SCHOLAR Study Guide CfE Higher Mathematics Assessment Practice 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 2017 by Heriot-Watt University SCHOLAR. Copyright 2017 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 Assessment Practice: CfE Higher Mathematics 1. CfE Higher Mathematics Course Code: C747 76
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1 Topic 7 Polynomials and quadratics Contents 7.1 Learning points............................................ 3 7.2 Assessment practice......................................... 5
2 TOPIC 7. POLYNOMIALS AND QUADRATICS Learning objective 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. By the end of this topic, you should have identified your strengths and areas for further revision.
TOPIC 7. POLYNOMIALS AND QUADRATICS 3 7.1 Learning points Read through the learning points before you attempt the assessments and go back to the course materials if you need to revise further. Polynomials and quadratics Polynomials A polynomial is an expression containing the sum or difference of algebraic terms with powers or the equivalent in factorised form. The degree of a polynomial is the value of the highest power. Synthetic division is a method for factorising a polynomial. (x a) is the divisor e.g. If a polynomial divided by (x a) has remainder 0 then (x a) is a factor. If (x a) is a factor then the remainder under division by (x a) is 0. When trying a divisor of the form (x a) it is usually a good idea to start with (x 1). If that does not work then think of the possible factors of the constant on the end of your polynomial. Be systematic and don't rub out any attempts that do not work. Solving a polynomial is best done in factorised form and allows you to identify the roots (i.e. the places where the graph of the polynomial crosses the x-axis). To determine the equation of a polynomial from its graph: use the roots to determine the factors e.g. roots a, b, c give factors (x a)(x b)(x c); remember the polynomial may have a scalar k e.g. y = k(x a)(x b)(x c); substitute the coordinates of the y-intercept to determine k; To determine points of intersection of lines and polynomials: make the equations equal and re-arrange to equal 0; factorise using synthetic division and trial and error; identify and interpret the roots; Equal roots are required for tangency. find any other points of intersection. The same method can be used to find the points of intersection of two polynomials
4 TOPIC 7. POLYNOMIALS AND QUADRATICS Quadratics Completing the square is used to: determine the coordinates of the turning point; to solve a quadratic equation which: does not factorise; is undefined using the quadratic formula; has a discriminant < 0, so the graph has no real roots; gives a parabola that lies entirely above or entirely below the x-axis e.g. a quadratic of the form ax 2 + bx + c can be written in the completed square form. Quadratic equations can be solved by: using the graph; factorising; using the quadratic formula: x = b ± b 2 4ac 2a (The formula is not given so learn it!) completing the square. To solve quadratic inequalities you need to: solve the related quadratic equation to find the roots; make a sketch of the graph of the quadratic function; identify the solution of the inequality: f(x) < 0 has solutions below the x-axis. f(x) > 0 has solutions above the x-axis. The discriminant of the quadratic equation ax 2 + bx + c = 0 is b 2 4ac. 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. To determine the relationship between a line and a parabola: make the equations equal and rearrange to equal 0; find and interpret the discriminant of the resulting quadratic equation: if b 2 4ac = 0 then the line is a tangent. if b 2 4ac > 0 then the line crosses the parabola in 2 places. if b 2 4ac < 0 then the line does not touch the parabola. Find the coordinates of any point(s) of contact.
TOPIC 7. POLYNOMIALS AND QUADRATICS 5 7.2 Assessment practice Make sure that you have read through the learning points and completed some revision before attempting these questions. Assessment practice: Polynomials and quadratics Go online SQA Past Paper: 2001 Paper 1 Q1: For what value of k does the equation x 2 5x +(k +6)=0have equal roots? 2 marks SQA Past Paper: 2002 Paper 2 The diagram shows part of the graph of the curve with equation y = 2x 3 7x 2 +4x +4. When 2x 3 7x 2 +4x +4is factorised it gives (kx + a)(x + b)(x + c), where b c. Q2: What is (kx + a)(x + b)(x + c)? 3 marks... Q3: What are the coordinates of A? 1 mark... Q4: What are the values of x for which 2x 3 7x 2 +4x +4< 0? 1 mark SQA Past Paper: 2001 Paper 1 Q5: Given f(x) = x 2 +2x 8, express f(x) in the form (x + a) 2 b. What are the values of a and b? 2 marks
6 TOPIC 7. POLYNOMIALS AND QUADRATICS SQA Past Paper: 2003 Paper 2 Q6: f(x) = 6x 3 5x 2 17x +6, show that (x 2) is a factor of f(x) by expressing f(x) in fully factorised form. 4 marks SQA Past Paper: 2004 Paper 2 Q7: What is the nature of the roots of the equation 2x 2 + px 3 = 0 for all real values of p? 3 marks SQA Past Paper: 2004 Paper 1 f(x) = x 3 x 2 5x 3 Q8: Is (x + 1) a factor of f(x)? 1 mark Q9: What is x 3 x 2 5x +3in fully factorised form? 3 marks... Q10: One of the turning points of the graph y = x 3 x 2 5x 3 lies on the x-axis. What are the coordinates of this turning point? 1 mark SQA Past Paper: 1999 Paper 1 Q11: What is f(x) = 2x 2 4x + 5 written in the form f(x) = a(x + b) 2 + c? 3 marks SQA Objective Question Q12: For what range of values of x is 2x 2 + x 6 > 0? 2 marks
TOPIC 7. POLYNOMIALS AND QUADRATICS 7 SQA Objective Question Q13: The diagram shows part of the graph of a cubic function y = ax 3 + bx 2 + cx + d. What is the equation of the function? 4 marks SQA Past Paper: 2002 Paper 1 Q14: The line with equation 5x + y +7intersects the parabola y = 3x 2 +4x 7 at two points. What are the coordinates of the points of intersection? 4 marks SQA Past Paper: 2001 Paper 2 Q15: Given that x + 2 is a factor of 2x 3 + x 2 + kx + 2, find the value of k and solve the polynomial. 5 marks
8 TOPIC 7. POLYNOMIALS AND QUADRATICS SQA Past Paper: 2001 Paper 2 The diagram shows a sketch of a parabola passing through (-1,0), (0,p) and (p,0). Q16: What is the equation of the parabola in the form y = c + bx x 2? 3 marks... Q17: For what value of p will the line y = x + p be a tangent to this curve? 3 marks SQA Past Paper: 2003 Paper 1 Q18: How many times does the line with equation y = 2x + 1 intersect the parabola with equation y = x 2 +3x +4? 5 marks
ANSWERS: UNIT 2 TOPIC 4 9 Answers to questions and activities Topic 4: Polynomials and quadratics Assessment practice: Polynomials and quadratics (page 5) Q1: Hints: For equal roots b 2 4ac = 0 Substitute for a, b and c in the discriminant to find the value of k. Steps: What are the values of a, b and c in the equation x 2 5x +(k +6)=0? a = 1, b = 5 and c = (k +6) Use these to substitute into b 2 4ac = 0 and solve for k. Answer: ( 5) 2 4(1)(k + 6) = 0 25 4(k + 6) = 0 25 4k 24 = 0 1 = 4k k = 1 4 Q2: Steps: What factor can you identify from the diagram? (x 2) Use this with synthetic division to factorise fully. Answer: (2x + 1)(x 2)(x 2) Q3: ( 1 2, 0) Q4: x < 1 2 Q5: Hints: ( x 2 + 2x ) 8 Half the coefficient of the middle term = (x + 1) 2 1 2 8 Answer: a = 1 and b = 9 Q6: Steps: What is the remainder when f(x) is divided by (x - 2)? 0 What is the quotient when f(x) is divided by (x - 2)? 6x 2 + 7x 3
10 ANSWERS: UNIT 2 TOPIC 4 Answer: (x 2)(3x 1)(2x +3) Q7: Hints: Find an expression for the discriminant in terms of p. Steps: What is an expression for the discriminant in terms of p? b 2 4ac = p 2 4(2)( 3) = p 2 + 24 Use this to find the nature of the roots of the equation for all real values of p. Answer: real and distinct b 2 4ac = p 2 4(2)( 3) = p 2 + 24 since p 2 0, p 2 + 24 24 hence b 2 4ac > 0 Q8: Yes, because the remainder under synthetic division is zero. Q9: (x +3)(x 1)(x 1) Q10: (1,0) Q11: Hints: 2x 2 4x + 5 = 2 [ x 2 2x ] + 5 [ = 2 (x 1) 2 1 2] + 5 Answer: f(x) = 2(x 1) 2 + 3 = 2(x 1) 2 2 + 5 Q12: Steps: What are the roots? -2 and 3 2 Make a sketch of the graph to determine the values of x where 2x 2 + x 6 > 0. Answer: x < 2 and x > 3 2 Q13: Hints: The equation of the graph takes the form y = k(x +3)(x 1)(x 2) The y-intercept has coordinates (0,12)
ANSWERS: UNIT 2 TOPIC 4 11 Let x = 0 and y = 12 to get 12 = k(3)( 1)( 2) so 6k = 12 and k = 2 Hence the equation is y = 2(x + 3)(x - 1)(x - 2) y = (2x + 6)(x 2-3x+2) Multiply out the brackets to identify a, b, c and d. Answer: y = 2x 3 14x +12 Q14: Hints: 5x + y +7=0 y = 5x 7 3x 2 + 4x 7 = 5x 7 3x 2 + 9x = 0 3x (x + 3) = 0 x = 0 or x = 3 Substitute the values of x in y = 5x 7 For x = 0, y = 7 For x = 3, y = 8 Answer: The points of intersection are (-3,8) and (0,-7) Q15: Steps: If x +2is a factor what is the remainder under synthetic division? 0 Form an expression for the remainder and make it equal 0 to find k. What is the value of k? k = 5 Factorise the quotient then solve the polynomial. Answer: x = 2 or x = 1 /2 or x = 1 Q16: Hints: The roots are x = 1 and x = p Hence the equation in factorised form is y = (x + 1)(p x) Notice that the parabola is a sad face in shape and hence x 2 is negative. (x + 1)(p x) = xp x 2 + p x = p + xp x x 2 =? Answer: y = p +(p 1)x x 2
12 ANSWERS: UNIT 2 TOPIC 4 Q17: Hints: At the point of intersection the equations are equal. x + p = p + xp x x 2 x 2 + 2x xp = 0 x 2 + (2 p)x = 0 Use the discriminant for equal roots to determine p. a = 1, b = 2 p and c = 0. b 2 4ac = 0 (2 p) 2 0 = 0 Answer: p = 2 2 p = 0 Q18: Steps: The equations are equal at any points of intersection. What is a simplified expression to represent this? x 2 + x +3=0 What is the value of b 2 4ac? -11 Interpret the value of the discriminant. Answer: 0, because b 2 4ac < 0