6 Chapter Complex numbers and polynomials Syllabus reference: 1.5, 1.8, 2.5, 2.6 cyan magenta yellow 95 100 50 Real quadratics with ¢ < 0 Complex numbers Real polynomials Zeros, roots, and factors Polynomial theorems Graphing real polynomials 75 25 0 5 95 A B C D E F 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Contents: black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\173IB_HL-3ed_06.cdr Thursday, 12 April 2012 12:24:44 PM BEN IB_HL-3ed 174 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) A REAL QUADRATICS WITH ¢ < 0 In Chapter 1, we determined that: If ax2 + bx + c = 0, a 6= 0 and a, b, c 2 R , then the solutions or roots are found using the p ¡b § ¢ where ¢ = b2 ¡ 4ac is known as the discriminant. formula x = 2a ² ¢ > 0 we have two real distinct solutions ² ¢ = 0 we have two real identical solutions ² ¢ < 0 we have no real solutions. We also observed that if: However, it is in fact possible to write down two solutions for the case where ¢ < 0. To do this we need imaginary numbers. p In 1572, Rafael Bombelli defined the imaginary number i = ¡1. It is called ‘imaginary’ because we cannot place it on a number line. With i defined, we can write down solutions for quadratic equations with ¢ < 0. They are called complex solutions because they include a real and an imaginary part. Any number of the form a + bi where a and b are real and i = is called a complex number. Self Tutor Example 1 2 a x = ¡4 Solve the quadratic equations: a x2 = ¡4 p ) x = § ¡4 p p ) x = § 4 ¡1 ) x = §2i 2 b z +z+2= 0 p ¡1, If the coefficient of i is a square root, we write the i first. b z 2 + z + 2 has a = 1, b = 1, c = 2 p ¡1 § 12 ¡ 4(1)(2) ) z= 2(1) p ¡1 § ¡7 ) z= 2 p p ¡1 § i 7 ) z= = ¡ 12 § 27 i 2 In Example 1 above, notice that ¢ < 0 in both cases. In each case we have found two complex solutions of the form a + bi, where a and b are real. Self Tutor Example 2 Write as a product of linear factors: cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 b x2 + 11 100 50 75 25 0 5 95 100 50 75 25 0 5 a x2 + 4 black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\174IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:56:20 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 175 Self Tutor Example 3 Solve for x: a x2 + 9 = 0 b x3 + 2x = 0 Self Tutor Example 4 Solve for x: a x2 ¡ 4x + 13 = 0 b x4 + x2 = 6 EXERCISE 6A 1 Write in terms of i: p p a ¡9 b ¡64 c q ¡ 14 d p ¡5 e p ¡8 2 Write as a product of linear factors: a x2 ¡ 9 e 4x2 ¡ 1 b x2 + 9 f 4x2 + 1 c x2 ¡ 7 g 2x2 ¡ 9 d x2 + 7 h 2x2 + 9 i x3 ¡ x j x3 + x k x4 ¡ 1 l x4 ¡ 16 3 Solve for x: a x2 ¡ 25 = 0 e 4x2 ¡ 9 = 0 b x2 + 25 = 0 f 4x2 + 9 = 0 c x2 ¡ 5 = 0 g x3 ¡ 4x = 0 i x3 ¡ 3x = 0 j x3 + 3x = 0 k x4 ¡ 1 = 0 cyan magenta yellow 95 100 50 75 c x4 + 5x2 = 36 f x4 + 2x2 + 1 = 0 25 b x4 = x2 + 6 e x4 + 1 = 2x2 0 95 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 5 Solve for x: a x4 + 2x2 = 3 d x4 + 9x2 + 14 = 0 c x2 + 14x + 50 = 0 5 d 2x2 + 5 = 6x l x4 = 81 b x2 + 6x + 25 = 0 p e x2 ¡ 2 3x + 4 = 0 100 4 Solve for x: a x2 ¡ 10x + 29 = 0 d x2 + 5 = 0 h x3 + 4x = 0 black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\175IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:56:25 AM BEN f 2x + 1 =1 x IB_HL-3ed 176 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) B COMPLEX NUMBERS Any number of the form a + bi where a, b 2 R and i = p ¡1, is called a complex number. Notice that all real numbers are complex numbers in the special case where b = 0. A complex number of the form bi where b 2 R , b 6= 0, is called an imaginary number or purely imaginary. THE ‘SUM OF TWO SQUARES’ a2 + b2 = a2 ¡ b2 i2 fas i2 = ¡1g = (a + bi)(a ¡ bi) Notice that a2 ¡ b2 = (a + b)(a ¡ b) a2 + b2 = (a + bi)(a ¡ bi) Compare: fthe difference of two squares factorisationg fthe sum of two squares factorisationg If we write z = a + bi where a, b 2 R , then: For example: ² a is the real part of z and we write a = Re (z) ² b is the imaginary part of z and we write b = Im (z). then Re (z) = 2 and Im (z) = 3. p then Re (z) = 0 and Im (z) = ¡ 2. If z = 2 + 3i, p If z = ¡ 2i, OPERATIONS WITH COMPLEX NUMBERS Operations with complex numbers are identical to those with radicals, but with i2 = ¡1 rather than p p ( 2)2 = 2 or ( 3)2 = 3. For example: ² addition: ² multiplication: p p p p (2 + 3) + (4 + 2 3) = (2 + 4) + (1 + 2) 3 = 6 + 3 3 (2 + i) + (4 + 2i) = (2 + 4) + (1 + 2)i = 6 + 3i p p p p p p (2 + 3)(4 + 2 3) = 8 + 4 3 + 4 3 + 2( 3)2 = 8 + 8 3 + 6 (2 + i)(4 + 2i) = 8 + 4i + 4i + 2i2 = 8 + 8i ¡ 2 So, we can add, subtract, multiply, and divide complex numbers in the same way we perform these operations with radicals: (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) ¡ (c + di) = (a ¡ c) + (b ¡ d)i (a + bi)(c + di) = ac + adi + bci + bdi2 µ ¶µ ¶ ac ¡ adi + bci ¡ bdi2 a + bi c ¡ di a + bi = = c + di c + di c ¡ di c2 + d2 addition subtraction multiplication division cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Notice that in the division process, we use a multiplication technique to obtain a real number in the denominator. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\176IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:56:33 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 177 Self Tutor Example 5 If z = 3 + 2i and w = 4 ¡ i find: b z¡w a z+w c zw d z w You can use your calculator to perform operations with complex numbers. After solving the questions in the following exercise by hand, check your answers using technology. GRAPHICS CALCUL ATOR INSTRUCTIONS EXERCISE 6B.1 1 Copy and complete: Re (z) z Im (z) z 3 + 2i ¡3 + 4i 5¡i ¡7 ¡ 2i 3 ¡11i p i 3 0 Re (z) Im (z) 2 If z = 5 ¡ 2i and w = 2 + i, find in simplest form: a z+w e 2z ¡ 3w b 2z f zw c iw g w2 d z¡w h z2 3 For z = 1 + i and w = ¡2 + 3i, find in simplest form: b z2 f zw a z + 2w e w2 c z3 g z2w d iz h izw a Simplify in for n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and also for n = ¡1, ¡2, ¡3, ¡4, ¡5. 4 b Hence, simplify i4n+3 where n is any integer. 5 Write (1 + i)4 in simplest form. Hence, find (1 + i)101 in simplest form. 6 Suppose z = 2 ¡ i and w = 1 + 3i. Write in exact form a + bi where a, b 2 R : cyan magenta i z yellow w iz 95 100 50 75 25 0 5 95 c 100 50 25 0 5 95 100 50 75 b 75 z w 25 0 5 95 100 50 75 25 0 5 a black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\177IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:57:44 AM BEN d z ¡2 IB_HL-3ed 178 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 7 Simplify: i 1 ¡ 2i a i(2 ¡ i) 3 ¡ 2i b 1 2 ¡ 2¡i 2+i c 8 If z = 2 + i and w = ¡1 + 2i, find: a Im (4z ¡ 3w) c Im (iz 2 ) b Re (zw) d Re ³ ´ z w EQUALITY OF COMPLEX NUMBERS Two complex numbers are equal when their real parts are equal and their imaginary parts are equal. a + bi = c + di , a = c and b = d. Suppose b 6= d. Now if a + bi = c + di where a, b, c, and d are real, then bi ¡ di = c ¡ a ) i(b ¡ d) = c ¡ a Proof: ) i= c¡a b¡d fas b 6= dg This is false as the RHS is real but the LHS is imaginary. Thus, the supposition is false. Hence b = d and furthermore a = c. For the complex number a + bi, where a and b are real, a + bi = 0 , a = 0 and b = 0. Self Tutor Example 6 Find real numbers x and y such that: a (x + yi)(2 ¡ i) = ¡i b (x + 2i)(1 ¡ i) = 5 + yi EXERCISE 6B.2 1 Find exact real numbers x and y such that: a 2x + 3yi = ¡x ¡ 6i b x2 + xi = 4 ¡ 2i c (x + yi)(2 ¡ i) = 8 + i d (3 + 2i)(x + yi) = ¡i cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 d (x + yi)(2 + i) = 2x ¡ (y + 1)i 0 c (x + i)(3 ¡ iy) = 1 + 13i 5 b (x + 2i)(y ¡ i) = ¡4 ¡ 7i 95 a 2(x + yi) = x ¡ yi 100 50 75 25 0 5 2 Find exact x, y 2 R such that: black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\178IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:58:31 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 179 3 The complex number z satisfies the equation 3z + 17i = iz + 11. Write z in the form a + bi p where a, b 2 R and i = ¡1. ³ ´2 4 4 Express z in the form a + bi where a, b 2 Z , if z = + 7 ¡ 2i . 1+i 5 Find the real values of m and n for which 3(m + ni) = n ¡ 2mi ¡ (1 ¡ 2i). 6 Express z = p 3i 2¡i + 1 in the form a + bi where a, b 2 R are given exactly. 7 Suppose (a + bi)2 = ¡16 ¡ 30i where a, b 2 R and a > 0. Find the possible values of a and b. HISTORICAL NOTE 18th century mathematicians enjoyed playing with these new ‘imaginary’ numbers, but they were regarded as little more than interesting curiosities until the work of Gauss (1777 - 1855), the German mathematician, astronomer, and physicist. For centuries mathematicians had attempted to find a method of trisecting an angle using a compass and straight edge. Gauss put an end to this when he used complex numbers to prove the impossibility of such a construction. By his systematic use of complex numbers, he was able to convince mathematicians of their usefulness. Carl Friedrich Gauss Early last century, the American engineer Steinmetz used complex numbers to solve electrical problems, illustrating that complex numbers did have a practical application. Complex numbers are now used extensively in electronics, engineering, and physics. COMPLEX CONJUGATES Complex numbers a + bi and a ¡ bi are called complex conjugates. If z = a + bi we write its conjugate as z ¤ = a ¡ bi. We saw on page 176 that the complex conjugate is important for division: z z w¤ zw¤ = = ¤ w w w ww¤ which makes the denominator real. Quadratics with real coefficients are called real quadratics. This does not necessarily mean that their zeros are real. ² If a quadratic equation has rational coefficients and an irrational root of the form p p c + d n, then the conjugate c ¡ d n is also a root of the quadratic equation. ² If a real quadratic equation has ¢ < 0 and c + di is a complex root, then the complex conjugate c ¡ di is also a root. cyan magenta 95 yellow Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\179IB_HL-3ed_06.cdr Monday, 23 April 2012 3:34:02 PM BEN 100 50 75 ¢ = (¡2)2 ¡ 4(1)(5) = ¡16 the solutions are x = 1 + 2i and 1 ¡ 2i ¢ = 02 ¡ 4(1)(4) = ¡16 the solutions are x = 2i and ¡2i 25 0 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 ² x2 + 4 = 0 has and has and 5 ² x2 ¡ 2x + 5 = 0 For example: black IB_HL-3ed 180 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) If c + di and c ¡ di are roots of a quadratic equation, then the quadratic equation is a(x2 ¡ 2cx + (c2 + d2 )) = 0 for some constant a 6= 0. Theorem: The sum of the roots = 2c and the product = (c + di)(c ¡ di) = c2 + d2 Proof: ) x2 ¡ (sum)x + (product) = 0 ) x2 ¡ 2cx + (c2 + d2 ) = 0 In general, a(x2 ¡ 2cx + c2 + d2 ) = 0 for some constant a 6= 0. ² the sum of complex conjugates c + di and c ¡ di is 2c which is real Notice that: ² the product is (c + di)(c ¡ di) = c2 + d2 which is also real. Self Tutor Example 7 Find all real quadratic equations having 1 ¡ 2i as a root. Example 8 p Find exact values for a and b if 2 + i is a root of x2 + ax + b = 0, a, b 2 R . Self Tutor EXERCISE 6B.3 1 Find all quadratic equations with real coefficients and roots of: a 3§i p e 2§ 3 b 1 § 3i c ¡2 § 5i p g §i 2 f 0 and ¡ 23 d p 2§i h ¡6 § i 2 Find exact values for a and b if: a 3 + i is a root of x2 + ax + b = 0, where a and b are real p b 1 ¡ 2 is a root of x2 + ax + b = 0, where a and b are rational cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 c a + ai is a root of x2 + 4x + b = 0, where a and b are real. [Careful!] black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\180IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:59:19 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 181 PROPERTIES OF CONJUGATES INVESTIGATION 1 PROPERTIES OF CONJUGATES The purpose of this investigation is to discover any properties that complex conjugates might have. What to do: 1 Given z1 = 1 ¡ i and z2 = 2 + i find: b z2¤ a z1¤ c (z1¤ )¤ d (z2¤ )¤ h z1¤ ¡ z2¤ p (z2¤ )3 e (z1 + z2 )¤ f z1¤ + z2¤ i (z1 z2 )¤ j z1¤ z2¤ g (z1 ¡ z2 )¤ ³ ´¤ z1 k n (z1¤ )2 o (z23 )¤ m (z12 )¤ z2 l z1¤ z2¤ 2 Repeat 1 with z1 and z2 of your choice. 3 Examine your results from 1 and 2, and hence suggest some rules for complex conjugates. From the Investigation you should have discovered the following rules for complex conjugates: ² (z ¤ )¤ = z ² (z1 + z2 )¤ = z1¤ + z2¤ and (z1 ¡ z2 )¤ = z1¤ ¡ z2¤ µ ¶¤ z1 z¤ and = 1¤ , z2 6= 0 z2 z2 ² (z1 z2 )¤ = z1¤ £ z2¤ ² (z n )¤ = (z ¤ )n ² z + z¤ for positive integers n and zz ¤ are real. Self Tutor Example 9 Show that for all complex numbers z1 and z2 : cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 b (z1 z2 )¤ = z1¤ £ z2¤ 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 a (z1 + z2 )¤ = z1¤ + z2¤ black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\181IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:59:26 AM BEN IB_HL-3ed 182 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) CONJUGATE GENERALISATIONS Notice that (z1 + z2 + z3 )¤ = (z1 + z2 )¤ + z3¤ = z1¤ + z2¤ + z3¤ ftreating z1 + z2 as one complex numberg .... (1) Likewise (z1 + z2 + z3 + z4 )¤ = (z1 + z2 + z3 )¤ + z4¤ = z1¤ + z2¤ + z3¤ + z4¤ ffrom (1)g There is no reason why this process cannot continue for the conjugate of 5, 6, 7, .... complex numbers. We can therefore generalise the result: (z1 + z2 + z3 + :::: + zn )¤ = z1¤ + z2¤ + z3¤ + :::: + zn¤ The process of proving the general case using the simpler cases when n = 1, 2, 3, 4, .... requires mathematical induction. Formal proof by the Principle of Mathematical Induction is discussed in Chapter 9. EXERCISE 6B.4 1 Show that (z1 ¡ z2 )¤ = z1¤ ¡ z2¤ for all complex numbers z1 and z2 . 2 Simplify the expression (w¤ ¡ z)¤ ¡ (w ¡ 2z ¤ ) using the properties of conjugates. 3 It is known that a complex number z satisfies the equation z ¤ = ¡z. Show that z is either purely imaginary or zero. 4 Suppose z1 = a + bi and z2 = c + di are complex numbers. ³ ´¤ z ¤ z z1 a Find 1 in the form X + iY . b Show that = 1¤ z2 z2 ³ 5 a An easier way of proving ´ z¤ z1 ¤ = 1¤ z2 z2 z2 ³ is to start with for all z1 and z2 6= 0. ´ z1 ¤ £ z2¤ . z2 Show how this can be done, remembering we have already proved that “the conjugate of a product is the product of the conjugates” in Example 9. b Let z = a + bi be a complex number. Prove the following: i If z = z ¤ , then z is real. ii If z ¤ = ¡z, then z is purely imaginary or zero. 6 Prove that for all complex numbers z and w: a zw¤ + z ¤ w b zw¤ ¡ z ¤ w is always real is purely imaginary or zero. a If z = a + bi, find z 2 in the form X + iY . 7 b Hence, show that (z 2 )¤ = (z ¤ )2 for all complex numbers z. 3 c Repeat a and b but for z instead of z 2 . 8 Suppose w = z¡1 z¤ + 1 where z = a + bi. Find the conditions under which: a w is real b w is purely imaginary. a Assuming (z1 z2 )¤ = z1¤ z2¤ , explain why (z1 z2 z3 )¤ = z1¤ z2¤ z3¤ . 9 b Hence show that (z1 z2 z3 z4 )¤ = z1¤ z2¤ z3¤ z4¤ . c Generalise your results from a and b. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 d Given your generalisation in c, what is the result of letting all zi values be equal to z? black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\182IB_HL-3ed_06.cdr Thursday, 29 March 2012 9:59:37 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) C 183 REAL POLYNOMIALS Up to this point we have studied linear and quadratic functions at some depth, with perhaps occasional reference to cubic functions. These are part of a larger family of functions called the polynomials. A polynomial function is a function of the form P (x) = an xn + an¡1 xn¡1 + :::: + a2 x2 + a1 x + a0 , a1 , ...., an constant, an 6= 0. We say that: x is the variable a0 is the constant term an is the leading coefficient and is non-zero ar is the coefficient of xr for r = 0, 1, 2, ...., n n is the degree of the polynomial, being the highest power of the variable. In summation notation, we write P (x) = n P ar xr , r=0 which reads: “the sum from r = 0 to n, of ar xr ”. A real polynomial P (x) is a polynomial for which ar 2 R , r = 0, 1, 2, ...., n. The low degree members of the polynomial family have special names, some of which you are already familiar with. For these polynomials, we commonly write their coefficients as a, b, c, .... Polynomial function Degree Name ax + b, a 6= 0 1 linear ax + bx + c, a 6= 0 2 quadratic ax3 + bx2 + cx + d, a 6= 0 3 cubic 4 quartic 2 4 3 2 ax + bx + cx + dx + e, a 6= 0 ADDITION AND SUBTRACTION To add or subtract two polynomials, we collect ‘like’ terms. Self Tutor Example 10 If P (x) = x3 ¡ 2x2 + 3x ¡ 5 and Q(x) = 2x3 + x2 ¡ 11, find: magenta yellow 95 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 cyan 100 b P (x) ¡ Q(x) a P (x) + Q(x) black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\183IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:00:16 AM BEN IB_HL-3ed 184 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) SCALAR MULTIPLICATION To multiply a polynomial by a scalar (constant) we multiply each term by the scalar. Self Tutor Example 11 If P (x) = x4 ¡ 2x3 + 4x + 7 find: a 3P (x) b ¡2P (x) POLYNOMIAL MULTIPLICATION To multiply two polynomials, we multiply each term of the first polynomial by each term of the second polynomial, and then collect like terms. Self Tutor Example 12 If P (x) = x3 ¡ 2x + 4 and Q(x) = 2x2 + 3x ¡ 5, find P (x)Q(x). SYNTHETIC MULTIPLICATION (OPTIONAL) Polynomial multiplication can be performed using the coefficients only. multiplication. We call this synthetic coefficients of x3 + 2x ¡ 5 coefficients of 2x + 3 x2 magenta yellow 0 5 95 100 50 75 25 0 5 95 constants 100 75 x (x3 + 2x ¡ 5)(2x + 3) = 2x4 + 3x3 + 4x2 ¡ 4x ¡ 15. 95 x3 So 100 x4 ¡4 ¡15 50 4 cyan 2 2 75 3 50 2 25 0 6 ¡15 4 ¡10 0 3 0 5 2 95 0 £ 25 ¡5 3 1 100 50 75 25 0 5 For example, for (x3 + 2x ¡ 5)(2x + 3) we detach coefficients and multiply. It is different from the ordinary multiplication of large numbers because we sometimes have negative coefficients, and because we do not carry tens into the next column. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\184IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:00:43 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 185 EXERCISE 6C.1 1 If P (x) = x2 + 2x + 3 and Q(x) = 4x2 + 5x + 6, find in simplest form: a 3P (x) c P (x) ¡ 2Q(x) b P (x) + Q(x) d P (x)Q(x) 2 If f(x) = x2 ¡ x + 2 and g(x) = x3 ¡ 3x + 5, find in simplest form: a f(x) + g(x) b g(x) ¡ f (x) c 2f(x) + 3g(x) d g(x) + xf(x) e f (x) g(x) f [f (x)]2 a (x2 ¡ 2x + 3)(2x + 1) b (x ¡ 1)2 (x2 + 3x ¡ 2) c (x + 2)3 d (2x2 ¡ x + 3)2 e (2x ¡ 1)4 f (3x ¡ 2)2 (2x + 1)(x ¡ 4) 3 Expand and simplify: 4 Find the following products: a (2x2 ¡ 3x + 5)(3x ¡ 1) b (4x2 ¡ x + 2)(2x + 5) c (2x2 + 3x + 2)(5 ¡ x) d (x ¡ 2)2 (2x + 1) e (x2 ¡ 3x + 2)(2x2 + 4x ¡ 1) f (3x2 ¡ x + 2)(5x2 + 2x ¡ 3) g (x2 ¡ x + 3)2 h (2x2 + x ¡ 4)2 i (2x + 5)3 j (x3 + x2 ¡ 2)2 DIVISION OF POLYNOMIALS The division of polynomials is only useful if we divide a polynomial of degree n by another of degree n or less. DIVISION BY LINEARS Consider (2x2 + 3x + 4)(x + 2) + 7. If we expand this expression we obtain (2x2 + 3x + 4)(x + 2) + 7 = 2x3 + 7x2 + 10x + 15. Dividing both sides by (x + 2), we obtain 2x3 + 7x2 + 10x + 15 (2x2 + 3x + 4)(x + 2) + 7 = x+2 x+2 (2x2 + 3x + 4)(x + 2) 7 + x+2 x+2 7 = 2x2 + 3x + 4 + where x+2 = x + 2 is the divisor, 2 2x + 3x + 4 is the quotient, and 7 is the remainder. If P (x) is divided by ax + b until a constant remainder R is obtained, then P (x) R = Q(x) + ax + b ax + b where ax + b is the divisor, D(x), Q(x) is the quotient, and R is the remainder. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Notice that P (x) = Q(x) £ (ax + b) + R. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\185IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:00:55 AM BEN IB_HL-3ed 186 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) DIVISION ALGORITHM We can divide a polynomial by another polynomial using an algorithm similar to that used for division of whole numbers: What do we multiply x by to get 2x3 ? The answer is 2x2 , and 2x2 (x + 2) = 2x3 + 4x2 . | {z } Step 1: 2x2 + 3x + 4 x+2 Step 2: Subtract 2x3 + 4x2 from 2x3 + 7x2 . The answer is 3x2 . Step 3: Bring down the 10x to obtain 3x2 + 10x. 2x3 + 7x2 + 10x + 15 ¡ (2x3 + 4x2 ) 3x2 + 10x ¡ (3x2 + 6x) 4x + 15 ¡ (4x + 8) Return to Step 1 with the question: “What must we multiply x by to get 3x2 ?” 7 The answer is 3x, and 3x(x + 2) = 3x2 + 6x .... We continue the process until we are left with a constant. 2 The division algorithm can also be performed by leaving out the variable, as shown alongside. Either way, 2x3 1 2 + 7x2 + 10x + 15 7 = 2x2 + 3x + 4 + x+2 x+2 2 ¡ (2 3 4 7 10 15 4) 3 10 ¡ (3 6) 4 15 ¡ (4 8) 7 Self Tutor Example 13 Find the quotient and remainder for x3 ¡ x2 ¡ 3x ¡ 5 . x¡3 Hence write x3 ¡ x2 ¡ 3x ¡ 5 in the form Q(x) £ (x ¡ 3) + R. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Check your answer by expanding the RHS. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\186IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:01:55 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 187 Self Tutor Example 14 x4 + 2x2 ¡ 1 . x+3 Perform the division Hence write x4 + 2x2 ¡ 1 in the form Q(x) £ (x + 3) + R. Notice the insertion of 0x 3 and 0x. EXERCISE 6C.2 1 Find the quotient and remainder for the following, and hence write the division in the form P (x) = Q(x) D(x) + R, where D(x) is the divisor. a x2 + 2x ¡ 3 x+2 b x2 ¡ 5x + 1 x¡1 c 2x3 + 6x2 ¡ 4x + 3 x¡2 2 Perform the following divisions, and hence write the division in the form P (x) = Q(x) D(x) + R. a x2 ¡ 3x + 6 x¡4 b x2 + 4x ¡ 11 x+3 c 2x2 ¡ 7x + 2 x¡2 d 2x3 + 3x2 ¡ 3x ¡ 2 2x + 1 e 3x3 + 11x2 + 8x + 7 3x ¡ 1 f 2x4 ¡ x3 ¡ x2 + 7x + 4 2x + 3 3 Perform the divisions: a x2 + 5 x¡2 b 2x2 + 3x x+1 c 3x2 + 2x ¡ 5 x+2 d x3 + 2x2 ¡ 5x + 2 x¡1 e 2x3 ¡ x x+4 f x3 + x2 ¡ 5 x¡2 SYNTHETIC DIVISION (OPTIONAL) cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Click on the icon for an exercise involving a synthetic division process for the division of a polynomial by a linear. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\187IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:02:36 AM BEN PRINTABLE SECTION IB_HL-3ed 188 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) DIVISION BY QUADRATICS As with division by linears we can use the division algorithm to divide polynomials by quadratics. The division process stops when the remainder has degree less than that of the divisor, so If P (x) is divided by ax2 + bx + c then P (x) ex + f = Q(x) + 2 ax2 + bx + c ax + bx + c ax2 + bx + c is the divisor, where Q(x) is the quotient, ex + f is the remainder. ax2 + bx + c and The remainder will be linear if e 6= 0, and constant if e = 0. Self Tutor Example 15 x4 + 4x3 ¡ x + 1 . x2 ¡ x + 1 Find the quotient and remainder for Hence write x4 + 4x3 ¡ x + 1 in the form Q(x) £ (x2 ¡ x + 1) + R(x). EXERCISE 6C.3 1 Find the quotient and remainder for: a x3 + 2x2 + x ¡ 3 x2 + x + 1 b 3x2 ¡ x x2 ¡ 1 3x3 + x ¡ 1 x2 + 1 c d x¡4 x2 + 2x ¡ 1 2 Carry out the following divisions and also write each in the form P (x) = Q(x) D(x) + R(x): a x2 ¡ x + 1 x2 + x + 1 b x3 x2 + 2 c x4 + 3x2 + x ¡ 1 x2 ¡ x + 1 d 2x3 ¡ x + 6 (x ¡ 1)2 e x4 (x + 1)2 f x4 ¡ 2x3 + x + 5 (x ¡ 1)(x + 2) 3 Suppose P (x) = (x ¡ 2)(x2 + 2x + 3) + 7. Find the quotient and remainder when P (x) is divided by x ¡ 2. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 4 Suppose f(x) = (x ¡ 1)(x + 2)(x2 ¡ 3x + 5) + 15 ¡ 10x. Find the quotient and remainder when f(x) is divided by x2 + x ¡ 2. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\188IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:03:03 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) D 189 ZEROS, ROOTS, AND FACTORS A zero of a polynomial is a value of the variable which makes the polynomial equal to zero. ® is a zero of polynomial P (x) , P (®) = 0. The roots of a polynomial equation are the solutions to the equation. ® is a root (or solution) of P (x) = 0 , P (®) = 0. The roots of P (x) = 0 are the zeros of P (x) and the x-intercepts of the graph of y = P (x). P (x) = x3 + 2x2 ¡ 3x ¡ 10 P (2) = 23 + 2(2)2 ¡ 3(2) ¡ 10 = 8 + 8 ¡ 6 ¡ 10 =0 Consider ) An equation has roots. A polynomial has zeros. ² 2 is a zero of x3 + 2x2 ¡ 3x ¡ 10 ² 2 is a root of x3 + 2x2 ¡ 3x ¡ 10 = 0 ² the graph of y = x3 + 2x2 ¡ 3x ¡ 10 has the x-intercept 2. This tells us: If P (x) = (x + 1)(2x ¡ 1)(x + 2), then (x + 1), (2x ¡ 1), and (x + 2) are its linear factors. Likewise P (x) = (x + 3)2 (2x + 3) has been factorised into 3 linear factors, one of which is repeated. (x ¡ ®) is a factor of the polynomial P (x) , there exists a polynomial Q(x) such that P (x) = (x ¡ ®)Q(x). Self Tutor Example 16 a x2 ¡ 4x + 53 Find the zeros of: b z 3 + 3z EXERCISE 6D.1 magenta 95 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 cyan yellow 95 f z 4 = 3z 2 + 10 100 e z 3 + 5z = 0 d x3 = 5x 50 c ¡2z(z 2 ¡ 2z + 2) = 0 75 b (2x + 1)(x2 + 3) = 0 25 2 Find the roots of: a 5x2 = 3x + 2 0 c z 2 ¡ 6z + 6 f z 4 + 4z 2 ¡ 5 5 b x2 + 6x + 10 e z 3 + 2z 100 1 Find the zeros of: a 2x2 ¡ 5x ¡ 12 d x3 ¡ 4x black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\189IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:04:21 AM BEN IB_HL-3ed 190 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) Self Tutor Example 17 a 2x3 + 5x2 ¡ 3x Factorise: 3 Find the linear factors of: a 2x2 ¡ 7x ¡ 15 d 6z 3 ¡ z 2 ¡ 2z b z 2 + 4z + 9 b z 2 ¡ 6z + 16 e z 4 ¡ 6z 2 + 5 c x3 + 2x2 ¡ 4x f z4 ¡ z2 ¡ 2 4 If P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °) then ®, ¯, and ° are its zeros. Verify this statement by finding P (®), P (¯), and P (°). Self Tutor Example 18 Find all cubic polynomials with zeros 1 2 and ¡3 § 2i. Example 19 Find all quartic polynomials with zeros 2, ¡ 13 , p and ¡1 § 5. Self Tutor 5 Find all cubic polynomials with zeros: cyan magenta yellow d 2§ 95 p c § 3, 1 § i 100 d ¡1, ¡2 § 50 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 6 Find all quartic polynomials with zeros of: p p a §1, § 2 b 2, ¡1, §i 3 c 3, ¡1 § i 75 b ¡2, §i 25 a §2, 3 black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\190IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:04:31 AM BEN p 2 p 5, ¡2 § 3i IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 191 POLYNOMIAL EQUALITY Two polynomials are equal if and only if they have the same degree (order) and corresponding terms have equal coefficients. If we know that two polynomials are equal then we can equate coefficients to find unknown coefficients. For example, if 2x3 + 3x2 ¡ 4x + 6 = ax3 + bx2 + cx + d, where a, b, c, d 2 R , then a = 2, b = 3, c = ¡4, and d = 6. Self Tutor Example 20 Find constants a, b, and c given that: 6x3 + 7x2 ¡ 19x + 7 = (2x ¡ 1)(ax2 + bx + c) for all x. Self Tutor Example 21 cyan magenta yellow Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\191IB_HL-3ed_06.cdr Tuesday, 1 May 2012 5:12:59 PM BEN 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Find constants a and b if z 4 + 9 = (z 2 + az + 3)(z 2 + bz + 3) for all z. black IB_HL-3ed 192 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) Self Tutor Example 22 (x + 3) is a factor of P (x) = x3 + ax2 ¡ 7x + 6. Find a 2 R and the other factors. Self Tutor Example 23 (2x + 3) and (x ¡ 1) are factors of 2x4 + ax3 ¡ 3x2 + bx + 3. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Find constants a and b and all zeros of the polynomial. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\192IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:07:23 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 193 EXERCISE 6D.2 1 Find constants a, b, and c given that: a 2x2 + 4x + 5 = ax2 + [2b ¡ 6]x + c for all x b 2x3 ¡ x2 + 6 = (x ¡ 1)2 (2x + a) + bx + c for all x. 2 Find constants a and b if: a z 4 + 4 = (z 2 + az + 2)(z 2 + bz + 2) for all z b 2z 4 + 5z 3 + 4z 2 + 7z + 6 = (z 2 + az + 2)(2z 2 + bz + 3) for all z. 3 Show that z 4 + 64 can be factorised into two real quadratic factors of the form z 2 + az + 8 and z 2 + bz + 8, but cannot be factorised into two real quadratic factors of the form z 2 + az + 16 and z 2 + bz + 4. 4 Find real numbers a and b such that x4 ¡ 4x2 + 8x ¡ 4 = (x2 + ax + 2)(x2 + bx ¡ 2). Hence solve the equation x4 + 8x = 4x2 + 4. a (2z ¡ 3) is a factor of 2z 3 ¡ z 2 + az ¡ 3. Find a 2 R and all zeros of the cubic. 5 b (3z + 2) is a factor of 3z 3 ¡ z 2 + (a + 1)z + a. Find a 2 R and all the zeros of the cubic. a (2x + 1) and (x ¡ 2) are factors of P (x) = 2x4 + ax3 + bx2 ¡ 12x ¡ 8. Find constants a and b, and all zeros of P (x). 6 b (x + 3) and (2x ¡ 1) are factors of 2x4 + ax3 + bx2 + ax + 3. Find constants a and b, and hence determine all zeros of the quartic. a x3 + 3x2 ¡ 9x + c, c 2 R , has two identical linear factors. Prove that c is either 5 or ¡27, and factorise the cubic into linear factors in each case. b 3x3 + 4x2 ¡ x + m, m 2 R , has two identical linear factors. Find the possible values of m, and find the zeros of the polynomial in each case. 7 E POLYNOMIAL THEOREMS There are many theorems about polynomials, some of which we look at now. Some of the theorems are true for all polynomials, while others are true only for real polynomials. THE REMAINDER THEOREM Consider the cubic polynomial P (x) = x3 + 5x2 ¡ 11x + 3. If we divide P (x) by x ¡ 2, we find that x3 + 5x2 ¡ 11x + 3 9 = x2 + 7x + 3 + x¡2 x¡2 A real polynomial is a polynomial with real coefficients. remainder So, when P (x) is divided by x ¡ 2, the remainder is 9. Notice that P (2) = 8 + 20 ¡ 22 + 3 = 9, which is the remainder. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 By considering other examples like the one above, we formulate the Remainder theorem. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\193IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:08:01 AM BEN IB_HL-3ed 194 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) The Remainder Theorem When a polynomial P (x) is divided by x ¡ k until a constant remainder R is obtained, then R = P (k). P (x) = Q(x)(x ¡ k) + R P (k) = Q(k) £ 0 + R P (k) = R By the division algorithm, Letting x = k, ) Proof: When using the Remainder theorem, it is important to realise that the following statements are equivalent: ² P (x) = (x ¡ k)Q(x) + R ² P (k) = R ² P (x) divided by x ¡ k leaves a remainder of R. Self Tutor Example 24 Use the Remainder theorem to find the remainder when x4 ¡ 3x3 + x ¡ 4 is divided by x + 2. Self Tutor Example 25 2 2 cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 When P (x) is divided by x ¡ 3x + 7, the quotient is x + x ¡ 1 and the remainder R(x) is unknown. When P (x) is divided by x ¡ 2 the remainder is 29. When P (x) is divided by x + 1 the remainder is ¡16. Find R(x) in the form ax + b. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\194IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:08:25 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 195 EXERCISE 6E.1 1 For P (x) a real polynomial, write two equivalent statements for each of: a If P (2) = 7, then .... b If P (x) = Q(x)(x + 3) ¡ 8, then .... c If P (x) divided by x ¡ 5 has a remainder of 11 then .... 2 Without performing division, find the remainder when: a x3 + 2x2 ¡ 7x + 5 is divided by x ¡ 1 b x4 ¡ 2x2 + 3x ¡ 1 is divided by x + 2. 3 Find a 2 R given that: a when x3 ¡ 2x + a is divided by x ¡ 2, the remainder is 7 b when 2x3 + x2 + ax ¡ 5 is divided by x + 1, the remainder is ¡8. 4 When x3 + 2x2 + ax + b is divided by x ¡ 1 the remainder is 4, and when divided by x + 2 the remainder is 16. Find constants a and b. 5 2xn + ax2 ¡ 6 leaves a remainder of ¡7 when divided by x ¡ 1, and 129 when divided by x + 3. Find a and n given that n 2 Z + . 6 When P (z) is divided by z 2 ¡ 3z + 2 the remainder is 4z ¡ 7. Find the remainder when P (z) is divided by: a z¡1 b z ¡ 2. 7 When P (z) is divided by z + 1 the remainder is ¡8, and when divided by z ¡ 3 the remainder is 4. Find the remainder when P (z) is divided by (z ¡ 3)(z + 1). 8 If P (x) is divided by (x ¡ a)(x ¡ b), where a 6= b, a, b 2 R , prove that the remainder is: ³ ´ P (b) ¡ P (a) £ (x ¡ a) + P (a). b¡a THE FACTOR THEOREM For any polynomial P (x), k is a zero of P (x) , (x ¡ k) is a factor of P (x). k is a zero of P (x) , P (k) = 0 ,R=0 , P (x) = Q(x)(x ¡ k) , (x ¡ k) is a factor of P (x) Proof: fdefinition of a zerog fRemainder theoremg fdivision algorithmg fdefinition of a factorg cyan magenta 95 yellow Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\195IB_HL-3ed_06.cdr Monday, 23 April 2012 3:34:32 PM BEN 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 The Factor theorem says that if 2 is a zero of P (x) then (x ¡ 2) is a factor of P (x), and vice versa. black IB_HL-3ed 196 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) Self Tutor Example 26 Find k given that (x ¡ 2) is a factor of x3 + kx2 ¡ 3x + 6. Hence, fully factorise x3 + kx2 ¡ 3x + 6. EXERCISE 6E.2 1 Find the constant k and hence factorise the polynomial if: a 2x3 + x2 + kx ¡ 4 has the factor (x + 2) b x4 ¡ 3x3 ¡ kx2 + 6x has the factor (x ¡ 3). 2 Find constants a and b given that 2x3 + ax2 + bx + 5 has factors (x ¡ 1) and (x + 5). a Suppose 3 is a zero of P (z) = z 3 ¡ z 2 + (k ¡ 5)z + (k2 ¡ 7). Find the possible values of k 2 R and all the corresponding zeros of P (z). 3 b Show that (z ¡ 2) is a factor of P (z) = z 3 + mz2 + (3m ¡ 2)z ¡ 10m ¡ 4 for all values of m 2 R . For what value(s) of m is (z ¡ 2)2 a factor of P (z)? a Consider P (x) = x3 ¡ a3 4 where a is real. i Find P (a). What is the significance of this result? ii Factorise x3 ¡ a3 as the product of a real linear and a quadratic factor. b Now consider P (x) = x3 + a3 , where a is real. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 i Find P (¡a). What is the significance of this result? ii Factorise x3 + a3 as the product of a real linear and a quadratic factor. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\196IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:09:12 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) a Prove that “x + 1 is a factor of xn + 1, n 2 Z 5 197 , n is odd”. b Find the real number a such that (x ¡ 1 ¡ a) is a factor of P (x) = x3 ¡ 3ax ¡ 9. THE FUNDAMENTAL THEOREM OF ALGEBRA The theorems we have just seen for real polynomials can be generalised in the Fundamental Theorem of Algebra: a Every polynomial of degree n > 1 has at least one zero which can be written in the form a + bi where a, b 2 R . b If P (x) is a polynomial of degree n, then P (x) has exactly n zeros, some of which may be either irrational numbers or complex numbers. Gauss proved this theorem in 1799 as his PhD dissertation. Using the Fundamental Theorem of Algebra, the following properties of real polynomials can be established: ² Every real polynomial of degree n can be factorised into n complex linear factors, some of which may be repeated. ² Every real polynomial can be expressed as a product of real linear and real irreducible quadratic factors (where ¢ < 0). ² Every real polynomial of degree n has exactly n zeros, some of which may be repeated. ² If p + qi (q 6= 0) is a zero of a real polynomial then its complex conjugate p ¡ qi is also a zero. ² Every real polynomial of odd degree has at least one real zero. Self Tutor Example 27 Suppose ¡3 + i is a zero of P (x) = ax3 + 9x2 + ax ¡ 30 where a is real. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Find a and hence find all zeros of the cubic. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\197IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:09:51 AM BEN IB_HL-3ed 198 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) Self Tutor Example 28 One zero of ax3 + (a + 1)x2 + 10x + 15, a 2 R , is purely imaginary. Find a and the zeros of the polynomial. EXERCISE 6E.3 1 Find all real polynomials of degree 3 with zeros ¡ 12 and 1 ¡ 3i. 2 p(x) is a real cubic polynomial for which p(1) = p(2 + i) = 0 and p(0) = ¡20. Find p(x) in expanded form. 3 2 ¡ 3i is a zero of P (z) = z 3 + pz + q where p and q are real. a Using conjugate pairs, find p and q and the other two zeros. b Check your answer by solving for p and q using P (2 ¡ 3i) = 0. 4 3 + i is a root of z 4 ¡ 2z 3 + az 2 + bz + 10 = 0, where a and b are real. Find a and b and the other roots of the equation. 5 One zero of P (z) = z 3 + az 2 + 3z + 9, a 2 R , is purely imaginary. Find a and hence factorise P (z) into the product of linear factors. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 6 At least one zero of P (x) = 3x3 + kx2 + 15x + 10, k 2 R , is purely imaginary. Find k and hence factorise P (x) into the product of linear factors. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\198IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:10:22 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 199 SUM AND PRODUCT OF ROOTS THEOREM We have seen that for the quadratic equation ax2 + bx + c = 0, a 6= 0, the sum of the roots is ¡ and the product of the roots is b a c . a For the polynomial equation an xn + an¡1 xn¡1 + :::: + a2 x2 + a1 x + a0 = 0, an 6= 0 n P which can also be written as ar xr = 0, r=0 the sum of the roots is ¡an¡1 , and the product of the roots is an (¡1)n a0 . an We can explain this result as follows: Consider the polynomial equation an (x ¡ ®1 )(x ¡ ®2 )(x ¡ ®3 )::::(x ¡ ®n ) = 0 with roots ®1 , ®2 , ®3 , ...., ®n . Expanding the LHS we have an (x ¡ ®1 )(x ¡ ®2 )(x ¡ ®3 )(x ¡ ®4 )::::(x ¡ ®n ) | {z } = an (x2 ¡ [®1 + ®2 ]x + (¡1)2 ®1 ®2 )(x ¡ ®3 )(x ¡ ®4 )::::(x ¡ ®n ) | {z } = an (x3 ¡ [®1 + ®2 + ®3 ]x2 + :::: + (¡1)3 ®1 ®2 ®3 )(x ¡ ®4 )::::(x ¡ ®n ) fthe term of order x is no longer important as it will not contribute to either the xn¡1 term or the constant termg .. . = an (xn ¡ [®1 + ®2 + ®3 + :::: + ®n ]xn¡1 + :::: + (¡1)n ®1 ®2 ®3 :::: ®n ) = an xn ¡ an [®1 + ®2 + ®3 + :::: + ®n ]xn¡1 + :::: + (¡1)n ®1 ®2 ®3 :::: ®n an Equating coefficients, an¡1 = ¡an [®1 + ®2 + ®3 + :::: + ®n ] and a0 = (¡1)n ®1 ®2 ®3 :::: ®n an ) ¡ an¡1 = ®1 + ®2 + ®3 + :::: + ®n an (¡1)n a0 = ®1 ®2 ®3 :::: ®n an and Self Tutor Example 29 cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Find the sum and product of the roots of 2x3 ¡ 7x2 + 8x ¡ 1 = 0. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\199IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:10:49 AM BEN IB_HL-3ed 200 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) Self Tutor Example 30 A real polynomial has the form P (x) = 3x4 ¡ 12x3 + cx2 + dx + e. The graph of y = P (x) has y-intercept 180. It cuts the x-axis at 2 and 6, and does not meet the x-axis anywhere else. Suppose the other two zeros are m § ni, n > 0. Use the sum and product formulae to find m and n. If the other two zeros are m § ni, the sum of the zeros is EXERCISE 6E.4 1 Find the sum and product of the roots of: a 2x2 ¡ 3x + 4 = 0 c x4 ¡ x3 + 2x2 + 3x ¡ 4 = 0 e x7 ¡ x5 + 2x ¡ 9 = 0 b 3x3 ¡ 4x2 + 8x ¡ 5 = 0 d 2x5 ¡ 3x4 + x2 ¡ 8 = 0 f x6 ¡ 1 = 0 p 2 A real cubic polynomial P (x) has zeros 3 § i 2 and 2 3. It has a leading coefficient of 6. Find: b the coefficient of x2 a the sum and product of its zeros c the constant term. 3 A real polynomial of degree 5 has leading coefficient ¡1 and zeros of ¡2, 3 § i, and The y-intercept is 18. Find: p k § 1. b the coefficient of x4 . a k 4 A real polynomial of degree 5 has leading coefficient 2 and the coefficient of x4 is 3. When the p polynomial is graphed, the y-intercept is 5, and it cuts the x-axis at 12 and 1 § 2 only. Suppose the other two zeros are m § ni, n > 0. Use the sum and product formulae to find m and n. 5 A real quartic polynomial has leading coefficient 1 and zeros of the form a § i and 3 § a, where a 2 R . Its constant term is 25. What are the possible values that a may take? 6 x3 ¡ px2 + qx ¡ r = 0 has non-zero roots p, q, and r, where p, q, r 2 R . cyan magenta 1 r yellow 95 100 50 75 25 0 5 95 b Hence find p, q, and r. 100 50 75 25 0 and p = ¡ . 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 a Show that q = ¡r black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\200IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:11:00 AM BEN IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) F 201 GRAPHING REAL POLYNOMIALS Any polynomial with real coefficients can be graphed on the Cartesian plane. Use of a graphics calculator or the graphing package provided will help in this section. CUBIC POLYNOMIALS INVESTIGATION 2 CUBIC GRAPHS Every real cubic polynomial can be categorised into one of four types. In each case a 2 R , a 6= 0, and the zeros are ®, ¯, °. Type 1: Three real, distinct zeros: P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °) Type 2: Two real zeros, one repeated: P (x) = a(x ¡ ®)2 (x ¡ ¯) Type 3: One real zero repeated three times: P (x) = a(x ¡ ®)3 Type 4: One real and two complex conjugate zeros: P (x) = (x ¡ ®)(ax2 + bx + c), ¢ = b2 ¡ 4ac < 0. What to do: 1 Experiment with the graphs of Type 1 cubics. State the effect of changing both the size and sign of a. What is the geometrical significance of ®, ¯, and °? 2 Experiment with the graphs of Type 2 cubics. What is the geometrical significance of the squared factor? 3 Experiment with the graphs of Type 3 cubics. What is the geometrical significance of ®? GRAPHING PACKAGE 4 Experiment with the graphs of Type 4 cubics. What is the geometrical significance of ® and the quadratic factor which has complex zeros? From Investigation 2 you should have discovered that: ² If a > 0, the graph has shape or . If a < 0 it is or . ² All cubics are continuous smooth curves. ² Every cubic polynomial must cut the x-axis at least once, and so has at least one real zero. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 ² For a cubic of the form P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °), ®, ¯, ° 2 R , the graph has three distinct x-intercepts corresponding to the three distinct zeros ®, ¯, and °. The graph crosses over or cuts the x-axis at these points, as shown. ² For a cubic of the form P (x) = a(x ¡ ®)2 (x ¡ ¯), ®, ¯ 2 R , the graph touches the x-axis at the repeated zero ® and cuts it at the other x-intercept ¯, as shown. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\201IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:13:33 AM BEN ® ° ¯ ¯ ® x x IB_HL-3ed 202 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) ² For a cubic of the form P (x) = a(x ¡ ®)3 , x 2 R , the graph has only one x-intercept, ®. The graph is horizontal at this point, and the x-axis is a tangent to the curve even though the curve crosses over it. ® ² For a cubic of the form P (x) = (x ¡ ®)(ax2 + bx + c) where ¢ < 0, there is only one x-intercept, ®. The graph cuts the x-axis at this point. The other two zeros are complex and so do not appear on the graph. x ® x Self Tutor Example 31 Find the equation of the cubic with graph: a b y 2 y 4 6 x -1 -3 We -8 x 2 a The x-intercepts are ¡1, 2, 4. ) y = a(x + 1)(x ¡ 2)(x ¡ 4) But when x = 0, y = ¡8 ) a(1)(¡2)(¡4) = ¡8 ) a = ¡1 So, y = ¡(x + 1)(x ¡ 2)(x ¡ 4) When determining a polynomial function from a given graph, if we are not given all the zeros, or if some of the zeros are complex, we write a factor in general form. ² If an x-intercept is not given, use P (x) = (x ¡ k)2 (ax + b) . | {z } For example: most general form of a linear x 2 Using P (x) = a(x ¡ k) (x + b) is more complicated. ? k ² If there is clearly only one x-intercept and that is given, use P (x) = (x ¡ k) (ax2 + bx + c) . | {z } x k most general form of a quadratic ¢<0 cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Either graph is possible. black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\202IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:15:10 AM BEN IB_HL-3ed 203 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) Self Tutor Example 32 Find the equation of the cubic which cuts the x-axis at 2 and ¡3, cuts the y-axis at ¡48, and which passes through the point (1, ¡40). EXERCISE 6F.1 1 For a cubic polynomial P (x), state the geometrical significance of: a a single real linear factor such as (x ¡ ®), ® 2 R b a squared real linear factor such as (x ¡ ®)2 , ® 2 R c a cubed real linear factor such as (x ¡ ®)3 , ® 2 R . 2 Find the equation of the cubic with graph: a b y c y 12 - Qw y -4 6 Qw -3 3 x x x -1 2 3 -12 d e y f y -3 9 -5 5 y x -2 x -2 - Qw -12 -5 3 -4 x 3 Find the equation of the cubic whose graph: a cuts the x-axis at 3, 1, and ¡2, and passes through (2, ¡4) b cuts the x-axis at ¡2, 0, and 12 , and passes through (¡3, ¡21) c touches the x-axis at 1, cuts the x-axis at ¡2, and passes through (4, 54) cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 d touches the x-axis at ¡ 23 , cuts the x-axis at 4, and passes through (¡1, ¡5). black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\203IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:15:42 AM BEN IB_HL-3ed 204 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 4 Match the given graphs to the corresponding cubic function: a y = 2(x ¡ 1)(x + 2)(x + 4) b y = ¡(x + 1)(x ¡ 2)(x ¡ 4) c y = (x ¡ 1)(x ¡ 2)(x + 4) d y = ¡2(x ¡ 1)(x + 2)(x + 4) e y = ¡(x ¡ 1)(x + 2)(x + 4) f y = 2(x ¡ 1)(x ¡ 2)(x + 4) A B y C y y x 2 8 -4 16 x 1 2 D -1 x -4 1 E y -8 2 F y y 16 8 -2 x -2 -4 4 x -2 1 -4 1 x -4 1 -16 5 Find the equation of a real cubic polynomial which: a cuts the x-axis at 1 2 and ¡3, cuts the y-axis at 30, and passes through (1, ¡20) b cuts the x-axis at 1, touches the x-axis at ¡2, and cuts the y-axis at (0, 8) c cuts the x-axis at 2, cuts the y-axis at ¡4, and passes through (1, ¡1) and (¡1, ¡21). cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 QUARTIC POLYNOMIALS black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\204IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:16:20 AM BEN IB_HL-3ed 205 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) From Investigation 3 you should have discovered that: ² For a quartic polynomial in which a is the coefficient of x4 : I If a > 0 the graph opens upwards. I If a < 0 the graph opens downwards. ² If a quartic with a > 0 is fully factorised into real linear factors, then: I I for a single factor (x ¡ ®), the for a square factor (x ¡ ®)2 , the graph cuts the x-axis at ® graph touches the x-axis at ® ® y ® y x x for a cubed factor (x ¡ ®)3 , the graph cuts the x-axis at ® and is ‘flat’ at ® I y I ® y x or for a quadruple factor (x ¡ ®)4 , the graph touches the x-axis and is ‘flat’ at that point. ® y ® x x I I If a quartic with a > 0 has one real quadratic factor with ¢ < 0 we could have: If a quartic with a > 0 has two real quadratic factors both with ¢ < 0 we have: y y x x The graph does not meet the x-axis at all. Self Tutor Example 33 Find the equation of the quartic with graph: y -1 3 x -3 cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 -3 black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\205IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:19:20 AM BEN IB_HL-3ed 206 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) Self Tutor Example 34 Find the quartic which touches the x-axis at 2, cuts the x-axis at ¡3, and also passes through (1, ¡12) and (3, 6). EXERCISE 6F.2 1 Find the equation of the quartic with graph: a b y c y -3 2 x -1 y -2 -1 x -6 x 1 2 -1 We -16 d e y f y (-3, 54) 4 x -1 -3 y x Ew -1 3 x -2 -9 3 -16 2 Match the given graphs to the corresponding quartic functions: a y = (x ¡ 1)2 (x + 1)(x + 3) b y = ¡2(x ¡ 1)2 (x + 1)(x + 3) c y = (x ¡ 1)(x + 1)2 (x + 3) d y = (x ¡ 1)(x + 1)2 (x ¡ 3) e y = ¡ 13 (x ¡ 1)(x + 1)(x + 3)2 f y = ¡(x ¡ 1)(x + 1)(x ¡ 3)2 B y x x -1 magenta yellow 50 75 25 0 5 95 100 50 75 25 0 1 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 cyan -3 1 -3 95 -1 C y 100 A black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\206IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:20:10 AM BEN y -3 -1 1 x IB_HL-3ed 207 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) D E y F y 1 -1 -1 1 y x 3 3 1 x -3 x -1 3 Find the equation of the quartic whose graph: a cuts the x-axis at ¡4 and 12 , touches it at 2, and passes through the point (1, 5) b touches the x-axis at c cuts the x-axis at § 12 2 3 and ¡3, and passes through the point (¡4, 49) and §2, and passes through the point (1, ¡18) d touches the x-axis at 1, cuts the y-axis at ¡1, and passes through (¡1, ¡4) and (2, 15). ACTIVITY Click on the icon to run a card game for graphs of cubic and quartic functions. CARD GAME GENERAL POLYNOMIALS We have already seen that every real cubic polynomial must cut the x-axis at least once, and so has at least one real zero. If the exact value of the zero is difficult to find, we can use technology to help us. We can then factorise the cubic as a linear factor times a quadratic, and if necessary use the quadratic formula to find the other zeros. This method is particularly useful if we have one rational zero and two irrational zeros that are radical conjugates. DISCUSSION GENERAL POLYNOMIALS Consider the general polynomial P (x) = an xn + an¡1 xn¡1 + :::: + a1 x + a0 , an 6= 0. ² Discuss the behaviour of the graph as x ! ¡1 and x ! 1 depending on I I the sign of an whether n is odd or even. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 ² Under what circumstances is P (x) an: I I odd function even function? black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\207IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:23:31 AM BEN IB_HL-3ed 208 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) Self Tutor Example 35 Find exactly the zeros of P (x) = 3x3 ¡ 14x2 + 5x + 2. Self Tutor Example 36 3 2 Find exactly the roots of 6x + 13x + 20x + 3 = 0. For a quartic polynomial P (x) we first need to establish if there are any x-intercepts at all. If there are not then the polynomial must have four complex zeros. If there are x-intercepts then we can try to identify linear or quadratic factors. EXERCISE 6F.3 1 Find exactly all zeros of: a x3 ¡ 3x2 ¡ 3x + 1 c 2x3 ¡ 3x2 ¡ 4x ¡ 35 e 4x4 ¡ 4x3 ¡ 25x2 + x + 6 b x3 ¡ 3x2 + 4x ¡ 2 d 2x3 ¡ x2 + 20x ¡ 10 f x4 ¡ 6x3 + 22x2 ¡ 48x + 40 2 Find exactly the roots of: cyan magenta 95 yellow Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\208IB_HL-3ed_06.cdr Wednesday, 9 May 2012 9:19:03 AM BEN 100 50 75 25 0 5 95 100 50 b 2x3 + 3x2 ¡ 3x ¡ 2 = 0 d 2x3 + 18 = 5x2 + 9x f 2x4 ¡ 13x3 + 27x2 = 13x + 15 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 a x3 + 2x2 + 3x + 6 = 0 c x3 ¡ 6x2 + 12x ¡ 8 = 0 e x4 ¡ x3 ¡ 9x2 + 11x + 6 = 0 black IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 3 Factorise into linear factors with exact values: a x3 ¡ 3x2 + 4x ¡ 2 c 2x3 ¡ 9x2 + 6x ¡ 1 e 4x3 ¡ 8x2 + x + 3 g 2x4 ¡ 3x3 + 5x2 + 6x ¡ 4 209 x3 + 3x2 + 4x + 12 x3 ¡ 4x2 + 9x ¡ 10 3x4 + 4x3 + 5x2 + 12x ¡ 12 2x3 + 5x2 + 8x + 20 b d f h 4 The following cubics will not factorise neatly. Find their zeros using technology. a x3 + 2x2 ¡ 6x ¡ 6 b x3 + x2 ¡ 7x ¡ 8 5 A scientist is trying to design a crash test barrier with the characteristics shown graphically below. The independent variable t is the time f(t) ¡ after impact, measured in milliseconds, 120 100 such that 0 6 t 6 700. The dependent variable is the distance 80 60 the barrier is depressed during the 40 impact, measured in millimetres. 20 t (ms) 200 400 600 800 a The equation for this graph has the form f (t) = kt(t ¡ a)2 , 0 6 t 6 700. Use the graph to find a. What does this value represent? b If the ideal crash barrier is depressed by 85 mm after 100 milliseconds, find the value of k. Hence find the equation of the graph given. 6 Last year, the volume of water in a particular reservoir could be described by the model V (t) = ¡t3 + 30t2 ¡ 131t + 250 ML, where t is the time in months. The dam authority rules that if the volume falls below 100 ML, irrigation is prohibited. During which months, if any, was irrigation prohibited in the last twelve months? Include in your answer a neat sketch of any graphs you may have used. 7 A ladder of length 10 metres is leaning against a wall so that it is just touching a cube of edge length one metre as shown. What height up the wall does the ladder reach? xm 10 m 1m REVIEW SET 6A NON-CALCULATOR 1 Find real numbers a and b such that: a a + ib = 4 b (1 ¡ 2i)(a + bi) = ¡5 ¡ 10i c (a + 2i)(1 + bi) = 17 ¡ 19i 2 If z = 3 + i and w = ¡2 ¡ i, find in simplest form: cyan magenta yellow 95 100 50 75 c z3 25 0 5 95 100 50 75 25 0 5 95 z¤ w b 100 50 75 25 0 5 95 100 50 75 25 0 5 a 2z ¡ 3w black Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\209IB_HL-3ed_06.cdr Thursday, 29 March 2012 10:24:31 AM BEN IB_HL-3ed 210 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 3 Find exactly the real and imaginary parts of z if z = p 3 p + 3. i+ 3 4 Find a complex number z such that 2z ¡ 1 = iz ¡ i. Write your answer in the form z = a + bi where a, b 2 R . 5 Prove that zw¤ ¡ z ¤ w 6 Given w = z+1 z¤ + 1 is purely imaginary or zero for all complex numbers z and w. where z = a + bi, a, b 2 R , write w in the form x + yi, x, y 2 R . Hence determine the conditions under which w is purely imaginary. a (3x3 + 2x ¡ 5)(4x ¡ 3) 7 Expand and simplify: 8 Carry out the following divisions: a b (2x2 ¡ x + 3)2 x3 x+2 b x3 (x + 2)(x + 3) 9 Find the sum and product of the zeros of: a 3x4 ¡ 4x3 + 3x2 + 8 b 2x6 + 2x4 ¡ x3 + 7x ¡ 10 10 State and prove the Remainder theorem. 11 ¡2 + bi is a solution to z 2 + az + (3 + a) = 0. Find constants a and b given that they are real. 12 P (x) has remainder 2 when divided by x ¡ 3, and remainder ¡13 when divided by x + 2. Find the remainder when P (x) is divided by x2 ¡ x ¡ 6. p p 13 Find all quartic polynomials with real, rational coefficients, having 2 ¡ i 3 and 2 + 1 as two of the zeros. 14 If f(x) = x3 ¡ 3x2 ¡ 9x + b has (x ¡ k)2 as a factor, show that there are two possible values of k. For each of these two values of k, find the corresponding value for b, and hence solve f (x) = 0. p p 15 Find all real quartic polynomials with rational coefficients, having 3 ¡ i 2 and 1 ¡ 2 as two of the zeros. 16 When P (x) = xn + 3x2 + kx + 6 is divided by (x + 1) the remainder is 12. When P (x) is divided by (x ¡ 1) the remainder is 8. Find k and n given that 34 < n < 38. 17 If ® and ¯ are two of the roots of x3 ¡ x + 1 = 0, show that ®¯ is a root of x3 + x2 ¡ 1 = 0. Hint: Let x3 ¡ x + 1 = (x ¡ ®)(x ¡ ¯)(x ¡ °). REVIEW SET 6B ³ 1 If z = CALCULATOR ´3 5 ¡ 3 ¡ 2i , x, y 2 Z , express z in the form z = x + yi. 2¡i 2 Without using a calculator, find z if z 2 = 5 ¡ 12i. Check your answer using a calculator. 3 Prove that if z is a complex number then both z + z ¤ and zz ¤ are real. 4 If z = 4 + i and w = 3 ¡ 2i find 2w¤ ¡ iz. magenta 95 yellow Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\210IB_HL-3ed_06.cdr Monday, 23 April 2012 3:36:16 PM BEN 100 50 75 25 0 5 95 100 50 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 cyan 75 2 ¡ 3i = 3 + 2i. 2a + bi 5 Find rationals a and b such that black IB_HL-3ed COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 211 6 a + ai is a root of x2 ¡ 6x + b = 0 where a, b 2 R . Explain why b has two possible values. Find a in each case. 7 Find the remainder when x47 ¡ 3x26 + 5x3 + 11 is divided by x + 1. 8 Factorise 2z 3 + z 2 + 10z + 5 as a product of linear factors with exact terms. 9 A quartic polynomial P (x) has graph y = P (x) which touches the x-axis at (¡2, 0), cuts the x-axis at (1, 0), cuts the y-axis at (0, 12), and passes through (2, 80). Find an expression for P (x) in factored form, and hence sketch the graph of y = P (x). 10 Find all zeros of 2z 4 ¡ 5z 3 + 13z 2 ¡ 4z ¡ 6. 11 Factorise z 4 + 2z 3 ¡ 2z 2 + 8 into linear factors. 12 Find the general form of all real polynomials of least degree which have zeros 2 + i and ¡1 + 3i. 13 3 ¡ 2i is a zero of z 4 + kz 3 + 32z + 3k ¡ 1, where k is real. Find k and all zeros of the quartic. 14 Find all the zeros of the polynomial z 4 + 2z 3 + 6z 2 + 8z + 8 given that one of the zeros is purely imaginary. 15 When a polynomial P (x) is divided by x2 ¡ 3x + 2 the remainder is 2x + 3. Find the remainder when P (x) is divided by x ¡ 2. 16 Find possible values of k if the line with equation y = 2x + k equation x2 + y2 + 8x ¡ 4y + 2 = 0. does not meet the circle with p 17 P (x) = 2x4 ¡ 8x3 + ax2 + bx ¡ 110, a, b 2 R , has zeros m § 2i and 1 § n 3 where m, n 2 R . a Find m and n. b Sketch the graph of y = P (x). REVIEW SET 6C 1 Find real numbers x and y such that (3x + 2yi)(1 ¡ i) = (3y + 1)i ¡ x. 2 Solve the equation: z 2 + iz + 10 = 6z. 3 Prove that zw¤ + z ¤ w is real for all complex numbers z and w. 4 Find real x and y such that: c (x + iy)2 = x ¡ iy b (3 ¡ 2i)(x + i) = 17 + yi a x + iy = 0 5 z and w are non-real complex numbers with the property that both z + w and zw are real. Prove that z ¤ = w. 6 Find the sum and product of the roots of: a 2x3 + 3x2 ¡ 4x + 6 = 0 7 Find z if p z= b 4x4 = x2 + 2x ¡ 6 2 + 2 + 5i. 3 ¡ 2i cyan magenta 95 yellow Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\211IB_HL-3ed_06.cdr Monday, 23 April 2012 3:37:19 PM BEN 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 8 Find the remainder when 2x17 + 5x10 ¡ 7x3 + 6 is divided by x ¡ 2. black IB_HL-3ed 212 COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6) 9 5 ¡ i is a zero of 2z 3 + az 2 + 62z + (a ¡ 5), where a is real. Find a and the other two zeros. 10 Find, in general form, all real polynomials of least degree which have zeros: p b 1 ¡ i and ¡3 ¡ i. a i 2 and 12 11 P (x) = 2x3 + 7x2 + kx ¡ k is the product of 3 linear factors, 2 of which are identical. a Show that k can take 3 distinct values. b Write P (x) as the product of linear factors for the case where k is greatest. 12 Find all roots of 2z 4 ¡ 3z 3 + 2z 2 = 6z + 4. 13 Suppose a and k are real. For what values of k does z 3 + az 2 + kz + ka = 0 have: a one real root b 3 real roots? 14 (3x + 2) and (x ¡ 2) are factors of 6x3 + ax2 ¡ 4ax + b. Find a and b. 15 Find the exact values of k for which the line y = x¡k (x ¡ 2)2 + (y + 3)2 = 4. is a tangent to the circle with equation 16 Find the quotient and remainder when x4 + 3x3 ¡ 7x2 + 11x ¡ 1 is divided by x2 + 2. Hence, find constants a and b for which x4 + 3x3 ¡ 7x2 + (2 + a)x + b is exactly divisible by x2 + 2. cyan magenta 95 yellow Y:\HAESE\IB_HL-3ed\IB_HL-3ed_06\212IB_HL-3ed_06.cdr Monday, 23 April 2012 3:37:54 PM BEN 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 17 P (x) is a real polynomial of degree 5 with leading coefficient 1 and zeros m § 2i, 1 § mi, and 2 (m 2 R ). The graph of y = P (x) cuts the y-axis at ¡56. p a Show that m = § 3. b Find the coefficient of x4 . black IB_HL-3ed