOMIALS AND QUADRATICS Example Problem: Write down the equation of the curve in the form y = kx 2. The curve passes through (1, 5) substitute x = 1 and y = 5 into y = kx 2. 5 = k 1 2 5 = k k = 5 The equation is y = 5x 2. Note we can pick the coordinates of any point where the curve passes through the corner of a grid square. Here we could have chosen (-1,5). The value for k would still be the same. The only point we cannot use is (0,0). Quadratic functions of the form y = kx 2 practice Go online Q23: Write down the equation of the curve in the form y = kx 2.

TOPIC 1. POLYNOMIALS AND QUADRATICS 23 Q24: Write down the equation of the curve in the form y = kx 2. Quadratic functions of the form y = (x - a) 2 +b This is a graph of the quadratic function y = (x a) 2 + b. You may recognise this as completed square form. The curve is called a parabola. If we change the values of a and b notice what happens to the turning point. Go online If we set a = 1and b = 1we will achieve the following quadratic. Notice what happens to the turning point. If we set a = 2 and b = 3 we will achieve the following this quadratic. Notice what happens to the turning point.

24 TOPIC 1. POLYNOMIALS AND QUADRATICS Quadratic functions of the form y = -(x - a) 2 +b Go online This is a graph of the quadratic function y = (x a) 2 + b. The curve is called a parabola. If we change the values of a and b notice what happens to the turning point. If we set a = 1 and b = 1 we will achieve the following this quadratic. Notice what happens to the turning point. If we set a = 3 and b = 3 we will achieve the following this quadratic. Notice what happens to the turning point.

TOPIC 1. POLYNOMIALS AND QUADRATICS 25 Key point If the equation of a quadratic function is of the form y = (x a) 2 + b or y = (x a) 2 + b then the coordinates of the turning point will be (a,b). Example Problem: This is the graph of a quadratic function whose equation is of the form y = (x a) 2 + b. Find the values of a and b and write down the equation of the function. By inspection the turning point is (-5,-1). Thus a = 5 and b = 1. The equation is y = (x ( 5)) 2 +( 1) or simply y = (x +5) 2 1. Quadratic functions of the form y = ±(x - a) 2 + b practice Q25: This is the graph of a quadratic function whose equation is of the form y = (x a) 2 + b. Find the values of a and b and write down the equation of the function. Go online

26 TOPIC 1. POLYNOMIALS AND QUADRATICS Q26: This is the graph of a quadratic function whose equation is of the form y = a) 2 + b. Find the values of a and b and write down the equation of the function. (x Go online 1.1.6 Sketching quadratic functions Sketching a quadratic graph from its equation (1) This is a graph of the quadratic function y = k(x a) 2 + b. When k = 1, a = 0 and b = 0 the parabola has y-intercept (0,0). Notice that the point (a,b) is a minimum turning point, in this case, (0,0) and the parabola has an axis of symmetry x = 0 If we set k = 1, a = 3 and b = 5 the parabola has y-intercept (0,4).

TOPIC 1. POLYNOMIALS AND QUADRATICS 27 Notice that the point (a,b) is a minimum turning point, in this case, (3,-5) and th parabola has an axis of symmetry x = 3 If we set k = 1, a = 4 and b = 2 the parabola has y-intercept (0,-14). Notice that the point (a,b) is a maximum turning point, in this case, (-4,2) and th parabola has an axis of symmetry x = 4. Key point If the equation of a quadratic function is of the form y = (x a) 2 + b then: the turning point at (a,b) will be a minimum; the axis of symmetry will have equation x = a; the y-intercept occurs when x = 0. Substitute x = 0into y = (x a) 2 + b to find the y coordinate of the y-intercept.

28 TOPIC 1. POLYNOMIALS AND QUADRATICS Key point If the equation of a quadratic function is of the form y = (x a) 2 + b then: the turning point at (a,b) will be a maximum; the axis of symmetry will have equation x = a; the y-intercept occurs when x = 0. Substitute x = 0 into y = (x a) 2 + b to find the y coordinate of the y-intercept. Example Problem: A quadratic function has the equation y = (x 3) 2 +1. a) Write down the coordinates of the turning point. b) Say whether it is a maximum or minimum. c) Give the axis of symmetry of the parabola. d) Write down the coordinates of the y-intercept. e) Sketch the graph a) The turning point is (3, 1). x = 3 is the value of x which makes the bracket zero. (i.e. x 3 = 0 gives x = 3) y = 1 is the value of y when the bracket is worth zero. b) It is a minimum turning point. The x 2 term is positive so we have a smiley face. c) The axis of symmetry is x = 3. The equation of the vertical line passing through the turning point. d) The y-intercept is (0,10). Any point on the y axis has an x coordinate of zero. If we substitute x = 0 we get: y = (0 3) 2 + 1 = ( - 3) 2 + 1 = 9 + 1 = 10

TOPIC 1. POLYNOMIALS AND QUADRATICS 29 e) Sketching a quadratic graph from its equation practice (1) Q27: A quadratic function has the equation y = (x +2) 2 +3. Go online a) Write down the coordinates of the turning point. b) Say whether it is a maximum or minimum. c) Give the axis of symmetry of the parabola. d) Write down the coordinates of the y-intercept. e) Sketch the graph Sketching a quadratic graph from its equation (2) This is a graph of the quadratic function y = (x m)(x n). The x-coordinate of the turning point is x = (m + n)/2. The y-coordinate of the turning point can be found by substituting x = (m + n)/2 into the equation of the function y = (x m)(x n). The coordinates of the minimum turning point are (0,0). The axis of symmetry x = (m + n) 2, exactly in the middle of m and n. When m = 0 and n = 0 notice that the equation of the function is y = x 2. The root, or x-intercept, the turning point and y-intercept are all at the same point and have coordinates (0,0). Go online

30 TOPIC 1. POLYNOMIALS AND QUADRATICS When m = 4 and n = 2 notice that the roots, or x-intercept, are (0,4) and (0,-2), the parabola has y-intercept (0,-8) and the axis of symmetry is x = 1. When m = 2 and n = 2 notice that the roots, or x-intercept, are (0,-2) and (0,2), the parabola has y-intercept (0,-4) and the axis of symmetry is x = 0.

TOPIC 1. POLYNOMIALS AND QUADRATICS 31 Key point If the equation of a quadratic function is of the form y = (x m)(x n) then: the roots have coordinates (m,0) and (n,0); the y-intercept has x coordinate = 0. We find the y-coordinate by substituting x = 0 into y = (x m)(x n); the axis of symmetry is a vertical line half way between the roots; the equation of the axis of symmetry is x = (m + n)/2; the x-coordinate of the turning point is (m + n)/2. Substitute x = (m + n)/2 into y = (x m)(x n) to find the y-coordinate of the turning point. Top tip If the equation of the quadratic function is of the form y = ax 2 + bx + c then: factorise to help find the roots. the coordinates of the y-intercept are (0,c). Examples 1. Problem: A quadratic function has the equation y = (x 2)(x 6). a) Write down the coordinates of the roots. b) Write down the coordinates of the y-intercept. c) Write down the equation of the axis of symmetry. d) State the nature of the turning point. e) Write down the coordinates of the turning point. f) Sketch the graph a) The roots are (2,0) and (6,0). The value(s) of x which make each bracket zero. b) The coordinates of y-intercept is (0,12). The value of y when x = 0. If we substitute x = 0 we get: y = (x 2) (x 6) = (0 2) (0 6) = ( - 2) ( - 6) = 12 c) The axis of symmetry is x = 4. The exact middle of 2 and 6 is (2 + 6) / 2 = 8 / 2 = 4

32 TOPIC 1. POLYNOMIALS AND QUADRATICS d) It is a minimum turning point. (x 2)(x 6) = x 2 8x + 12 so since x 2 is positive the shape is a smiley face and a minimum turning point. e) The minimum turning point is (4,-4). Substitute x = 4 into y = (x 2)(x 6) we get: y = (4 2) (4 6) = 2 ( - 2) = - 4 f) 2. Problem: A quadratic function has the equation y = x 2 2x 15. a) Write down the coordinates of the roots. b) Write down the coordinates of the y-intercept. c) Write down the equation of the axis of symmetry. d) State the nature of the turning point. e) Write down the coordinates of the turning point. f) Sketch the graph a) The roots are (-3,0) and (5,0). If we factorise x 2 + 2x 15 we get (x 5)(x +3). The value(s) of x which make each bracket zero. b) The coordinates of y-intercept is (0,-15). The value of y when x = 0. If we substitute x = 0 we get: y = (0 5) (0 + 3) = ( - 5) 3 = - 15 c) The axis of symmetry is x = 1. The exact middle of -3 and 5 is ((-3) + 5) / 2 = 2 / 2 = 1 d) It is a minimum turning point. Since x 2 is positive in the original equation the shape is a smiley face and a minimum turning point.

TOPIC 1. POLYNOMIALS AND QUADRATICS 33 e) The minimum turning point is (1,-16). Substitute x = 1 into y = (x 5)(x +3)we get: y = (1 5) (1 + 3) = ( - 4) 4 = - 16 f) Sketching a quadratic graph from its equation practice (2) Q28: A quadratic function has the equation y = (x + 3)(x 1). Go online a) Write down the coordinates of the roots. b) Write down the coordinates of the y-intercept. c) Write down the equation of the axis of symmetry. d) State the nature of the turning point. e) Write down the coordinates of the turning point. f) Sketch the graph Q29: A quadratic function has the equation y = x 2 +5x +6. a) Write down the coordinates of the roots. b) Write down the coordinates of the y-intercept. c) Write down the equation of the axis of symmetry. d) State the nature of the turning point. e) Write down the coordinates of the turning point. f) Sketch the graph

34 TOPIC 1. POLYNOMIALS AND QUADRATICS 1.1.7 Quadratic equations: Solving graphically Quadratic equations are solved to find the value or values of x where the graph crosses the x-axis. Solving quadratic equations graphically Go online Given a graph of a quadratic function you should be able to find the solutions of the function, that is, the values of x when y = 0. Use the following examples to see how this is done.

TOPIC 1. POLYNOMIALS AND QUADRATICS 35 Key point The solutions of the equation ax 2 + bx + c = 0 will be the x-coordinates of the points where the graph of y = ax 2 + bx + c cuts the x-axis. Example Problem: The diagram shows the graph of the function y = x 2 +2x 8. Use the graph to solve the equation x 2 +2x 8 = 0. The curve cuts the x-axis at - 4 and 2. So the solutions to the equation are x = 4 and x = 2. Q30: The diagram shows the graph of the function y = x 2 +2x 3. Use the graph to solve the equation x 2 +2x 3 = 0.

36 TOPIC 1. POLYNOMIALS AND QUADRATICS Go online 1.1.8 Quadratic equations: Solving by factorising Solving quadratic equations by factorisation x 2 +3x +2is a simple quadratic expression. It can be factorised as (x + 2)(x +1). The coefficient of x is 3... this comes from 2+1. The constant term is 2... this comes from 2 1. Key point The general factorised form is (x + m)(x + n)...... where the coefficient of x is m + n... and the constant term is m n. Examples 1. Problem: Factorise x 2 +4x 12. Step 1: List the pairs of numbers that multiply to make -12. Step 2: Work out the sum of each pair. Step 3: Pick the pair that sum to + 4 which is -2 and +6. Step 4: Write down the factors of the expression: (x 2)(x + 6). 2. Problem: Solve x 2 +4x 12 = 0. Step 1 First factorise the equation to give (x 2)(x +6)=0. Step 2 (x 2)(x +6)=0will be true if either bracket is equal to zero. (x 2) = 0 or (x + 6) = 0 Step 3 Solve x 2 = 0 or x +6=0to give x = 2 or x = 6. Therefore the solution to x 2 +4x 12 = 0 is x = 2 or x = 6.

TOPIC 1. POLYNOMIALS AND QUADRATICS 37 Solving quadratic equations by factorisation practice Q31: Factorise x 2 +2x 15. (Remember:You can check whether your answer is correct by multiplying out the brackets. If you are correct, the result should give the original quadratic.) Go online Q32: Solve x 2 + x 6 = 0. Q33: Solve 2x 2 +9x 18 = 0. Remember as we now have 2x 2 finding the factorised form requires more thought. Q34: Solve 1 2x x 2 = 0. 1.1.9 Quadratic equations: Solving using the quadratic formula Sometimes it is not possible to factorise the quadratic and the graph has not been provided. In this situation you need the quadric formula to be able to solve the quadratic equation. Key point For a quadratic equation of the form ax 2 + bx + c = 0. x = b ± b 2 4ac 2a i.e. the solutions to the equation ax 2 + bx + c = 0 are: x = b + b 2 4ac 2a x = b b 2 4ac 2a. and Top tip The Quadratic Formula is on the National 5 Formula Sheet which will be given to you in the exam. Usually, the quadratic formula would be used to solve ax 2 + bx + c = 0 if it cannot be easily factorised but a big hint is given when you are asked to solve the quadratic equation to a number of decimal places or significant figures. Example Problem: Solve x 2 +3x 5 = 0. Comparing the equation x 2 +3x 5 = 0with the standard one of ax 2 + bx + c = 0 we see that a = 1, b = 3 and c = 5. Substituting these values into the quadratic formula gives:

38 TOPIC 1. POLYNOMIALS AND QUADRATICS x = b ± b 2 4ac 2a x = 3 ± 3 2 4 1 ( 5) 2 1 = 3 ± 9 + 20 2 x = 3 + 29 or x = 3 29 2 2 x = 1 19 and x = 4 19 (to 3 s.f.) Top tip If after substituting into the Quadratic Formula you find you have a negative value underneath the square root you have made a mistake. Go back and check the steps in your working. It is not possible to find the square root of a negative number. Solving quadratic equations using the quadratic formula practice Go online Q35: Solve x 2 +2x 7 = 0. Q36: Solve 2x 2 5x +1=0giving your solutions correct to 1 d.p. Solving quadratic equations using the quadratic formula exercise Go online Q37: Solve x 2 x 4 = 0 using the quadratic formula. Q38: Solve x 2 6x 8 = 0 using the quadratic formula, give your answer correct to 1 d.p. Q39: Solve 2x 2 2x 4 = 0 using the quadratic formula, give your answer correct to 1 d.p. 1.1.10 Determining and interpreting the nature of the roots using the discriminant Key point The discriminant of the quadratic equation ax 2 + bx + c = 0 is b 2 4ac. Notice b 2 4ac is the expression underneath the square root in the Quadratic Formula.

TOPIC 1. POLYNOMIALS AND QUADRATICS 39 Key point The following conditions on the discriminant hold: If b 2 4ac < 0, there are no real roots. If b 2 4ac = 0, the roots are real and equal. If b 2 4ac > 0, the roots are real and distinct. If the discriminant is a positive square number then the roots are real and rational. If the discriminant is positive but not a square number then the roots are real and irrational. These conditions can be related to the position of the graph of a quadratic with respect to the x-axis. Roots of an equation Go online No Real Roots The discriminant is less than zero. The graph never crosses the x-axis. Real and Equal Roots The discriminant is equal to zero. The graph only touches the x-axis at one point.

40 TOPIC 1. POLYNOMIALS AND QUADRATICS Real and Distinct Roots The discriminant is greater than zero. The graph crosses the x-axis at two distinct points. Here are some examples of both the graphs and the expressions in relation to the discriminant. Roots and the discriminant Go online An equation with no real roots has a discriminant with a negative value. Therefore the graphs do not cross the x-axis. An equation with real and equal roots has a discriminant equal to zero. Therefore the graph touches the x-axis at only one point. An equation with real and distinct roots has a discriminant with a positive value. Therefore the graphs cross the x-axis at two distinct points. Example : Real and distinct roots Problem: Determine the nature of the roots of the quadratic y = x 2 2x 2 From the equation of the quadratic we get a = 1, b = 2 and c = 1. Using the discriminant this gives b 2 4ac = ( 2) 2 4 1 ( 1) = 8 Since the discriminant is greater than zero the roots are real and distinct. i.e. there are 2 solutions to x 2 2x 1 = 0 The shape of the graph is a smiley face and looks like this,

TOPIC 1. POLYNOMIALS AND QUADRATICS 41 Examples 1. Equal real roots Problem: Determine the nature of the roots of the quadratic y = x 2 +6x +9 From the equation of the quadratic we get a = 1, b = 6 and c = 9. Using the discriminant this gives b 2 4ac = 6 2 4 1 9 = 0 Since the discriminant is equal to zero the roots are real and equal. i.e. there are really 2 solutions to x 2 +6x + 9 = 0 but they are the same so we think of this quadratic as having 1 solution. The shape of the graph is a smiley face and looks like this, 2. No real roots Problem: Find the roots of the quadratic y = x 2 x +4 From the equation of the quadratic we get a = 1, b = 1 and c = 4. Using the discriminant this gives b 2 4ac = ( 1) 2 4 1 4 = 15 Since the discriminant is less than zero there are no real roots. i.e. there are no solutions to x 2 x + 4 = 0. This is because we cannot find the square root of a negative number. The shape of the graph is a smiley face and looks like this,

42 TOPIC 1. POLYNOMIALS AND QUADRATICS Determining and interpreting the nature of the roots using the discriminant exercise Go online Q40: Examine the discriminant of these quadratics, determine the nature of the roots and state whether they are: real and distinct real and equal no real roots a) 7x 2 3x 8 b) 7x 2 3x + 8 c) 2x 2 8x + 8 d) x 2 +6x +24 Q41: Identify the position of the following graphs. Do they touch; avoid; cross the x-axis? real and distinct real and equal no real roots a) x 2 +8x +32 b) 7x 2 3x 4 c) 4x 2 8x + 4 d) 7x 2 3x + 4 The discriminant can tell us more about the roots. Example Problem: Prove that the roots of the equation 5x 2 +2x 7 = 0 are real and rational. There are two ways of proving this. 1. Factorise the quadratic equation. 5x 2 + 2x 7 = 0 (5x + 7)(x 1) = 0 5x + 7 = 0 or x 1 = 0 5x = 7 or x = 1 The solutions 7 5 and 1 are both rational numbers.

TOPIC 1. POLYNOMIALS AND QUADRATICS 43 2. Find the discriminant. b 2 4ac = 2 2 4 5 ( 7) = 4 + 140 = 144 Since 144 is a positive square number (i.e. 122 = 144) we know that the roots are real and rational. This is because the discriminant is the part of the quadratic formula under the square root. If we were to find the solutions using the quadratic formula we would get, x = 2 ± 144 2 5 2 + 12 2 12 x = or x = 10 10 again we find that both solutions are rational. Note: It is quicker to calculate the discriminant to determine whether the roots are rational or irrational. Key point If the discriminant is a positive square number then the roots are real and rational. Example Problem: Prove that the roots of the equation 3x 2 b 2 4ac = ( 8) 2 4 3 2 = 64 24 = 40 8x +2=0are real and irrational. Since 40 is positive but not a square number (i.e. 40 is an irrational number) we know that the roots are real and irrational. Key point If the discriminant is positive but not a square number then the roots are real and irrational. Rational and irrational roots practice Q42: Prove that the roots of the equation 2x 2 +3x 1 = 0 are real and irrational. Go online

44 TOPIC 1. POLYNOMIALS AND QUADRATICS Q43: Prove that the roots of the equation 3x 2 x 2 = 0 are real and rational. Rational and irrational roots exercise Go online Q44: Prove that the roots of the equation 6x 2 +5x +1=0are real and rational. Q45: Prove that the roots of the equation 3x 2 +6x 4 = 0 are real and irrational. Q46: Are the roots of the equation 5x 2 +6x 3 = 0 are real and rational or real and irrational? Q47: Are the roots of the equation 2x 2 13x + 3 = 0 are real and rational or real and irrational? 1.2 Polynomials A polynomial is an expression containing the sum or difference of algebraic terms with powers or the equivalent in factorised form e.g. 5x 4 + 2x 3 3x 2 6x +1or (x 2)(x +1)(x 4). The powers must be whole numbers. The powers cannot be negative or fractional. The degree of a polynomial is the value of the highest power. 5x 4 +2x 3 3x 2 6x +1 is a 4th degree polynomial because 5x 4 has the highest power. (x 2)(x +1)(x 4) = x 3 5x 2 x+ 8and is therefore a 3rd degree polynomial. In the polynomial 5x 4 +2x 3 3x 2 6x +1the coefficient of x 4 is 5. The coefficient of x 2 is 3. 1 is the constant term. Degrees of polynomials Go online x +3has degree 1. x 2 2x +3has degree 2. 4x 3 +2x 2 5 has degree 3. x 4 x has degree 4. and of course a constant such as 7 has degree 0. A root of a polynomial f(x) is a solution to the equation f(x) = 0.

TOPIC 1. POLYNOMIALS AND QUADRATICS 45 Examples 1. Problem: Which of the following expressions are polynomials and which are not? a) 5x 3 2x + 4 b) 3x 2 1 c) x 5 + 3x 4 2x 1 2 d) 2 3 x 2 + 7x a) Is a polynomial. It is a third degree polynomial with 5x 3 2x 1 + 4x 0 All powers are whole numbers. b) Is not a polynomial because the first term, 3x 2 1, has a negative power. c) Is not a polynomial because the last term, 2x 1 2, has a fractional negative power. d) Is not a polynomial because the first term 2 3 x 2 = 2x 3 2, has a fractional power. 2. Solution of an equation Problem: Find the solutions of the equation f(x) = x 2 +4x 5 where f(x) = 0. Factorise the left hand side to give: (x +5)(x 1) = 0 (x +5)=0or (x 1) = 0 x = 5 or x = 1 There are other methods of solving quadratic equations for example the quadratic formula but we also need a method which will allow us to solve higher degree polynomials. Determining polynomials exercise Q48: What is the degree of the polynomial 2x 6 +3x 4 5x + 2? Go online Q49: What is the degree of the polynomial 3 2x x 3? Q50: What is the degree of the polynomial 7x 9? Q51: What is the coefficient of 3x 5?

46 TOPIC 1. POLYNOMIALS AND QUADRATICS Q52: Is 4x -3 3x 2 a polynomial? Q53: Is 17x 5 a polynomial? Q54: Is x 5 + x 4 x 3 x -2 + x +5apolynomial? Q55: Is x 7 + x a polynomial? Q56: Is x2 + x 3 x 7 a polynomial? 1.2.1 Synthetic division Synthetic division is a method of performing polynomial long division, with less writing, fewer calculations and simpler arithmetic. Synthetic division for linear divisors is also called division through Ruffini s rule. Ruffini s rule is a technique for dividing a polynomial by a binomial of the form x a and is a special case of synthetic division when the divisor is a linear factor. Synthetic division Go online If f(x) = 4x 3 +5x 2 x +3. Substitute x = 2 into f(x). f (2) = 4 2 3 + 5 2 2 2 + 3 = 32 + 20 2 + 3 f (2) = 53 This substitution method can take some time especially if the value for x is large. We can shorten this method using synthetic division. Let s look at how to do synthetic division. If f(x) = 4x 3 +5x 2 x +3find f(2). First, layout the expression...... and then form the division.

TOPIC 1. POLYNOMIALS AND QUADRATICS 47 Bring down the first term and multiply by 2, from f(2). Add the 5 and 8, multiply the answer by 2. Repeat this process along the division. So when f(x) = 4x 3 +5x 2 x +3then f(2) = 53. Synthetic division practice Try out the method of synthetic division. It is really easy and will help when we factorise polynomials. Go online Q57: If f(x) = 2x 3 + x 2 3x +7, what is f( 2)? Q58: If f(x) = x 5 2x 4 +5x 3 2x 2 +3x 6, find f(3). Synthetic division exercise Q59: What is f(4) if f(x) = x 3 + 2x 2 5x 15? Go online Q60: What is f( 2) if f(x) = 2x 4 +3x 3 2x 2 +8x +7? Q61: What is f(2) if f(x) = 2x 5 +4x 3 6x +8? Q62: What is f( 5) if f(x) = x 5 +6x 4 + x 2 +9?

48 TOPIC 1. POLYNOMIALS AND QUADRATICS 1.2.2 Division by (x - a) We can divide polynomials using the method of synthetic division too. Key point The Remainder Theorem If a polynomial f(x) is divided by (x a) the remainder is f(a). Example Problem: f(x) = 3x 3 +5x +4 Use synthetic division to find (3x 3 +5x +4) (x 1). (x a) is the divisor and in this example a = 1. Using synthetic division we will be able to find the quotient and the remainder. Also, notice that we do not have a squared term so we must interpret this in a polynomial as 0x 2. The polynomial should be interpreted as 3x 3 +0x 2 +5x +4. So under division by (x 1) the quotient is 3x 2 +3x +8and the remainder is 12. We can express the function as f(x) = (x 1)( 3x 2 +3x + 8 ) + 12. Key point When f(x) is divided by (x a) we can say f(x) = (x a)q(x) + R where Q(x) is the quotient and R is the remainder. Examples 1. Problem: If f(x) = 2x 4 + x 2 x +1, divide f(x) by (x +1). We must interpret the function as 2x 4 +0x 3 + x 2 x + 1. We can only divide by (x a) so we interpret the divisor as (x ( 1)). This gives,

TOPIC 1. POLYNOMIALS AND QUADRATICS 49 So the quotient is 2x 3 2x 2 +3x 4 and the remainder is 5. Hence f(x) = (x + 1)( 2x 3 2x 2 +3x 4) + 5. 2. Problem: If f(x) = 2x 3 + 5x 2 x 1, divide f(x) by (2x 1). We can only divide by (x a) so we interpret the divisor as 2(x 1 /2). This gives, So the quotient is 2x 2 +6x +2and the remainder is 0 So f(x) = (x 1 /2)( 2x 2 + 6x +2)but we have to take the common factor of 2 out of the quotient and put it back into the divisor. ( x 1 (2x 2) 2 + 6x + 2 ) ( = x 1 ) 2 ( x 2 + 3x + 1 ) 2 ( = 2 x 1 ) (x 2 + 3x + 1 ) 2 giving f(x) = (2x 1)(x 2 + 3x + 1). Division by (x - a) practice Q63: What is the remainder when (4x 2 10x +2) (x 3)? Go online Q64: What is the remainder when (5x 3 7x 2 + 14) (x 3)? Q65: What is the remainder when (3x 2 8x +4) (x +3)? Q66: Express (6x 3 + 7x 2 1) (3x 1) in the form (3x 1)Q(x) + R, where Q(x) is the quotient and R is the remainder.

50 TOPIC 1. POLYNOMIALS AND QUADRATICS Polynomial long division You could also use algebraic long division to divide a polynomial. Example Problem: Divide 3x 3 2x 2 +6by x +4. The polynomial has a missing term and should be interpreted as 3x 3 2x 2 +0x +6. The divisor is x +4. x + 4 ) 3x 3 2x 2 + 0x + 6 Set up long division. Divide the first term of the polynomial by the first term of the divisor. It is often easier to identify what you would multiply the first term of the divisor by to get the first term of the polynomial. Put the answer at the top. This is the first term of the quotient. Multiply the divisor by the first term of the quotient. Subtract to form a new polynomial. Divide the first term of the new polynomial by the first term of the divisor and put it at the top. Multiply the divisor by the second term of the quotient and subtract to form a new polynomial.

TOPIC 1. POLYNOMIALS AND QUADRATICS 51 Divide the first term of the new polynomial by the first term of the divisor and put it at the top. Then multiply the divisor by the third term of the quotient and subtract to form the remainder. So 3x 3 2x 2 + 6 x + 4 gives 3x 2 14x + 56 as the quotient and 218 as the remainder. We can now express 3x 3 2x 2 +6as (x + 4)(3x 2 14x 2 + 56) 218. Q67: Divide x 4 +2x 3 7x 2 8x + 12 by x 1. Q68: Divide 3x 3 + x 2 3x 1 by x 2 1. Division by (x - a) exercise Q69: What is the remainder when 3x 2 x + 7 is divided by (x 2)? Go online Q70: What is the remainder when x 3 4x 2 +11is divided by (x 1)? Q71: What is the remainder when 2x 3 + x 2 +3x 4 is divided by (x +3)? Q72: What is the remainder when 4x 4 +2x 2 5x +1is divided by (x + 1)?

52 TOPIC 1. POLYNOMIALS AND QUADRATICS 1.2.3 Factor theorem The Factor Theorem states that if f(a) = 0 then (x a) is a factor of f(x) and if (x a) is a factor of f(x) then f(a) = 0. Key point If a polynomial divided by (x a) has remainder 0 then (x a) is a factor and if (x a) is a factor then the remainder under division by (x a) = 0. Example Problem: Is (x 4) a factor of f(x) where f(x) = x 3 + 7x 2 26x 72. Since the remainder is 0, (x 4) is a factor of f(x) and f(x) = (x 4)(x 2 +11x + 18). Notice that we can factorise the quotient and f(x) = (x 4)(x + 2)(x + 9) which gives the function in its fully factorised form. Factor theorem practice Go online Q73: Is (x 1) a factor of f(x) where f(x) = x 4 +2x 3 7x 2 8x +12? Q74: Is (x 2) a factor of f(x) where f(x) = x 3 +3x 2 4x 12? Factor theorem exercise Go online Q75: Is (x 1) a factor of 3x 3 + x 2 3x 1? Q76: Is (x +4)afactor of 2x 3 4x 2 + x 8? Q77: Is (x +1)afactor of x 3 +3x 2 + 6x + 4? Q78: Is (x 2) a factor of x 4 +4x 3 + 3x 2 + x +2?