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Textbook Fundamental Mathematics This book is available for studying in Science-Based Technology Vocational College (Chonburi) and Suranaree Technical College only Name........................................................................... NO. ..................................

1Chapter Calculation of numbers and expressions 1 Calculation of polynomials 1 1 Addition / subtraction of polynomials An expression represented by a product of numbers or letters, such as x4 and −3x2, is called a monomial. The number in a monomial is called a coefficient and the multiplied number of letters in the monomial is called a degree. If the monomial has no letter the degree is 0. For example, for the monomials x4 and −3x2, the coefficients are 1 and −3 and the degrees are 4 and 2 respectively. An expression represented by the sum or difference of monomials is called a multinomial, and together with a monomial, they are called polynomials. The monomials that form a polynomial are called terms. Note that each term includes a sign. As shown in the 4-th term of the expression x3 + 4x2 − 2x − 5, a term with 0 degrees is called a constant term. The greatest degree among the terms is called the degree of the polynomial. If the degree is n, it is called n-th degree polynomial. Degree Coefficient x 3 + 4 x2 − 2 x − 5 Term Constant term Ex1 x3 + 4x2 − 2x − 5 is a cubic polynomial and the constant term is −5.

2 Chapter1 Calculation of numbers and expressions The following calculation laws are established for the calculation of poly- nomials. Likewise in the case of numbers. The three laws of calculation A + B = B + A AB = BA (Commutative law) (A + B) + C = A + (B + C) (AB)C = A(BC) (Associative law) A(B + C) = AB + AC (A + B)C = AC + BC (Distributive law) The terms that contain the same letters are called similar terms. The similar terms can be combined into a single term using the laws above. Ex2 For a polynomial x + 3x2 − 5 − 3x, combine the similar terms x and −3x into a single term and arrange the terms in order from the greatest to the smallest (Descending order of powers). x + 3x2 − 5 − 3x = −2x + 3x2 − 5 Combine the similar = 3x2 − 2x − 5 terms togther Note There is a case where the terms are arranged in order from the smallest to the greatest (Ascending order of powers). Q. 1 Arrange the following expressions in descending order. (1) 5x2 + 2 − 3x2 − x − 1 + 3x (2) 5x − 4x2 + 1 − 2x + 7x2 − 6 Once a polynomial is arranged in descending order, addition and subtraction can be performed as follows. Ex3 When A = −4x2 + 3x + 2 B = 2x3 + x2 − 1, A + B = 2x3 − 3x2 + 3x + 1, A − B = −2x3 − 5x2 + 3x + 3. − 4x2 + 3x + 2 − 4x2 + 3x + 2 +) 2x3 + x2 −1 −) 2x3 + x2 −1 2x3 − 3x2 + 3x + 1 − 2x3 − 5x2 + 3x + 3

§ 1 Calculation of polynomials 3 Q. 2 Calculate A + B and A − B for the following sets of polynomials. (1) A = 3x2 + 2x + 1 B = x2 − 7x + 2 (2) A = x3 − x2 + 4 B = 2x4 + x2 − 3 If the polynomial contains multiple letters, you can focus on one letter by combining the similar terms into a single term, then, arrange them in descending order of powers. Ex4 x2 − 2xy2 + y3 − 3y + 1 is a cubic polynomial of x and y. Put the focus on the letter x and arrange them in descending order. x2 − 2y2x + (y3 − 3y + 1) Then, this is a quadratic polynomial with a constant term of y3 − 3y + 1. When A = 4x2 + 3ax + 2a2 and B = 2x3 + ax2 − a, calculate A + B and A − B and arrange them in descending order of x. A + B = 2x3 + (a + 4)x2 + 3ax + (2a2 − a) A − B = −2x3 + (−a + 4)x2 + 3ax + (2a2 + a) +) 2x3+ 4x2 + 3ax + 2a2 ax2 − a 2x3+ (a + 4)x2 + 3ax + (2a2 − a) Q. 3 Arrange the following polynomials in descending order of x. (1) 3ax2 + bx − ax2 − 3bx + c (2) 3x2 + 4xy + y2 + 3x − x2 − y + 7 Q. 4 Calculate A + B and A − B for the following sets of polynomials, and arrange them in descending order of letters in the ( ). (1) A = x3 + ax2 + 4a3 B = 2x4 + a2x2 − 3x (x) (2) A = x2 + 2xy + y2 B = −3x2 + 7xy + 2y2 (y)

4 Chapter1 Calculation of numbers and expressions 1 2 Multiplication of polynomials When multiplying a by nth times, assuming that n is a positive integer, it is called n-th power of a and denoted as an. a1 = a a2 = a · a a3 = a · a · a (1) All of these are called the power of a, and n is called an exponent or power. Note As you see in (1), the product of a and b can be written as a · b. The power can be calculated as shown below. a2 · a3 = a · a · a · a · a = a5 (a2)3 = a2 · a2 · a2 = a6 (ab)3 = ab · ab · ab = a3b3 In general, assuming that m and n are positive integers, the following equa- tions are established. Exponential law aman = am+n, (am)n = amn, (ab)n = anbn Multiplication of polynomials is performed based on the three laws of cal- culation and the exponential law. Ex5 (1) (3a2b)2 (−2ab3)3 = (9a4b2)(−8a3b9) = −72a4+3b2+9 = −72a7b11 (2) (x2 − 2x + 4)(2x − 3) = (x2 − 2x + 4) · 2x + (x2 − 2x + 4) · (−3) = (2x3 − 4x2 + 8x) + (−3x2 + 6x − 12) = 2x3 − 7x2 + 14x − 12 ×) x2 − 2x + 4 The calculation above can be done 2x − 3 −3x2 + 6x −12 as shown on the right. 2x3 − 4x2 + 8x Note From (−a)2 = (−a) · (−a) = a2, 2x3 − 7x2 +14x − 12 when a =\\ 0, −a2 =\\ (−a)2.

§ 1 Calculation of polynomials 5 Q. 5 Calculate the following expressions. (1) (−5)2 (2) −52 (3) −a2 (−b)3 (4) (−3a2b)3 (−2ab3)2 (5) ab3(a2 − 5b2) (6) (x2 + 5x − 2)(x + 2) When the products of several polynomials are expressed by the sums of monomials, it is called an expansion of polynomials. For the expansion of the polynomials, the following equations are applied. Expansion equations I (a + b)2 = a2 + 2ab + b2 (a − b)2 = a2 − 2ab + b2 II (a + b)(a − b) = a2 − b2 III (x + a)(x + b) = x2 + (a + b)x + ab IV (ax + b)(cx + d) = acx2 + (ad + bc)x + bd V (a + b)3 = a3 + 3a2b + 3ab2 + b3 (a − b)3 = a3 − 3a2b + 3ab2 − b3 Proof Prove the first equation of V . // (a + b)3 = (a + b)2(a + b) = (a2 + 2ab + b2)(a + b) = a3 + 2a2b + ab2 + a2b + 2ab2 + b3 = a3 + 3a2b + 3ab2 + b3 Ex6 (2x − 3y)2 = (2x)2 − 2 · 2x · 3y + (3y)2 = 4x2 − 12xy + 9y2 (2x + 1)(3x + 5) = 2 · 3 x2 + (2 · 5 + 1 · 3) x + 1 · 5 = 6x2 + 13x + 5 (a + 2b)3 = a3 + 3a2 · 2b + 3a · (2b)2 + (2b)3 = a3 + 6a2b + 12ab2 + 8b3 Q. 6 Calculate the following expressions using the expansion equations. (1) (x + 3)(x + 5) (2) (x + 5y)(x + 2y) (3) (3x − 1)(2x + 5) (4) (2x − 3y)(2x + 3y)

6 Chapter1 Calculation of numbers and expressions Q. 7 Prove the second equation of V by transforming into a−b = a+(−b), and calculate the following expressions. (1) (2a + b)3 (2) (3a − 2b)3 Q. 8 Prove the following equations. (1) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (2) (a + b)(a2 − ab + b2) = a3 + b3 (3) (a − b)(a2 + ab + b2) = a3 − b3 Q. 9 Calculate the following expressions. (1) (2a + b − 3c)2 (2) (x + 2y)(x2 − 2xy + 4y2) Exercise1 Calculate the following expressions. (1) (x + y + 1)(x + y + 2) (2) (a + b − c)(a − b + c) Solution (1) Let X = x + y, // The given expression = (X + 1)(X + 2) = X2 + 3X + 2 = (x + y)2 + 3(x + y) + 2 = x2 + 2xy + y2 + 3x + 3y + 2 (2) Since −b + c = −(b − c), The given expression = {a + (b − c)}{a − (b − c)} Therefore, by letting X = b − c, The given expression = (a + X)(a − X) = a2 − X2 = a2 − (b − c)2 = a2 − b2 + 2bc − c2 Q. 10 Calculate the following expressions. (1) (x + 2y − 1)(x + 2y − 3) (2) (x + y + z)(x − y − z) (3) (a + 2b + c)(a − 2b + c) (4) (a − b + 3c)(b − a + 3c)

§ 1 Calculation of polynomials 7 1 3 Factorization When polynomial P is expressed with products of several first and higher order polynomials A, B, · · · , it is called factorization of P , and the poly- nomials A, B · · · are referred to as the factors of P . Ex7 x2 − 4 = (x + 2)(x − 2) Factors are x + 2 and x − 2 xy − x2 = x(y − x) = −x(x − y) Factors are x and x − y) By looking at the expansion equations in reverse order, the following factoring formulas are obtained. Factoring formulas I ma + mb = m(a + b) an + bn = (a + b)n II a2 + 2ab + b2 = (a + b)2 a2 − 2ab + b2 = (a − b)2 III a2 − b2 = (a + b)(a − b) IV a3 + b3 = (a + b)(a2 − ab + b2) a3 − b3 = (a − b)(a2 + ab + b2) For the left side of each formula in I , m and n are called the common factors and the transformation is called factoring out the common factors. Exercise2 Factor the following expressions. (1) x4y − xy4 (2) a + b + ab + 1 Solution (1) x4y − xy4 Factor out the common factor xy Formula IV = xy(x3 − y3) = xy(x − y)(x2 + xy + y2) Arrange them with respect to a (2) a + b + ab + 1 = ab + a + b + 1 = a(b + 1) + (b + 1) The common factor b + 1 = (a + 1)(b + 1) //

8 Chapter1 Calculation of numbers and expressions Q. 11 Factor the following expressions. (1) a3 − 6a2b + 9ab2 (2) 2a − 3b + 3ab − 2 (3) 2x2 − 18y2 (4) a2 − b2 + 4bc − 4c2 (5) a3 + 8 (6) x2 − y2 + x3 − y3 The following equation is applied, when factoring a quadratic equation, where the coefficient of quadratic term is 1. x2 + (a + b)x + ab = (x + a)(x + b) Q. 12 Factor the following expressions. (1) x2 + 11x + 24 (2) x2 − 5x − 6 The following equation is applied, when factoring a quadratic equation, where the coefficient of quadratic term is not 1. Formula for factoring a quadratic equation acx2 + (ad + bc)x + bd = (ax + b)(cx + d) Exercise3 Factor the following expressions. (1) 3x2 + 10x + 8 (2) 6x2 − 7x − 3 Solution (1) You need to find the numbers for a, b, c and d, such that ac = 3, bd = 8, ad + bc = 10. In order to find the numbers, the following diagrams can be used. (Crossing multiplication diagram) a b bc 344 c d ad 126 ac bd ad + bc 3 8 10 The possible combination for ac = 3 is 3 and 1. So, place 3 and 1 on the left side of diagram, then, find the appropriate numbers for the rest of spaces. Thus, b = 4 and d = 2. Therefore, 3x2 + 10x + 8 = (3x + 4)(x + 2)

§ 1 Calculation of polynomials 9 (2) There are two possible combinations of numbers for ac = 6. However, for the combination 6 and 1, there is no set of numbers that can form ad + bc = −7, as you can see in the left diagram. 6 33 31 2 −9 1 −1 −6 2 −3 −7 6 −3 −3 6 −3 For the combination 3 and 2, b = 1, d = −3 can be found, as you can see in the right diagram. 6x2 − 7x − 3 = (3x + 1)(2x − 3) // Q. 13 Factor the following expressions. (1) 5x2 + 13x + 6 (2) 6x2 + x − 2 Exercise4 Factor the following expressions. (1) x4 + 5x2 − 6 (2) 2x2 + 7xy + 3y2 + x − 7y − 6 Solution (1) Let x2 = X, x4 + 5x2 − 6 = X2 + 5X − 6 = (X − 1)(X + 6) = (x2 − 1)(x2 + 6) = (x + 1)(x − 1)(x2 + 6) (2) Arrange the equation with respect to letter x before factoring. The expression = 2x2 + (7y + 1)x + (3y2 − 7y − 6) Factor the last term = 2x2 + (7y + 1)x + (3y + 2)(y − 3) Factor the whole = (2x + y − 3)(x + 3y + 2) // 2 y−3 y−3 1 3y + 2 6y + 4 2 (y − 3)(3y + 2) 7y + 1 Factor the following expressions. Q. 14 (1) x4 − 13x2 + 36 (2) (a + b)2 − 2(a + b) − 3 (3) x2 + 2xy + y2 − x − y − 2 (4) 2x2 + 5xy + 2y2 + 5x + y − 3

10 Chapter1 Calculation of numbers and expressions 1 4 Division of polynomials Division of polynomials can determine the quotient and the remainder, as shown on the right. When the result of dividing 27 by 6 is expressed with an equality, the following is obtained. 4 Quotient 27 = 6 × 4 + 3 Remainder Quotient Remainder 6 ) 27 24 3 For division of polynomials, arrange the given polynomial in descending order, then, divide them in the same way as division of integral numbers. For example, suppose that A = 3x3 − 4x2 + 12x + 16 and B = x2 − 2x + 5. Then, the calculation of A ÷ B is done as follow. 3x + 2 )B x2 − 2x + 5 3x3 − 4x2 + 12x + 16 A 3x3 − 6x2 + 15x 3xB 2x2 − 3x + 16 A − 3xB 2x2 − 4x + 10 2B x+ 6 A − 3xB − 2B In this calculation, subtract 3xB and 2B from A successively until the remaining degree becomes smaller than the degree of B. As a result, you can see that the following relation is established. A − 3xB − 2B = x + 6 Then, A = B(3x + 2) + (x + 6) At the time, the quotient is 3x + 2, and the remainder is x + 6. Equality of division When polynomial A is divided by B, let the quotient be Q and the remainder be R. A = BQ + R (Note that the degree of R < the degree of B) Quotient Remainder When R = 0, A = BQ. At this time, A is divisible by B.

§ 1 Calculation of polynomials 11 Exercise5 Find the quotients and remainders when polynomial A is divided by B, and express with equalities. (1) A = 4x2 + 7x + 5, B = 2x + 3 (2) A = x3 − 3, B = x + 2 Solution (1) When the calculation of A B is 2x + 1 )2 done as shown on the right, 2x + 3 4x2 + 7x + 5 The quotient is 2x + 1 4x2 + 6x 2 x+ 5 7 and the remainder is 2 . x+ 3 2 Thus, ( 1)+ 7 4x2 + 7x + 5 = (2x + 3) 2x + 7 2 22 (2) Be sure to match the degrees of terms, ) x2 − 2x + 4 and calculate as shown on the right. x + 2 x3 −3 The quotient is x2 − 2x + 4 x3 + 2x2 and the remainder is −11. Thus, − 2x2 x3 − 3 = (x + 2)(x2 − 2x + 4) − 11 − 2x2 − 4x 4x − 3 4x + 8 // −11 Q. 15 Find the quotients and remainders when polynomial A is divided by B, and express with equalities. (1) A = 3x2 + 5x − 10 B = x + 3 (2) A = 6x2 − 7x + 15 B = 2x − 5 (3) A = x2 + 1 B = 3x + 2 Q. 16 When a certain polynomial is divided by x − 3, the quotient is x2 + x + 6 and the remainder is 14. Find the polynomial.

12 Chapter1 Calculation of numbers and expressions When polynomial A is divisible by B, B is a divisor of A, and A is a multiple of B. When A = BC, B is a divisor of A, and A is a multiple of B. Ex8 (1) ab is a divisor of 3a3b2c. (2) (x − 2)(x + 3) is a multiple of (x − 2). The divisors that are in common among two or more polynomials are called common devisors, and the largest degree among them is called the greatest common measure (or greatest common devisor). The greatest common measure is the multiples of all the common divisors. On the other hand, the multiples that are in common among two or more polynomials are called common multiples, and the smallest degree among them is called the least common multiple. The least common multiple is the divisors of all the common multiples. The polynomial A and B, which have common divisors that are fixed num- bers only, are referred to as relatively prime. Exercise6 Find the greatest common measure and the least common multiple of the expressions 6a3b2d 8ab4c3 12a2b3c2. Solution The greatest common measure can be found by taking out the smallest exponents among the common factors of the expressions, and forming a product as shown below. The least common multiple can be found by taking out all the factors and forming a product with the largest exponents. The greatest common measur The least common multiple 6 a3 b2 d 6 a3 b2 d 8 a b4 c3 8 a b4 c3 12 a2 b3 c2 12 a2 b3 c2 2 a b2 24 a3 b4 c3 d Therefore, the greatest common measure is 2ab2 and the least common multiple is 24a3b4c3d. // Q. 17 Find the greatest common measures and the least common multi- ples of the following sets of polynomials. (1) ab, bc (2) 4a2bc3 6a3b2cd (3) 2x2(x − 1)3(x + 3) 6x(x − 1)2(x + 2)2

§ 1 Calculation of polynomials 13 1 5 Remainder theorem and factor theorem In order to define clearly that polynomial P is a polynomial of letter x, you can write as P (x). For example, P (x) = x3 + x2 − 2 x − 5, Q(x) = 2x3 + 5x2 + x − 2 When x is substituted with a specific value a, the value is expressed as P (a). Ex9 For P (x) and Q(x) from above, P (x) = x3 + x2 − 2 x − 5 Polynomial Substituted P (2) = 23 + 22 − 2 · 2 − 5 = 3 Value Likewise, Q(−1) = −2 + 5 − 1 − 2 = 0 Q. 18 For P (x) = 2x3 − 3x2 + 5x + 4 and Q(x) = −x3 + x2 − 2x + 2, find the following polynomials and values. Note that a is a specific value. (1) P (x) + Q(x) (2) 3 P (x) − 2 Q(x) (3) P (1) (4) Q(0) (5) Q(a) (6) P (−a) When P (x) = x3 + x2 − 2x − 5 is divided ) x2 + 3x + 4 by x − 2, the quotient is x2 + 3x + 4 and x − 2 x3 + x2 − 2x − 5 the remainder is 3. The remainder 3 is equal to the value of P (2) in Ex.9. This can be x3 − 2x2 explained as follows. 3x2 − 2x 3x2 − 6x When the polynomial P (x) is expressed by 4x − 5 the quotient and remainder, 4x − 8 3 P (x) = (x − 2)(x2 + 3x + 4) + 3. By substituting x with 2, P (2) = (2 − 2)(22 + 3 · 2 + 4) + 3 = 0 · (22 + 3 · 2 + 4) + 3 = 3 Thus, you will see that the value P (2) are equal to the remainder 3. In general, when the polynomial P (x) is divided by x − a, let the quotient be Q(x) and the remainder be R, P (x) = (x − a)Q(x) + R. By substituting x by a, the following remainder theorem is proved.

14 Chapter1 Calculation of numbers and expressions Remainder theorem The remainder is equal to P (a), when the polynomial P (x) is divided by x − a. Ex10 When the polynomial P (x) = 2x3 + 3x2 + 5x + 10 is divided by x − 3, the remainder is, P (3) = 2 · 33 + 3 · 32 + 5 · 3 + 10 = 106. When divided by x + 3, the remainder is P (−3) = 2 · (−3)3 + 3 · (−3)2 + 5 · (−3) + 10 = −32 with consideration of x + 3 = x − (−3). Q. 19 Find the remainders, when the following polynomial A(x) is divided by B(x). (1) A(x) = x3 − 2x2 + x + 3 B(x) = x − 1 (2) A(x) = x4 + x3 − 2x2 + 5x − 1 B(x) = x + 1 Find the remainder R, when the polynomial P (x) is divided by the linear polynomial ax − b (a =\\ 0). By letting the quotient be Q(x), when polynomial P (x) is divided by ax−b, P (x) = (ax − b)Q(x) + R. By substituting x with b a ( )( )( ) b = a× b −b Q b +R P a a a =R ( b ) a , Thus, the remainder is equal to P when P (x) is divided by ax − b. Q. 20 Find the remainders, when the polynomial x3 − 2x2 + 4x + 3 is divided by 2x − 1 and 2x + 3 respectively. For the polynomial Q(x) = 2x3 + 5x2 + x − 2 from Ex.9 on page 13, since Q(−1) = 0, the remainder is 0 according to the remainder theorem, when Q(x) is divided by x − (−1) = x + 1. Therefore, Q(x) is divisible by x + 1. In general, the following factor theorem is established.

§ 1 Calculation of polynomials 15 Factor theorem As for the polynomial P (x), if P (a) = 0, then P (x) is divisible by x − a. In reverse, If P (x) is divisible by x − a, then P (a) = 0 is established. Q. 21 By which linear expression, x − 1, x − 2 or x − 3, the polynomial P (x) = x3 − 3x2 + 4 can be divided without remainder? Q. 22 Find the constant value of k, such that the polynomial x3 + 5x2 + kx + 2 is divisible by x + 2. The polynomials can be factored using the factor theorem. Exercise7 Factor P (x) = x3 + 5x2 − 2x − 24. Solution P (1) = 1 + 5 − 2 − 24 = −20, · · · , P (2) = 8 + 20 − 4 − 24 = 0 Thus, P (x) is divisible by x − 2. If you actually divide them, the quotient x2 + 7x + 12 is obtained. P (x) = (x − 2)(x2 + 7x + 12) = (x − 2)(x + 4)(x + 3) // Note By substituting the divisors 1, −1, 2, −2, · · · · · · of a constant term −24 successively, you will find the value of a, where P (a) = 0. Q. 23 Factor the following expressions. (1) x3 + x2 − 3x + 1 (2) x3 + 2x2 − 11x − 12 (3) 2x3 − 7x2 + 7x − 2 (4) x4 − x3 − 6x2 + 4x + 8

16 Chapter1 Calculation of numbers and expressions Column Synthetic division There is a method called synthetic division, which enables you to find the quotient and the remainder easily, when a polynomial is divided by a linear expression x − a. For example, when P (x) = x3 + 5x2 − 2x 18 is divided by x + 3, the calculation can be done as shown on the right diagram using synthetic division. x2 + 2x − 8 Synthetic division x + 3 ) x3 + 5x2 − 2x − 18 −3 1 5 − 2 − 18 − 3 − 6 24 x3 + 3x2 2x2 − 2x 1 2 −8 6 2x2 + 6x Quotient x2 + 2x − 8 Remainder 6 − 8x − 18 − 8x − 24 6 Let me explain the system of synthetic division. First, arrange P (x) in descending order. Secondly, write −3 from x+3 = x−(−3) and the coefficients 1, 5, −2, −18 of P (x) on the top line, as shown on the diagram above.Leave one line space under the coefficients and draw a horizontal line.Then, carry out synthetic division in the order shown below. (1) First, write the coefficient on the left, 1, under the horizontal line. (2) Multiply 1 by the number from top left corner, −3, and write the product, −3, under 5. (3) Add 5 and −3 and write the sum, 2, under the line. (4) Write the product of 2 and −3, −6, under −2 and the write the sum, −8, under the line. (5) Repeat the same. 6 is the remainder and 1, 2, −8 are the coefficients of quotients. Factor P (x) = x3 − 7x − 6 using synthetic division. Since P (−1) = (−1)3 − 7(−7) + 6 = 0, −1 1 0 − 7 − 6 P (x) is divisible by x + 1. −1 1 6 From the synthetic division, 1 −1 −6 0 P (x) = (x + 1)(x2 − x − 6) = (x + 1)(x + 2)(x − 3)

§ 1 Calculation of polynomials 17 Exercise Problem 1-A 1. When A = 3a2 + 2ab − 4b2, B = a2 − ab + 3b2, and C = 2a2 + 3ab − b2, calculate the following expressions. (1) A + B + C (2) 3A − (2B + 5C) (3) AB − BC 2. Expand the following expressions. (1) (a + b)2(a − b)2 (2) (6x − 5)(7x + 8) (3) (3a + 2b − 5)(3a + 2b + 1) (4) (a − b)(a2 + ab + b2)(a3 + b3) (5) (x − 4y)3 (6) (x2 + 3xy − y2)(2x − 5y) 3. Factor the following expressions. (2) a4 − b4 (1) ax − by − bx + ay (4) x4 − 8x2 − 9 (3) 4a2 + 5a − 6 (6) x2 + 4xy + 3y2 − 8x − 6y − 9 (5) x2 + xy − 2y2 + 2x + 7y − 3 4. Find the quotient and the remainder, when polynomial A is divided by B, and express with equalities. (1) A = 2x4 + x3 + 6x2 + x + 10 B = 2x2 + 3x + 5 (2) A = x3 − 1 B = 2x + 1 5. Find the greatest common measures and the least common multiples of the following sets of polynomials. (1) ab3 a2bc a3b2c2 (2) x3 + 7x2 + 12x x2 − x − 20 (3) x4 − 5x2 + 4 x2 + x − 2 (4) x2 − 2x x2 − 3x + 2 x2 − 4x + 4 6. When a certain polynomial is divided by x2 − 1, the quotient is x4 + x2 + 1 and the remainder is x + 1. Find the quotient and the remainder when the polynomial is divided by x2 + 1. 7. When a certain polynomial is divided by (x + 1)(x − 3), the remainder is 3x + 1. Find the remainder, when the polynomial is divided by x − 3.

18 Chapter1 Calculation of numbers and expressions Exercise Problem 1-B 1. Expand the following expressions. (1) (2a + 6b − 1)(3a + 9b − 2) (2) (x + y − z)3 (3) (a + b + c)(a − b − c)(a − b + c)(a + b − c) (4) (x + 1)(x2 + x + 1)(x2 − x + 1)2 2. Factor the following expressions. (1) 3x3 − 2x2y − 5xy2 (2) a2 + b2 − c2 − d2 + 2ab − 2cd (3) xy2 − 2xyz + xz2 + y2z − yz2 (4) x6 − 7x3 − 8 (5) (x + y + 1)(x − 2y + 1) − 4y2 3. Factor the following expressions. (1) (b2 − c2)a + (c2 − a2)b + (a2 − b2)c (2) x5 + x4 + x3 + x2 + x + 1 (3) (x + y + z)3 − x3 − y3 − z3 4. For the polynomial of x, A = x2 − 4x − 5 and B, the greatest common measure is x + 1 and the least common multiple is x3 − 10x2 + 19x + 30. Find the polynomial B. 5. There is a quadratic polynomial and a cubic polynomial, in which the greatest common measure is 2x + 1 and the least common multiple is 4x4 + 3x2 − 1. Find the two polynomials. 6. When x4−1 is divided by the polynomial P (x), the quotient is x3−3x2+9x−27 and the remainder is 80. Find the polynomial P (x). 7. When the polynomial Q(x) is divided by x − 1 or x − 2, the remainder is 1 for both cases. Find the remainder, when Q(x) is divided by x2 − 3x + 2.

§ 2 Different types of numbers and expressions 19 2 Different types of numbers and expressions 2 1 Calculation of fractional expression When A and B are both polynomials and B =\\ 0, the expression represented by A is called a rational expression. If B is a constant number, it is a B polynomial. If B is not a constant number, it is a fractional expression, where A is the numerator and B is the denominator. If there are common divisors between the numerator and the denominator, you can simplify by dividing them by the common divisor, which is called reduction. A fractional expression that has been reduced to the smallest term is called an irreducible fractional expression. x2 + 2x − 2 = 1 x2 + x − 1 Polynomial Rational 2 2 Fractional expression x2 + 2x − 2 expression x+2 Common (x − 1) (x + 1) x + 1 Irreducible divisor = fractional expression (x − 1) (x + 2) x + 2 Ex1 a5 = a×a×a×a×a = a × a × a = a3 a2 a×a a3 = a×a×a =1 a3 a×a×a a2b2 = a×a×b×b = b = b a5b a×a×a×a×a×b a×a×a a3 In general, when m and n are positive integers and a =\\ 0, the following equations are established. Division law of exponents I when m > n, am = am−n an II When m = n, am =1 an III When m < n, am = 1 an an−m

20 Chapter1 Calculation of numbers and expressions Ex2 (2ab)5 = 25a5b5 = 22a2 = 4a2 (−3ab)2 = 9a2b2 = 9a (2ab2)3 23a3b6 b b ab2 ab2 x2 + 5x + 6 = (x + 3) (x + 2) = x+2 x2 + 2x − 3 (x + 3) (x − 1) x−1 Q. 1 Rewrite the following fractional expressions into the irreducible frac- tional expressions. (1) 16xy2z (2) x2 − 2x − 3 a2 − (b + c)2 12x3yz4 x3 − 6x2 + 9x (3) (a + b)2 − c2 To change the denominators of two or more fractional expressions into the same value is called reduction to common denominator. In general, the least common multiple of the denominators are the common denominator. For addition and subtraction of the fractional expressions that have different denominators, you need to reduce to the common denominator before calcu- late them. In general, the result of calculation is expressed by an irreducible fractional expression or a polynomial. Ex3 x + 1 = x(x − 2) + 1 (x − 1)3 (x − 1)3(x − 2) (x − 1)3(x − 2) (x − 1)3(x − 2) = x2 − 2x + 1 = (x + 1)2 = 1 (x − 1)3(x − 2) (x − 1)3(x − 2) (x − 1)(x − 2) x − xy = x − xy = x(x + y) − xy = x2 x+y 1 x+y x+y x+y x+y Q. 2 Calculate the following fractional expressions. (1) 2 + 1 (2) x+5 − x+3 x+1 x−1 x2 + x − 2 x2 − 4x + 3 (3) x+ xy (4) a − b x−y ab − b2 a2 − ab For multiplication and division of two fractional expressions, you can calculate in the same way as with the calculation of fractions. A × C = AC , A ÷ C = A × D = AD B D BD B D B C BC Ex4 x+1 × x2 − 4x + 3 = x+1 × (x − 3)(x − 1) = − x − 1 3−x x2 − x − 2 −(x − 3) (x − 2)(x + 1) x − 2 x2 + 5x + 6 ÷ x2 + 3x = (x + 3)(x + 2) × (x − 2)(x + 2) x2 − 4x + 4 x2 − 4 (x − 2)2 x(x + 3) (x + 2)2 = x(x − 2)

§ 2 Different types of numbers and expressions 21 Q. 3 Calculate the following fractional expressions. (1) 3bc × 8a 2a2 9b2c (2) t2 − 3t ÷ t3 − 6t2 + 9t t−5 t2 − 11t + 30 (3) x2 + x × x2 − 3x ÷ x2 x2 − x − 6 x2 − 1 x2 + x − 2 (4) 10y2 × y−x x(x − y) 5y3 A fractional expression that contains the fractions in the numerator or de- nominator is called a complex fractional expression. Exercise1 Simplify the following complex fractional expressions. c x− 2 x+1 (1) ab (2) x+y (3) ab2c x2 + y2 4 x+1− x+1 yx Solution Multiply the numerator and the denominator by an appropriate ex- pression such that the fractions in the numerator and denominator will be canceled out. (1) Multiply the numerator and denominator by ab. c ×ab c 1 ab a2b3c a2b3 The expression = = = ab2c×ab (2) Multiply the numerator and denominator by xy. The expression = (x + y)xy = (x + y)xy = xy x3 + y3 (x + y)(x2 − xy + y2) x2 − xy + y2 (3) Multiply the numerator and denominator by x + 1. The expression = x(x + 1) − 2 = (x + 2)(x − 1) = x+2 // (x + 1)2 − 4 (x + 3)(x − 1) x+3 Q. 4 Simplify the following complex fractional expressions. bc 1+ 1 2 +1 x− t−2 (1) ad (2) x (3) (4) x−2 b2 1 2 −1 1+ 3 − 10 a x t+2 x x2

22 Chapter1 Calculation of numbers and expressions For the improper fraction 14 , since 14 = 5 × 2 + 4 using the quotient 2 5 and the remainder 4 of 14 divided by 5, it can be written as 14 = 5×2+4 = 5×2 + 4 =2+ 4 . 5 5 5 5 5 Likewise, when the degree of numerator A is equal or larger than the degree of denominator B, the fractional equation A can be written as B A =Q+ R BB using the quotient Q and the remainder R of A divided by B. (From the equality of division A = BQ + R on page 10) Ex5 When 6x2 − 7x + 5 is divided by 2x − 3, the quotient is 3x + 1 and the remainder is 8. 6x2 − 7x + 5 = 3x + 1 + 8 2x − 3 2x − 3 Q. 5 Convert the following fractional expressions as shown in Ex.5. (1) 4x2 + 3x − 1 (2) −5x3 + x2 + 2x − 9 x−2 x2 + x + 1 2 2 Real number Natural numbers 1, 2, 3, · · · , negative numbers −1, −2, −3, · · · , and 0 are all called integers. A number represented as m using the integers m n and n (n =\\ 0) is called a rational number. Since the integer m can be represented as m , it is also a rational number. 1 The rational number that is not an integer is called a fraction. When a fraction is represented in a decimal form, it will be either a finite decimal, such as 3 = 0.6, or a recurring infinite decimal (circulating decimal), 5 such as 1 = 0.333333333 · · · = 0.3˙ . Here, 0.3˙ that 3 is repeating infinitely. 3 0.6˙ = 0.6666 · · · , 0.4˙ 7˙ = 0.474747 · · · , 0.35˙ 82˙ = 0.3582582 · · · There are some numbers that are infi√nite decimals but do not repeat when represented by decimals. For example, 2 and π (the circumference of a circle) are repre√sented as follows. 2 = 1.41421356 · · · , π = 3.14159265 · · · They are known as the infinite decimals that do not repeat. These numbers are called the irrational numbers.

§ 2 Different types of numbers and expressions 23 Both rational number and irrational number are the real numbers. Natural 0 Negative number integer Expansion of numbers Integer n Non-integer Finite Circulating decimal decimal Rational number m Irrational number n (decimal fraction without circulation) Real number All real numbers can be represented by the points on a straight line. √ √ π −3 2 O −4 −3 −2 −1 0 1 2 3 4 A point corresponds to the real number 0 is called an origin and written as point O. When every points on the straight line correspond to the real numbers respectively, the line is called a number line. O |a| P The real number a, which corresponds point P 0a on the number line, is called a coordinate of point P. The distance (OP) between point P, where the coordinate is a, and origin O is called the absolute value of real number a and represented by a sign |a|. Ex6 2 = 2 − 2 = 2 0 = 0 From the definition of absolute value. | − 2| |2| When x = 2, x = 2 x = −2 −2 0 2 Note x = 2 and x = −2 can be written as x = ±2. The sign is called a double sign and is used to express two equations in one.

24 Chapter1 Calculation of numbers and expressions a is equal to a, if a is a positive number or 0, and is equal to a with an opposite sign, if a is a negative number. It can be expressed as shown below. Absolute value { (if a ≧ 0) (if a < 0) |a| = a −a The absolute values have the following properties. Properties of absolute value I −a = a II b − a = a − b III ab = a b IV a = a (Note b =\\ 0) bb Proof I Divide into the cases of a > 0, a = 0 and a < 0. When a > 0, −a < 0. Thus, − a = −(−a) = a = a When a = 0, −a = 0. Thus, − a = a = 0 When a < 0, −a > 0. Thus, − a = −a = a II From I , b − a = − (a − b) = a − b III When a ≧ 0 and b < 0, ab = −ab = a(−b) = a b The other cases can be proven likewise. IV This can be proven in the same way as III . // Q. 6 Find the values of x − 1 + x − 2 , when the values of x are as follows. (1) x = 0 (2) x = π (3) x= π 2 Note a 2 = a2 is established for the real number a.

§ 2 Different types of numbers and expressions 25 2 3 Square root When a is a positive number or 0, the number that becomes a after raised to the second power is called a square root of a. Every positive number has two square roots that are the same absolute √ values but opposite signs. The positive square root is denoted as a. The √ negative square root is denoted √as − a. Furthermore, the square root of 0 is √ only 0 and denoted as 0 = 0. is called a root sign (radical sign). √√ Ex7 Square root of 4 is 2 and −2, then, 4 = 2. 4 is not ±2. For the calculation of expression that contains the root signs, the following equations are established. Properties of radical sign I (√ )2= a (a ≧ 0) a √ II a2 = a √√ √ III a b = ab (a ≧ 0 b ≧ 0) IV √ =√a (a ≧ 0 b > 0) √a bb √ Proof I From the definition of square root, a will be a when raised to the second power. II Since a 2 = a2, a is a positive (or 0) square root of a2. √√ √√ √√ III Since a b ≧ 0 and ( a b )2 = ab, a b is a positive (or 0) square root of ab. ( )2 IV Since √ and √ = a, it is same as III . √a ≧ 0 √a b bb √√ Ex8 32 = 3 = 3, (−3)2 = − 3 = 3 √ √ √√ √ √ √ √ √ 18 − 50 = 9 2 − 25 2 = 3 2 − 5 2 = −2 2 √ √ √ √63 = 9=3 = 63 7 7

26 Chapter1 Calculation of numbers and expressions Q. 7 Simplify the following expressions. √√ √ √√ √ √ (1) 20 − 45 + 4 5 (2) 15 10 − 2 54 + 3 24 √ √√√ √√ (3) ( 3 − 3 2)(2 3 + 2) (4) (3 + 2 5)2 − (3 − 2 5)2 Exercise2 Calculate the following expressions. (1) √( − √5)2 √ 2 (2) a4 + 2a2b2 + b4 √ √√ Solution (1) The expression = 2 − 5 = −(2 − 5) = 5 − 2 √ (2) The expression = (a2 + b2)2 = |a2 + b2| = a2 + b2 [a2 + b2 ≧ 0 is used] // Q. 8 Calculate the following expressions. (1) √(√ − 2)2 √ 2 (2) π2 − 8π + 16 When an expression that contains a root sign in the denominator is con- verted into an expression that doesn’t contain a root sign in the denominator, it is called rationalization of denominator. √√ √3 √3 √2 32 Ex9 2 = 22 = 2 Exercise3 Rationalize the denominator of √ 5 . 3−1 Solution By using (√ − )(√ + ) = (√ )2 − 12 = 3 − 1, 3 13 1 3 (√ ) (√ ) (√ ) (√ 5 3 +√1 5 3+1 5 3+1 √5 = ) ) = 3−1 = 2 // 3−1 3−1 ( 3+1 Q. 9 Rationalize the denominators of the following expressions. √√ (1) 1√4 (2) 1√ (3) √ 3√ (4) 3 − 2√2 5 7 3+ 5 5− 2 3+2 2

§ 2 Different types of numbers and expressions 27 2 4 Complex number When a real number is raised to the second power, it will be either a positive number or 0. Therefore, the square root of a negative number can’t be found in the range of real numbers. However, let’s define that the square root of a negative number can be found once the range of numbers is extended. Create the square root of −1 and represent it by i, which is called an imaginary unit. i2 = −1 When a and b are real numbers, the number α represented below is called a complex number. α = a+bi a is a real part of complex number α. b is an imaginary part of complex number α. complex number a + b i real part imaginary part Ex10 The complex number α = 3 + 2 i, where the real part is 3 and the imaginary part is 2. The complex number β = 5 − 7 i = 5 + (−7) i, where the real part is 5 and the imaginary part is −7. When b = 0, α is equal to the real number a. Thus, the real number is considered as a complex number. When b =\\ 0, α is called an imaginary number, and when a = 0, b =\\ 0, α is called a purely imaginary number. For example, 3 + 2 i is an imaginary number and 5 i is a purely imaginary number. Real number a Imaginary number a + bi (b =\\ 0) Purely imaginary b i Non-pure imaginary Complex number a + b i When a, b, c and d are real numbers, the two complex numbers α = a + b i and β = c + d i are equal if only a = c and b = d. Then, denoted as α = β.

28 Chapter1 Calculation of numbers and expressions Four arithmetic operations with complex numbers can be done in the same way as with the calculation of polynomials or fractional expressions with re- spect to letter i. Also note that i2 can be replaced by −1. Ex11 i3 = i2 × i = (−1)i = −i i4 = (i2)2 = (−1)2 = 1 1 = i = i = −i i i2 −1 Exercise4 For the complex numbers α = 4 − 3i and β = 3 + 2i, calculate the following. (1) α + β (2) α − β (3) αβ (4) α β Solution (1) α + β = (4 − 3i) + (3 + 2i) = 7 − i (2) α − β = (4 − 3i) − (3 + 2i) = 1 − 5i (3) αβ = (4 − 3i)(3 + 2i) = 12 − i − 6i2 = 12 − i + 6 = 18 − i (4) α = 4 − 3i = (4 − 3i)(3 − 2i) = 12 − 17i + 6i2 β 3 + 2i (3 + 2i)(3 − 2i) 9 − 4i2 = 6 − 17i = 6 − 17 i // 13 13 13 Commutative law, associative law and distributive law is established for the calculation of complex numbers. Likewise in the case of real numbers. The calculation result is a complex number. Q. 10 Calculate the following expressions. (1) (1 − 2i)(3 + 4i) (2) i+ 1 (3) 1 − 2i i 3 + 4i (4) 1 − 5i + 1 + 5i 1 + 5i 1 − 5i Determine the square roots of negative numbers using the complex numbers. By letting k be a positive number, the following is obtained. (√ki)2 = ki2 = −k (√ i)2 −k = ki2 = −k √√ Therefore, there are two numbers k i and − k i that are the square roots √√ of −k. k i shall be represented as −k.

§ 2 Different types of numbers and expressions 29 Square root of a negative number When k > 0, √√ −k = k i √√ The square root of −k is ± −k = ± k i. √√ √√ √√ √ √ Ex12 −2 = 2i −2 −3 = 2i 3i = 6i2 = − 6 √√ √ Note In general, a b = ab (Refer to the formula on page 25) is not established. Q. 11 Calculate the following. √√ √√ (2) −4 − −9 (1) −4 −9 A complex number can be represented as a y(imaginary axis) point in a plane, when the complex number b α = a+bi α = a + bi corresponds to the point (a, b) on a coordinate plane. The coordinate plane a x(real axis) is called a complex plane and the x-axis O and the y-axis is called the real axis and the imaginary axis respectively. The real −b α = a−bi number a is represented by a point (a, 0) and the purely imaginary number b i is represented by a point (0, b). For the complex number α = a + bi, the complex number a − bi is called a conjugate complex number of α and y 4+2i denoted as α. When α and α are plotted on 2 2i a complex plane, they are symmetrical with 1 1 3x respect to the real axis. Furthermore, α = α. 4 O Ex13 4 + 2i = 4 − 2i −1 4 − 2i = 4 + 2i, −2i = 2i 3 = 3 −2 −2 i 4−2i Q. 12 Plot the following complex numbers on a complex plane. (1) 2 − 3i (2) −2 + 3i (3) 3i (4) −4 Q. 13 Find the conjugate complex numbers for the complex numbers above and plot on a complex plane.

30 Chapter1 Calculation of numbers and expressions Q. 14 Calculate the following. (1) 3 + i + 3 + i (2) (2 − 5i)(2 − 5i) On a complex plane, the distance between the complex number α and origin O is called the absolute value of α and represented by a sign α . When α = a+bi, the following is established y(imaginary axis) with respect to Pythagorean proposition. b α = a + bi √ |α| α = |a + bi| = a2 + b2 Especially, when α is a real number, O a x(real axis) √ y 3+4i 4 α = a2 = a . ( Because b = 0 ) x O 3 It is equal to the absolute value of real number. −1 √ −3 Ex14 |3 + 4 i| = 32 + 42 = 5 −3 √ | − 3 − 2i| = (−3)2 + (−2)2 √ = 13 √ |2 i| = 02 + 22 = 2 | − 3| = 3 −3 − 2 i −2 −2 i Q. 15 Find the absolute values of the following complex numbers. (1) 5i (2) 4 + i (3) 4 − i (4) −4 − i Regarding the absolute value of a complex number, the following properties are established. Properties of absolute value of a complex number I −α = α α = α (β =\\ 0) II αβ = α β ββ

§ 2 Different types of numbers and expressions 31 Proof Prove the first equation of II . Suppose that α = a + b i, β = c + d i, αβ = (a + bi)(c + di) = (ac − bd) + (ad + bc)i Therefore, |αβ|2 = (ac − bd)2 + (ad + bc)2 = a2c2 − 2abcd + b2d2 + a2d2 + 2abcd + b2c2 = a2c2 + b2d2 + a2d2 + b2c2 Rearrange with consideration of a = a2(c2 + d2) + b2(c2 + d2) = (a2 + b2)(c2 + d2) = |α|2 |β|2 From |αβ|2 = (|α| |β|)2 and |αβ| ≧ 0 |α| |β| ≧ 0 , |αβ| = |α| |β| The others can be proven likewise. // Q. 16 Find the absolute values of the following complex numbers. (1) (2 + 3i)(3 − 2i) (2) 1 2+i Note For the complex number α, the following properties are established. |α| = |α|, αα = |α|2

32 Chapter1 Calculation of numbers and expressions Exercise Problem 2-A 1. Simplify the following expressions. (3xy3)2 (1) (−2x2y)3 (2) x + y − x2 + y2 x + y x − y x2 − y2 (3) 1 − y − z x x(x + y) (x + y)(x + y + z) (4) a2 − a − 6 × a2 − 16 ÷ a − 4 a2 + a − 12 a2 − 4 a − 2 a2 + 1 −1 a2 − 1 (5) a−1 − a+1 a+1 a−1 √ √ 2. Find the values of following expressions, when x = √3 + 1 y = √3 − 1 . 3−1 3+1 (1) x + y (2) xy (3) x2 + y2 (4) x3 + y3 3. Simplify the following expressions. √√ √ √ √√ √ √ (2) 1 − √3 − √3 + 1 2√+ 3 √ 3 − 1 (1) ( 5 + 3 − 2)( 5 − 3 + 2) (4) √2 + i − √2 − i √√ 2−i 2+i (3) 2 + √−3 + 2 − √−3 2 − −3 2 + −3 4. Find the values of the following expressions. √ (1) 5 −√3 √√ −2 + 5 (2) | − 2 + 2i | − | − 5 − i | √ (3) 2 +√3i 2 − 3i

§ 2 Different types of numbers and expressions 33 Exercise Problem 2-B 1. Simplify the following expressions. (1) 2a2 + a − b 4a2 − b2 b − 2a (2) 1 − 1 − 2 − 4 a − 1 a + 1 a2 + 1 a4 + 1 (3) x3 x+ 1 1 x− x (4) x+2 − x+2 1− 1 1+ 1 x+3 x+1 (5) 2a 1 − 1 1− 1 1+ 1 aa 2. Simplify the following expressions. (1) √ 1 √ + √ 1 √ + √ 1 √ 2+ 1 3+ 2 4+ 3 (2) √1 √ 1+ 2+ 3 √√ 3. Find the value of a2 − x2, when a ≧ 1 and x = 2 a − 1 4. For two complex numbers α and β, prove the following. (1) α + β = α + β (2) α − β = α − β (3) αβ = α β (4) ( α ) = α (β =\\ 0) ββ

Chapter2 Equations and Inequalities 1 Equations 1 1 Quadratic equations If an equation holds only when certain values are substituted for x, then we say this is an equation for x. The values of x that satisfy the equation are called the solutions of an equation, and finding the solutions is called solving its equation. In addition, the letter x is called an unknown. By simplifying an equation using transposition, ax2 + bx + c = 0 (Note that a, b and c are constants with a =\\ 0.) (1) The equation in the form of the above is called a quadratic equation. Here- inafter, let a, b and c be real numbers. Exercise1 Solve the equation 6x2 − x − 2 = 0. Solution By factoring the left side, (3x − 2)(2x + 1) = 0 From this, 3x − 2 = 0 or 2x + 1 = 0 Therefore, x= 2 x = − 1 ( be written as, x= 2 − 1) // 3 2 It can 3 2 Note If AB = 0, then either A = 0 or B = 0.

§ 1 Equations 35 Q. 1 Solve the following equations. (1) x2 + 2x = 18 − x (2) x2 = 64 (3) 9x2 + 12x + 4 = 0 (4) 7x2 + 24x + 9 = 0 Let us derive the formula representing the solution of a quadratic equation ax2 + bx + c = 0. ax2 + bx + c = 0 Divide both sides by a (=\\ 0) Transform the left side x2 + b x + c = 0 Simplify using transposition aa Take the square root Simplify using transposition ( b )2 ( b )2 c =0 x+ − + 2a 2a a ( b )2 b2 − 4ac x+ = 4a2 2a √ x + b = ± b2 − 4ac 2a 2a √ x = −b ± b2 − 4ac 2a This is called the formula giving roots of a quadratic equation. The formula giving roots of a quadratic equation The solutions of a quadratic equation ax2 + bx + c = 0 are, √ x = −b ± b2 − 4ac 2a Let’s determine how the solutions change depending on the expression inside the radical sign, b2 − 4ac. If b2 − 4ac > 0 √ Since b2 − 4ac is a positive real number, the equation has two distinct real solutions.

36 Chapter2 Equations and Inequalities Ex1 The solutions of 2x2 + 3x − 4 = 0 are √ √√ x = −3 ± 32 − 4 × 2 × (−4) −3 ± 9 + 32 −3 ± 41 2×2 = 4 = 4 Q. 2 Solve the following equations. (1) x2 + 5x + 2 = 0 (2) x2 − 1 x − 5 = 0 22 (3) x2 − 3x − 9 = 0 (4) 3x2 + 6x + 1 = 0 If b2 − 4ac = 0 Although this indicates only one real solution x = − b , 2a in this case we consider that the two real solutions overlap each other. We call it a double root. Ex2 The solutions of x2 − 8x + 16 = 0 are √√ x = −(−8) ± (−8)2 − 4 × 1 × 16 8± 64 − 64 8±0 2×1 = 2 = 2 =4 Q. 3 Solve the following equations. (1) 4x2 + 12x + 9 = 0 (2) x2 − 1 x + 1 = 0 2 16 If b2 − 4ac < 0 √ Since b2 − 4ac is an imaginary number, the equation has two distinct imaginary solutions. Ex3 The solutions of2x2 − 3x + 5 = 0 are √ √√ x= 3± 9 − 40 3± −31 3 ± 31 i 4 = 4 = 4 Q. 4 Solve the following equations. (1) x2 − 3x + 5 = 0 (2) x2 + 16 = 0 (3) 2x2 + 3x + 4 = 0 (4) 4x2 − 4x + 5 = 0 If b2 − 4ac is D, then we say this is the discriminant of a quadratic equation ax2 + bx + c = 0, and the formula giving roots is represented as follows. √ −b ± D x= 2a

§ 1 Equations 37 Besides, depending on the sign of D, the discriminant can discriminate whether the solutions of an equation have real solutions, a double root or imaginary solutions. Relationship between solutions and the discriminant When the discriminant of a quadratic equation ax2 + bx + c = 0 is D = b2 − 4ac, I If D > 0, then the quadratic equation has two distinct real solutions. II If D = 0, then the quadratic equation has a double root of real numbers. III If D < 0, then the quadratic equation has two distinct imaginary solutions. Q. 5 Find the discriminants of the following quadratic equations. (1) x2 − 3x + 4 = 0 (2) 2x2 − x − 6 = 0 (3) 9x2 − 6x + 1 = 0 Exercise2 Determine the value of constant k such that the quadratic equation x2 + (k − 1)x − 2k − 1 = 0 has a double root, and also find the double root. Solution First, determine the value of k for which the discriminant is 0. D = (k − 1)2 − 4(−2k − 1) = k2 + 6k + 5 = (k + 1)(k + 5) = 0 From this, k = −1 k = −5 √ Then the solutions are, x= −(k − 1) ± D = − k − 1 2 2 Therefore, the double roots are, x = 1 when k = −1 and x = 3 when k = −5. Q. 6 Determine the value of constant k such that each of the following quadratic equations has a double root and also find the double root. (1) x2 + 6x + (2k + 1) = 0 (2) x2 − (k + 2)x + 3k + 1 = 0 Note When the coefficient b is an even number for a quadratic equation ax2 + bx + c = 0, and if b = 2b′, then the formula giving roots and the discriminant D ar√e given as follows. x = −b′ ± b′2 − ac , a D = 4(b′2 − ac)

38 Chapter2 Equations and Inequalities 1 2 Relationship between solutions and coefficients If the two solutions of a quadratic equation ax2 + bx + c = 0 are α and β, then the sum of the solutions α + β and the product of the solutions αβ can simply be represented using coefficients. By using the formula giving roots, √√ α + β = −b + b2 − 4ac + −b − b2 − 4ac = − 2b = − b √ 2a √ 2a 2a a αβ = −b + b2 − 4ac × −b − b2 − 4ac 2a√ 2a (−b)2 − ( b2 − 4ac)2 b2 − (b2 − 4ac) = 4a2 = 4a2 = 4ac = c 4a2 a From this, the following relationship between solutions and coefficients is established. Relationship between solutions and coefficients If the two solutions of a quadratic equation ax2 + bx + c = 0 are α and β, α + β = − b αβ = c a a Exercise3 Find the value of each expression, if α and β are the two solutions of the quadratic equation 3x2 + 5x − 7 = 0. (1) α2β + αβ2 (2) α2 + β2 Solution From the relationship between solutions and coefficients, α + β = − 5 αβ = − 7 3 3( = αβ(α =− 7 ) ( 5 ) 35 − (1) Given equation + β) 3( )32 = =− (9 5 2− 7 ) 67 (2) Given equation = (α + β)2 − 2αβ − = // 3 39 Q. 7 Find the value of each expression, if α and β are the two solutions of the quadratic equation x2 − 7x + 9 = 0. (1) (α − 2)(β − 2) (2) β + α (3) α3 + β3 α β

§ 1 Equations 39 Next, a quadratic expression ax2 + bx + c is always factorable by using the two solutions α and β of a quadratic equation ax2 + bx + c = 0. From the relationship between solutions and coefficients, b = −a(α + β), c = aαβ Then, ax2 + bx + c = ax2 − a(α + β)x + aαβ = a{x2 − (α + β)x + αβ} = a(x − α)(x − β) Therefore, by expanding the range of numbers to complex numbers, the following formula is established. Factorization of a quadratic expression If the two solutions of a quadratic equation ax2 + bx + c = 0 are α and β, ax2 + bx + c = a(x − α)(x − β) Exercise4 Factor the following quadratic expressions. (1) x2 + 6x + 7 (2) 2x2 − 2x + 5 Solution (1) The solutions of x2 + 6x + 7 = 0 are, √ √ x= −6 ± 8 = −6 ± 2 2 √ = −3 ± 2 2 2√ √ ∴ x2 + 6x + 7 = {x − (−3 + 2)}{x − (−3 − 2)} √√ = (x + 3 − 2)(x + 3 + 2) (2) The solution√s of 2x2 − 2x + 5 = 0 are, x = 2 ± 4 − 40 = 2 ± 6i = 1 ± 3i 4( 4 (2 2x2 − 2x + 5 = 2 x − 1+ 3i ) x− 1 − 3i ) ∴ 22 // Q. 8 Factor the following quadratic expressions. (1) x2 − 5x + 3 (2) x2 − 4x + 1 (3) 3x2 + 2x + 4 (4) 4x2 + 8x − 1

40 Chapter2 Equations and Inequalities 1 3 Various types of equations Higher degree equations If p(x) is a polynomial of n-th degree, then the equation in the form of P (x) = 0 is called an equation of n-th degree. When n ≧ 3, then there are no simple formulas such as giving roots of a quadratic equation. Here, we deal with higher degree equations that can be solved by using transformation and factorization. Exercise5 Solve the following equations. (1) x4 − 2x2 − 3 = 0 (2) x3 − 2x + 1 = 0 Solution (1) If x2 = X, then the given equation is, X2 − 2X − 3 = 0 By solving for X = 3 −1 When X = 3, √ From x2 = 3, x = ± 3 When X = −1, From x2 = −1, x = ± i √ Therefore, x = ± 3 ± i (2) If P (x) = x3 − 2x + 1, then, P (1) = 0. Since it is divisible by x − 1 from the factor theorem, we use factorization to find the quotient. (x − 1)(x2 + x − 1) = 0 Then, x − 1 = 0 x2 + x − 1 = 0 // √ From this, x = 1 −1 ± 5 2 Q. 9 Solve the following equations. (1) x4 − 2x2 − 35 = 0 (2) x5 − 5x3 + 4x = 0 Q. 10 Solve the following equations using the factor theorem. (1) 2x3 − 7x2 + 2x + 3 = 0 (2) x4 + x3 − x2 + x − 2 = 0

§ 1 Equations 41 Simultaneous equations A set of two or more equations is called simultaneous equations, and a set of values of the unknowns that satisfies all the equations simultaneously is called the solutions of simultaneous equations. Additionally, the simulta- neous equations consisting of only linear equations are called simultaneous linear equations. Exercise6 Solve the following simultaneous linear equations. (1) x + 2y + z = 3 (2) 2x + 3y − z = −4 (3) 4x − y + 3z = 0 Solution Eliminate the unknowns one by one. First, eliminate the unknown z. From (1) (2), 3x + 5y = −1 (4) From (1)×3 − (3), −x + 7y = 9 (5) From (4)+(5)×3, 26y = 26 ∴ y=1 By substituting into (5), x = −2 By substituting the values for x and y into (1), z=3 ∴ x = −2 y = 1 z = 3 // Q. 11 Solve each of the following sets of simultaneous linear equations. x+y+z = 3 3x − y = 0 (1) x + 2y + 3z = 2 (2) x − y + 2z = 6 2x + 3y − 2z = −1 2x − 3y + z = 9

42 Chapter2 Equations and Inequalities Exercise7 Solve the following simultaneous equations. (1) 3x + y = 1 (2) 6x2 − y2 − 2y = 3 Solution Eliminate the unknowns using the substitution method. From (1), y = 1 − 3x By substituting into (2), 6x2 − (1 − 3x)2 − 2(1 − 3x) = 3 By simplifying, x2 − 4x + 2 = 0 // √ From this, √ x=2± 2 √√ When x = 2 + 2, √ y = 1 − 3(2 + 2) = −5 − 3 2 √√ When x = 2 − 2, √ y = 1 − 3(2 − 2) = −5 + 3 2 √ x=2+ 2 ∴√ x=2− 2 √ y = −5 − 3 2 y = −5 + 3 2 Note The solutions can also be written as follows. √ x=2± 2 √ (Double sign in same order) y = −5 ∓ 3 2 Double sign in same order means that, for double sign with x and y, if x takes the sign of the above, then y also takes the sign of the above, and if x takes the sign of the below, then y also takes the sign of the below. Q. 12 Solve each of the following sets of simultaneous equations. 2x + y = 3 x−y = 2 (1) (2) 2x2 − y2 − 3y + 2 = 0 x2 + xy + y2 = 5

§ 1 Equations 43 Other types of equation Exercise8 Solve the following equation. 2x + 3 = 11 Solution This is, 2x + 3 = ±11 ∴ x = −7, 4 // When 2x + 3 = 11, x = 4 When 2x + 3 = −11, x = −7 Q. 13 Solve the following equations. (1) |2x − 1| = 1 (2) |2x − 5| = |x| Exercise9 Solve the following equations. 1 − x = 4 (1) x−2 x+2 x2 − 4 Solution Since the denominator of a fractional expression is not 0, x =\\ ±2. By multiplying both sides by (x + 2)(x − 2) to clear the denominators, (x + 2) − x(x − 2) = 4 (2) From this, x2 − 3x + 2 = 0 ∴ x = 1, 2 Since x =\\ ±2, the solution is, x = 1 // Note In the exercise above, x = 2 is the solution of (2), but not the solution of (1). This is because we need to divide both sides by (x+2)(x−2) in order to derive (1) from (2), but when (x + 2)(x − 2) = 0, we cannot divide both sides. In this manner, an apparent solution of an equation that appears in the transformation process is called an extraneous root. Q. 14 Solve the following equations. (1) 6 + 1 = 4x x+3 x−2 (x + 3)(x − 2) (2) x − 3 = 2x2 x+2 x−2 x2 − 4

44 Chapter2 Equations and Inequalities Exercise10 Solve the following equation. (1) √ 7 − 2x = x − 2 Solution By squaring both sides, 7 − 2x = (x − 2)2 From this, x2 − 2x − 3 = 0 ∴ x = 3, x = −1 Among these solutions, x = 3 satisfies (1). As for x = −1, since the left side of (1) is positive and the right side is negative, it is not the solution of (1). Therefore, the solution of the equation is, x = 3 // √ Note x = −1 is the solution of − 7 − 2x = x − 2 and is an extraneous root. Q. 15 Solve the following equations. √√ (1) x + 1 = x − 5 (2) 25 − x2 = x − 1 1 4 Identical equations No matter what values are substituted into the letters in the equations such as x2 − 4 = (x + 2)(x − 2) and (x + y)2 = x2 + 2xy + y2, the equation that holds for any values of the letters is called an identical equation. Here, if a, b, c, a′, b′ and c′ are constants, ax2 + bx + c = a′x2 + b′x + c′ (1) let us derive the conditions for (1) to be identical for x. Since the equation will hold no matter what values are substituted into x of (1), By substituting x = 0c, = c′ By substituting x = 1a, + b = a′ + b′ (2) By substituting x = −a1−, b = a′ − b′ (3) From (2) + (3), 2a = 2a′ Where, a = a′ From (2) − (3), 2b = 2b′ Where, b = b′ Therefore, a = a′ b = b′ c = c′ (4)

§ 1 Equations 45 If (4) holds on the other hand, then (1) is obviously an identical equation. Conditions for an identical equation If a, b, c, a′, b′ and c′ are constants, ax2 + bx + c = a′x2 + b′x + c′, the conditions for the equation to be identical for x is, a = a′ b = b′ c = c′. The equation holds only under the conditions. Ex4 If x2 + bx − 15 = ax2 + 2x + cis identical for x, a = 1, b = 2, c = −15 Q. 16 Determine the values of constants a and b that will make the fol- lowing equations identical for x. (1) ax2 + 6 = −2x2 + b (2) x2 − 1 = (x − 1)(ax + b) The same applies to the conditions for an identical equation of n-th degree. Exercise11 Determine the values of constants a, b and c that will make the following equation identical for x. 2x3 + 3x2 + ax + 3 = (x2 + 2x + b)(2x + c) (1) Solution Since the right side = 2x3 + (c + 4)x2 + (2b + 2c)x + bc, (1) is given as follows. 2x3 + 3x2 + ax + 3 = 2x3 + (c + 4)x2 + (2b + 2c)x + bc (2) The conditions for (2) to be identical are, c + 4 = 3 2b + 2c = a bc = 3 By solving these, a = −8 b = −3 c = −1 // Q. 17 Determine the values of constants a, b and c that will make the following equations identical for x. (1) 3x2 + 2x + 5 = a + b(x − 1) + c(x − 1)(x − 2) (2) x3 − 2x2 − 3x + 5 = (x + 1)3 + a(x + 1)2 + b(x + 1) + c

46 Chapter2 Equations and Inequalities Exercise12 Determine the values of constants a and b that will make the following equation identical. x+5 = a + b (1) (x + 2)(x − 1) x+2 x−1 Solution Since the right side = a + b = a(x − 1) + b(x + 2) , x+2 x−1 (x + 2)(x − 1) by comparing the numerators on the left side with the right side, x + 5 = a(x − 1) + b(x + 2) Where, x + 5 = (a + b)x + (−a + 2b) By making this identical, a + b = 1, −a + 2b = 5 By solving these, a = −1, b = 2 Q. 18 Determine the values of constants a, b and c that will make the following equations identical. (1) 1 = a + b (x − 3)(x − 1) x−3 x−1 (2) 2x = a + bx + c (x − 1)(x2 + 1) x−1 x2 + 1 Note If a fractional expression is transformed into the sum of simpler frac- tional expressions whose denominators have lower degree than the original, then it is called the decomposition into partial fractions. 1 5 Proof of equations In the following exercises, let us show some processes for proving that an equation holds. Exercise13 Prove that the following equation holds. (ac + bd)2 + (ad − bc)2 = (a2 + b2)(c2 + d2) Proof The left side = a2c2 + 2abcd + b2d2 + a2d2 − 2abcd + b2c2 = a2c2 + a2d2 + b2c2 + b2d2

§ 1 Equations 47 // Likewise, the right side = a2c2 + a2d2 + b2c2 + b2d2 ∴ The left side = The right side Q. 19 Prove the following equations. (1) x4 + 4y4 = {(x + y)2 + y2}{(x − y)2 + y2} (2) (a2 − b2)(c2 − d2) = (ac + bd)2 − (ad + bc)2 Exercise14 If a + b + c = 0 where abc =\\ 0, prove the following equation. b+c + c+a + a+b = −3 a b c Proof From the given condition, c = −a − b. By substituting this into the left side of the equation, The left side = −a + −b + a+b a b −(a + b) = −1 − 1 − 1 = −3 = The right side // Q. 20 If x + y + z = 0, prove x3 + y3 + z3 = 3xyz. If a, b, c and d are not 0, but a = c , then it should be written as bd a : b = c : d. Exercise15 If a : b = c : d, prove that the following equation holds. a+b = c+d b d Proof If a = c = k, a = bk and c = dk b d By substituting these into both sides of the equation, The left side = bk + b = b(k + 1) = k + 1 bb The right side = dk + d = d(k + 1) =k+1 d d ∴ The left side = The right side // Q. 21 If a : b = c : d, prove that each of the following equations holds. (1) 2a + 3b = 2c + 3d (2) a−b = c−d b d a+b c+d

48 Chapter2 Equations and Inequalities Exercise Problem 1-A 1. Solve the following equations. (2) 3x2 − 2x + 5 = 0 (1) 2x2 + 3x − 1 = 0 (3) (x + 1)2 + (x − 2)2 = 0 (4) 3x3 − 2x2 − 6x − 1 = 0 √ (5) 1 − 1 =2 x−2 x−4 (6) 5 − x2 = 2x + 5 2. Solve each of the following sets of simultaneous equations. 2x − y + 3z = 7 x+y = 3 (1) x + y − z = 4 (2) 2x + 3y − 4z = 8 x2 − 2xy − 2y2 = 0 3. Solve the following simultaneous equations. (1) x + y − 2 = 2x − y = x − 2y + 4 (2) x + 2y + 4 = 2x − y + 7 = 2y − x (3) 2x + 3y − 5z − 3 = x − y + z = 3x − 6y + 2z + 7 = 0 4. Determine the value of constant k such that the following quadratic equation has a double root. x2 + 4x − 4 = k(x + 5) 5. Find the value of each expression, if α and β are the two solutions of the quadratic equation 2x2 − 8x − 3 = 0. (1) α2 + β2 (2) α3 + β3 (3) α4 + β4 6. Factor the following expressions. (2) 8x2 − 12x + 5 (1) 15x2 + 22x + 8 7. There is a rectangle of land with length 30 m and width 50 m. Suppose you want to build a road of constant width along the inside perimeter of the land such that the road will have an area of 200 m2. How many meters (m) should the width of the road be? 8. Determine the values of constants a, b, c and d that will make the following equation identical for x. 6x3 − 16x2 − 5 = a + b(x − 2) + c(x − 2)2 + d(x − 2)3

§ 1 Equations 49 Exercise Problem 1-B 1. Solve the following equations. (1) 3x4 + 10x2 − 8 = 0 (2) 2x4 + 2x3 − 13x2 + 12x − 3 = 0 (3) 2(x + 1) − 1 = x x2 + 2x − 15 x + 5 x − 3 (4) 2x − 3 = 3x − 2 √√ (5) x + 5 = x − 3 + 2 2. Solve each of the following sets of simultaneous equations. x+y = 2 x + 2y + 3z = 10 (1) (2) x = y = z x3 + y3 = 26 432 3. Find the values of constants a, b and c that will make the following equation identical for x. 2x − 1 = a + bx + c x(x2 − x + 1) x x2 − x + 1 4. Determine the values of constants a and b such that the polynomial x3 + 2x2 + ax + b is divisible by (x − 1)2. 5. If abc = 1, prove the following equation. a + b + c =1 ab + a + 1 bc + b + 1 ca + c + 1 6. If x = y = a z b , find the value of ax + by + cz. b−c c−a − 7. Find the length of the three sides of a right triangle, if the length of the perimeter is 24 cm and the area is 24 cm2.