More about Polynomials 10 More about Polynomials Case Study 10.1 Polynomials and Their Operations 10.2 Division of Polynomials 10.3 Remainder Theorem 10.4 Factor Theorem and Its Applications 10.5 The G.C.D. and the L.C.M. of Polynomials 10.6 Algebraic Fractions Chapter Summary

Case Study If the volume of each pillar is 22 500p cm2, how can we find the radius of the spherical part? We need to use a polynomial to represent the volume of the pillar. By solving the equation obtained, we can find the radius. Since we know the total volume of the pillar, we can then set up an equation containing r, where r is the radius of the sphere. However, equations involving volumes in terms of radii are usually of degree 3, which are not easy to solve. We are going to learn a method to solve such kind of equations.

10.1 Polynomials and Their Operations A. Polynomials in One Variable In junior forms, we discovered that an algebraic expression, like x3  2x2  3x  5, is an example of polynomials in one variable. The degree of the polynomial is the same as the degree of the highest degree term. We usually write the polynomials in descending order of the degrees of terms. The general form of a polynomial in one variable of degree n is as follows: where (i) n is a non-negative integer, (ii) the coefficients an, an – 1, ... , a0 are real numbers and an  0, (iii) a0 is called the constant term of the polynomial.

10.1 Polynomials and Their Operations B. Addition, Subtraction and Multiplication of Polynomials Let us revise the fundamental operations of polynomials including addition, subtraction and multiplication in this section.

Example 10.1T 10.1 Polynomials and Their Operations Solution: B. Addition, Subtraction and Multiplication of Polynomials Example 10.1T Add 7x3  11x2  6x  1 and 5x4  4x3  4x2  12. Solution: Group the like terms in the same column and leave a space for the missing term. 7x3  11x2  6x  1 ) 5x4  4x3  4x2  12 5x4  3x3  7x2  6x  13 Alternative Solution:

Example 10.2T 10.1 Polynomials and Their Operations Solution: B. Addition, Subtraction and Multiplication of Polynomials Example 10.2T Subtract 3x4  10x3  14x  2 from x4  6x2  15x  8. Subtract A from B means B – A. Solution: x4  6x2  15x  8 ) 3x4  10x3  14x  2 2x4  10x3  6x2  x  6 Alternative Solution:

Example 10.3T 10.1 Polynomials and Their Operations Solution: B. Addition, Subtraction and Multiplication of Polynomials Example 10.3T Multiply x4  7x3  4 by 7  2x. Solution: x4  7x3  4 )  2x  7 2x5  14x4  8x )  7x4  49x3  28 2x5  7x4  49x3  8x  28 Alternative Solution:

10.2 Division of Polynomials A. Division of Polynomials by Monomials In junior forms, we learnt that when a polynomial is divided by a monomial, the result can be found by cancelling out the common factor. For example: The result 2x – 3 is called the quotient of the division.

10.2 Division of Polynomials B. Long Division of Polynomials In some cases, the method of cancelling out the common factor cannot be used in calculating the division of polynomials. For example, when 12x2  17x  8 is divided by 4x  7, the above method will not work. In this case, we have to use the method of long division. Let us first consider the division of numbers. Quotient Divisor Dividend Remainder Note that the remainder is always less than the divisor.

10.2 Division of Polynomials B. Long Division of Polynomials For the division of polynomials, we can perform the long division in a similar way. For example: To find the quotient and the remainder when 12x2  17x  8 is divided by 4x  7: The first term 3x in the quotient is obtained by dividing 12x2 (the leading term of the dividend) by 4x (in the divisor)  1 3x The second term 1 in the quotient is obtained by dividing 4x (the next leading term after the first subtraction) by 4x (in the divisor) 4x  7 12x2  17x  8 12x2  21x 12x2  21x is obtained by multiplying 4x  7 by 3x. 4x  8 17x  (21x) 4x  7  1 We can stop here because the degree of the constant 1 is less than that of the divisor. Hence we obtain the quotient 3x  1 and the remainder 1.

Example 10.4T 10.2 Division of Polynomials Solution: B. Long Division of Polynomials Example 10.4T Find the quotient and the remainder when 8x3  6x  9 is divided by 4x – 2. Solution: The missing term, 0x2, should be added into the dividend to avoid making mistakes in the calculation. 2x2  x  1 4x  2 8x3  0x2  6x  9 8x3  4x2 4x2  6x  9 0x2  (4x2) 4x2  2x  4x  9  4x  2  11

Example 10.5T 10.2 Division of Polynomials Solution: B. Long Division of Polynomials Example 10.5T Find the quotient and the remainder when 1  x  3x2 divides 4  22x2  12x3. A divides B means B  A. Solution:  4x  6 3x2  x  1 12x3  22x2  0x  4 12x3  4x2  4x  18x2  4x  4  18x2  6x  6 10x  10 Remainder of degree 1 Divisor of degree 2

Dividend  Divisor  Quotient  Remainder 10.2 Division of Polynomials C. Division Algorithm of Polynomials In the previous section, we discussed the division of two integers (987 ¸ 13). dividend divisor quotient remainder Actually, we have similar relationship in the division of polynomials. The dividend can always be expressed as the sum of the product of its divisor and quotient and its remainder: Dividend  Divisor  Quotient  Remainder The above relationship is called the division algorithm of polynomials.

Example 10.6T 10.2 Division of Polynomials Solution: C. Division Algorithm of Polynomials Example 10.6T If 4x3  8x2  7x  2 is divided by a polynomial, the quotient and the remainder are 2x2  7x  14 and –44 respectively. Find the polynomial. Solution: By long division, 2x  3 2x2  7x  14 4x3  8x2  7x  42 4x3  14x2  28x 6x2  21x  42 6x2  21x  42  The required polynomial is 2x  3.

10.3 Remainder Theorem The notations of function we learnt in Book 4, Chapter 3 like f(x), g(x), P(x) and Q(x) can also be used to denote a polynomial. For example, P(x)  x3  2x2  3x  1. When x  a, the value of the polynomial P(x) is denoted by P(a). For example, P(1)  13  2(1)2  3(1)  1  1  2  3  1  3 where 3 is the value of the polynomial when x  1.

10.3 Remainder Theorem Consider a polynomial P(x). When P(x) is divided by x  a, we have P(x)  (x  a) · Q(x)  R ... (*) where Q(x) is the quotient and R is the remainder. Since the divisor (x – a) is a linear polynomial of degree 1, the degree of the remainder must be zero, i.e., R must be a constant. In fact, (*) is an identity which means that the equality holds for all values of x. Substituting x  a into (*), we have P(a)  (a  a) · Q(a)  R  0 · Q(a)  R By using the remainder theorem, we are able to find a remainder without actually carrying out the long division of polynomials. However, we cannot find the quotient throughout the process.  R As a result, we have the remainder theorem: Remainder Theorem When a polynomial P(x) is divided by x – a, the remainder R is equal to P(a).

10.3 Remainder Theorem Similarly, when P(x) is divided by mx  n, we have where Q¢(x) is the quotient and R¢ is the remainder. So, by substituting , we have Therefore, the remainder theorem can be generalized as follows: Remainder Theorem When a polynomial P(x) is divided by mx – n, the remainder R is equal to .

Example 10.7T 10.3 Remainder Theorem Solution: Find the remainder when (x  2)(2x  1)(x  1)  3 is divided by (a) x  2, (b) 2x  1. Solution: Let P(x)  (x  2)(2x  1)(x  1)  3. It is not necessary to expand and simplify the polynomial when applying the remainder theorem. By the remainder theorem, we have (a) remainder  P(2)  (2  2)(2  2  1)(2  1)  3  0  3  3 (b) remainder  0  3  3

Example 10.8T 10.3 Remainder Theorem Solution: When the polynomial x3  kx2  x is divided by x  1, the remainder is 3. Find the value of k. Solution: Let P(x)  x3  kx2  x. By the remainder theorem, we have

Example 10.9T 10.3 Remainder Theorem Solution: When the polynomial 3x2  2x  5 is divided by x  a, the remainder is 3. Find the value(s) of a. Solution: Let P(x)  3x2  2x  5. By the remainder theorem, we have

10.4 Factor Theorem and Its Applications A. Factor Theorem According to the remainder theorem, when a polynomial P(x) is divided by x  a, the remainder is P(a). Thus, if P(a)  0, that is, the remainder is 0, it means that P(x) is divisible by x  a, that is, x  a is a factor of P(x). Actually, the above result is called the factor theorem of polynomials. Factor Theorem If P(x) is a polynomial and P(a)  0, then x  a is a factor of P(x). Conversely, if x  a is a factor of a polynomial P(x), then P(a)  0. The factor theorem can also be extended as follows. Factor Theorem If P(x) is a polynomial and  0, then mx  n is a factor of P(x). Conversely, if mx  n is a factor of a polynomial P(x), then  0.

Example 10.10T 10.4 Factor Theorem and Its Applications Solution: A. Factor Theorem Example 10.10T Let P(x)  x3  x2  8x  12. (a) Using the factor theorem, determine whether each of the following is a factor of P(x). (i) x  1 (ii) x  1 (iii) x  2 (b) Hence factorize P(x) completely. Solution: (a) (i) P(1) (iii) P(2)  x  1 is not a factor of P(x).  x  2 is a factor of P(x). (ii) P(1)  x  1 is not a factor of P(x).

Example 10.10T 10.4 Factor Theorem and Its Applications Solution: A. Factor Theorem Example 10.10T Let P(x)  x3  x2  8x  12. (a) Using the factor theorem, determine whether each of the following is a factor of P(x). (i) x  1 (ii) x  1 (iii) x  2 (b) Hence factorize P(x) completely. Solution: (b) From (a), x  2 is a factor of P(x). By the method of long division,

Example 10.11T 10.4 Factor Theorem and Its Applications Solution: A. Factor Theorem Example 10.11T If 3x  1 is a factor of P(x)  3x3  kx2  5x  2, (a) find the value of k. (b) Hence factorize P(x) completely. Solution: (a) Since 3x  1 is a factor of P(x), . (b) By long division, we have P(x)

10.4 Factor Theorem and Its Applications B. Applications of Factor Theorem Consider a cubic polynomial P(x)  2x3  3x2  8x  3. Let ax  b be a factor of P(x). We have P(x)  2x3  3x2  8x  3  (ax  b)(px2  qx  r), where a, b, p, q and r are integers and a  0. The possible values of a are 1 and 2. The possible values of b are –1, –3, 1 and 3.  All possible linear factors ax + b of P(x): x + 1 2x + 1 x – 1 2x – 1 x + 3 2x + 3 x – 3 2x – 3 Actually, we can make use of the factor theorem to verify which of the above are the factors of the polynomial P(x)  2x3  3x2  8x  3.

Example 10.12T 10.4 Factor Theorem and Its Applications Solution: B. Applications of Factor Theorem Example 10.12T Let P(x)  x3  10x2  31x  30. (a) Factorize P(x). (b) Solve the equation P(x)  0. Since the leading coefficient of P(x) is 1 and the constant term is 30, the possible linear factors of P(x) are x  1, x  2, x  3, x  5, x  6, etc. Do not show the unsuccessful trial in your calculation. Solution: (a) P(2)  By the factor theorem, x  2 is a factor of P(x). By long division, we have (b) P(x)  0 (x  2)(x  3)(x  5)  0 x  2, 3 or 5

Example 10.13T 10.4 Factor Theorem and Its Applications Solution: B. Applications of Factor Theorem Example 10.13T Let P(x)  3x3  5x2  8x  2. (a) Factorize P(x) into linear or quadratic factors with integral coefficients. (b) Solve the equation P(x)  0. Solution: (a)  By the factor theorem, 3x  1 is a factor of P(x). By long division, we have

Example 10.13T 10.4 Factor Theorem and Its Applications Solution: B. Applications of Factor Theorem Example 10.13T Let P(x)  3x3  5x2  8x  2. (a) Factorize P(x) into linear or quadratic factors with integral coefficients. (b) Solve the equation P(x)  0. Solution: (b) From (a), P(x)  (3x  1)(x2  2x  2). P(x)  0 (3x  1)(x2  2x  2)  0 3x  1  0 or x2  2x  2  0 x  or

10.4 Factor Theorem and Its Applications C. Limitation of the Use of Factor Theorem From the above example, we might infer that when a polynomial does not have any linear factors, or the coefficient of that factor is not rational, we cannot apply the factor theorem. Consider a polynomial P(x)  x4  3x2  2. All possible linear factors of P(x) are x  1, x  1, x  2 and x  2. By using the factor theorem, we can check that none of the above linear factors is a factor of P(x). This implies there is no linear factor in P(x). But this does NOT imply P(x) cannot be factorized. In fact, P(x) be factorized as P(x)  (x2)2  3(x2)  2  (x2  1)(x2  2) Therefore, if there is no linear factor in a polynomial, we cannot apply the factor theorem to factorize the polynomial.

10.5 The G.C.D. and the L.C.M. of Polynomials In junior forms, we learnt the G.C.D. and L.C.M. of numbers. G.C.D. is also known as H.C.F. (highest common factor). For example, consider the two numbers 30 and 42. ∵ 30  2  3  5 and 42  2  3  7 ∴ G.C.D. of 30 and 42  2  3  6 L.C.M. of 30 and 42  2  3  5  7  210 For two or more polynomials, we can also find their G.C.D. and L.C.M. Greatest Common Divisor (G.C.D.) For any two or more polynomials, the common factor of them with the highest degree is called the greatest common divisor (G.C.D.). Least Common Multiple (L.C.M.) For any two or more polynomials, the L.C.M. of them is the polynomial with the lowest degree which is divisible by each of the given polynomials.

10.5 The G.C.D. and the L.C.M. of Polynomials Consider P(x)  (x  2)(x  4)3 and Q(x)  (x  1)(x  2)2(x  4). G.C.D.  (x  2)(x  4) L.C.M.  (x  1)(x  2)2(x  4)3 Sometimes, we have to factorize the given polynomials before finding their G.C.D. or L.C.M. and factor theorem may be used if necessary.

Example 10.14T 10.5 The G.C.D. and the L.C.M. of Polynomials Solution: Given two polynomials, P(x)  x2  x  12 and Q(x)  x3  27. (a) Factorize P(x) and Q(x). (b) Find the G.C.D. and the L.C.M. of P(x) and Q(x). Solution: (a) P(x)  x2  x  12  (x  4)(x  3) Q(x)  x3  27  (x  3)(x2  3x  9) a3  b3  (a  b)(a2  ab  b2) (b) P(x)  (x  4)(x  3) Q(x)  (x  3)(x2  3x  9)  G.C.D.  (x  3) L.C.M.  (x  3)(x  4)(x2  3x  9)

Example 10.15T 10.5 The G.C.D. and the L.C.M. of Polynomials Solution: (a) Factorize P(x)  2x3  x2  13x  6 and Q(x)  x3  2x2  9x  18. (b) Hence find the G.C.D. and the L.C.M. of P(x) and Q(x). Solution: (a) P(3)  2(3)3  32  13(3)  6  54  9  39  6  0  By the factor theorem, x  3 is a factor of P(x). By long division, we have Q(x)  x3  2x2  9x  18 (b) G.C.D.  (x  2)(x  3)  x2(x  2)  9(x  2)  (x  2)(x2  9) L.C.M.  (x  2)(x  3)(2x  1)(x  3)  (x  2)(x  3)(x  3)

Example 10.16T 10.5 The G.C.D. and the L.C.M. of Polynomials Solution: Find the G.C.D. and the L.C.M. of a2  ab  2b2 and a2  4b2. Solution: ∵ a2  ab  2b2  (a  b)(a  2b) a2  4b2  (a  2b)(a  2b) ∴ G.C.D.  (a  2b) L.C.M.  (a  b)(a  2b)(a  2b)

10.6 Algebraic Fractions An algebraic fraction is a quotient of two polynomials where the denominator is not equal to zero, such as and . Remark: If a function R(x) is defined as the quotient of two non-zero polynomials P(x) and Q(x), i.e., , we call the function a rational function.

10.6 Algebraic Fractions A. Multiplication and Division of Algebraic The multiplication and division of algebraic fractions are just like the multiplication and division of fractions that we can just cancel the common factors. For example: (x  2) and (x  3) are the common factors of the numerator and denominator.

Example 10.17T 10.6 Algebraic Fractions Solution: A. Multiplication and Division of Algebraic Fractions Example 10.17T Simplify . Solution: First factorize the expressions. Then cancel the common factors.

Example 10.18T 10.6 Algebraic Fractions Solution: A. Multiplication and Division of Algebraic Fractions Example 10.18T Simplify . Solution:

10.6 Algebraic Fractions B. Addition and Subtraction of Algebraic Algebraic fractions can be added or subtracted. When the denominators of two algebraic fractions are equal, we can add or subtract them. However, when the denominators are not the same, we have to make the denominators equal which can be done by finding the L.C.M. of the denominators.

Example 10.19T 10.6 Algebraic Fractions Solution: B. Addition and Subtraction of Algebraic Fractions Example 10.19T Simplify . Solution:

Example 10.20T 10.6 Algebraic Fractions Solution: B. Addition and Subtraction of Algebraic Fractions Example 10.20T Simplify . Solution: The L.C.M. of (x  3)(x  3) and (x  3)(x  1) is (x  1)(x  3)(x  3). Expand and simplify. Factorize the numerator.

Chapter Summary 10.1 Polynomials and Their Operations A polynomial in one variable of degree n can be expressed as where an, an – 1, ... , a0 are real numbers and an  0.

Chapter Summary 10.2 Division of Polynomials 1. If a polynomial P(x) is divided by a polynomial Q(x), we can use the method of long division to find the quotient and the remainder. 2. When a polynomial P(x) is divided by a divisor, we get a quotient and a remainder. The division algorithm of polynomials is written as dividend  divisor  quotient  remainder.

Chapter Summary 10.3 Remainder Theorem 1. When a polynomial P(x) is divided by x  a, the remainder R is equal to P(a). 2. When a polynomial P(x) is divided by mx  n, the remainder R is equal to .

Chapter Summary 10.4 Factor Theorem and Its Applications 1. If P(x) is a polynomial and P(a)  0, then x  a is a factor of P(x). Conversely, if x – a is a factor of P(x), then P(a)  0. 2. If P(x) is a polynomial and , then mx  n is a factor of P(x). Conversely, if mx  n is a factor of P(x), then .

Chapter Summary 10.5 The G.C.D. and the L.C.M. of Polynomials 1. For any two or more polynomials, the common factor of them with the highest degree is called the greatest common divisor (G.C.D.). 2. For any two or more polynomials, the common multiple of them with the least degree is called the least common multiple (L.C.M.).

Chapter Summary 10.6 Algebraic Fractions An algebraic fraction is a quotient of two polynomials where the denominator is not equal to zero. We can perform addition, subtraction, multiplication and division for algebraic fractions.

Follow-up 10.1 10.1 Polynomials and Their Operations Solution: B. Addition, Subtraction and Multiplication of Polynomials Follow-up 10.1 Add 4x5  3x3  11x  6 and x5  8x3  9x2. Solution: 4x5  3x3  11x  6 ) x5  8x3  9x2 5x5  5x3  9x2  11x  6 Alternative Solution:

Follow-up 10.2 10.1 Polynomials and Their Operations Solution: B. Addition, Subtraction and Multiplication of Polynomials Follow-up 10.2 Subtract 3x4  x3  4x2  15 from 2x4  5x2  13x  3. Solution: 2x4  5x2  13x  3 ) 3x4  x3  4x2  15 x4  x3  9x2  13x  18 Alternative Solution:

Follow-up 10.3 10.1 Polynomials and Their Operations Solution: B. Addition, Subtraction and Multiplication of Polynomials Follow-up 10.3 Multiply 2x3  x2  7x  1 by 2x  3. Solution: 2x3  x2  7x  1 ) 2x  3 4x4  2x3  14x2  2x )  6x3  3x2  21x  3 4x4  4x3  17x2  23x  3 Alternative Solution:

Follow-up 10.4 10.2 Division of Polynomials Solution: B. Long Division of Polynomials Follow-up 10.4 Find the quotient and the remainder when 18x3  58x2  50 is divided by 2x  6. Solution: The missing term, 0x, should be added into the dividend to avoid making mistakes in the calculation. 9x2  6  2x 2x  6 18x3  58x2  0x  50 18x3  54x2 4x2  0x  50 4x2  12x  12x  50 0x  (12x)  12x  36  14

Follow-up 10.5 10.2 Division of Polynomials Solution: B. Long Division of Polynomials Follow-up 10.5 Find the quotient and the remainder when 2x2  x  7 divides 10x3  21  3x2. Solution:  4 5x 2x2  x  7 10x3  3x2  0x  21 10x3  5x2  35x  8x2  35x  21  8x2  4x  28  31x  7 Remainder of degree 1 Divisor of degree 2

Follow-up 10.6 10.2 Division of Polynomials Solution: C. Division Algorithm of Polynomials Follow-up 10.6 If x2  7x  5 is divided by a polynomial, the quotient and the remainder are x  5 and –5 respectively. Find the polynomial. Solution: By long division, x  2 x  5 x2  7x  10 x2  5x 2x  10 2x  10  The required polynomial is x  2.

Follow-up 10.7 10.3 Remainder Theorem Solution: Find the remainder when (x  2)(2x  1)  3 is divided by (a) x  2, (b) 2x  1. Solution: Let P(x)  (x  2)(2x  1)  3. By the remainder theorem, we have (a) remainder  P(2)  (2  2)(2  2  1)  3  0  3  3 (b) remainder  0  3  3

Follow-up 10.8 10.3 Remainder Theorem Solution: When the polynomial 2x2  kx  4 is divided by x  3, the remainder is 5. Find the value of k. Solution: Let P(x)  2x2  kx  4. By the remainder theorem, we have

Follow-up 10.9 10.3 Remainder Theorem Solution: When the polynomial 2x2  ax  4 is divided by x  a, the remainder is 5. Find the value(s) of a. Solution: Let P(x)  2x2  ax  4. By the remainder theorem, we have

Follow-up 10.10 10.4 Factor Theorem and Its Applications Solution: A. Factor Theorem Follow-up 10.10 Let P(x)  2x3  11x2  12x  36. (a) Using the factor theorem, determine whether each of the following is a factor of P(x). (i) x  2 (ii) x  2 (iii) x  4 (b) Hence factorize P(x) completely. Solution: (a) (i) P(2)  x  2 is not a factor of P(x). (ii) P(2)  x  2 is a factor of P(x). (iii)P(4)  x  4 is not a factor of P(x).

Follow-up 10.10 10.4 Factor Theorem and Its Applications Solution: A. Factor Theorem Follow-up 10.10 Let P(x)  2x3  11x2  12x  36. (a) Using the factor theorem, determine whether each of the following is a factor of P(x). (i) x  2 (ii) x  2 (iii) x  4 (b) Hence factorize P(x) completely. Solution: (b) From (a), x  2 is a factor of P(x). By the method of long division,

Follow-up 10.11 10.4 Factor Theorem and Its Applications Solution: A. Factor Theorem Follow-up 10.11 It is given that 4x  1 is a factor of P(x)  4x3  kx2  14x  3. (a) Find the value of k. (b) Hence factorize P(x) completely. Solution: (a) Since 4x  1 is a factor of P(x), . (b) By long division, we have P(x)

Follow-up 10.12 10.4 Factor Theorem and Its Applications Solution: B. Applications of Factor Theorem Follow-up 10.12 Let P(x)  x3  2x2  5x  6. (a) Factorize P(x). (b) Solve the equation P(x)  0. Since the leading coefficient of P(x) is 1 and the constant term is 6, the possible linear factors of P(x) are x  1, x  2, x  3 and x  6. Do not show the unsuccessful trial in your calculation. Solution: (a) P(1)  By the factor theorem, x  1 is a factor of P(x). By long division, we have (b) P(x)  0 (x  1)(x  2)(x  3)  0 x  2, 1 or 3

Follow-up 10.13 10.4 Factor Theorem and Its Applications Solution: B. Applications of Factor Theorem Follow-up 10.13 Let P(x)  4x3  5x  2. (a) Factorize P(x) into linear or quadratic factors with integral coefficients. (b) Solve the equation P(x)  0. Solution: (a)  By the factor theorem, 2x  1 is a factor of P(x). By long division, we have (b) P(x)  0, (2x  1)(2x2  x  2)  0 2x  1  0 or 2x2  x  2  0 x  or

Follow-up 10.14 10.5 The G.C.D. and the L.C.M. of Polynomials Given two polynomials, P(x)  x2  2x  8 and Q(x)  x2  8x  16. (a) Factorize P(x) and Q(x). (b) Find the G.C.D. and the L.C.M. of P(x) and Q(x). Solution: (a) P(x)  x2  2x  8  (x  4)(x  2) Q(x)  x2  8x  16  (x  4)2 (b) P(x)  (x  4)(x  2) Q(x)  (x  4)2  G.C.D.  (x  4) L.C.M.  (x  4)2(x  2)

Follow-up 10.15 10.5 The G.C.D. and the L.C.M. of Polynomials (a) Factorize P(x)  x3  5x2  7x  3 and Q(x)  2x3  9x2  10x  3. (b) Hence find the G.C.D. and the L.C.M. of P(x) and Q(x). Solution: (a) P(1)  13  5(1)2  7(1)  3  1  5  7  3  0  By the factor theorem, x  1 is a factor of P(x). By long division, we have Q(1)  2(1)3  9(1)2  10(1)  3  0  By the factor theorem, x  1 is a factor of Q(x).  Q(x) By long division. (b) G.C.D.  (x  1)(x  3) L.C.M.  (x  1)2(x  3)(2x  1)

Follow-up 10.16 10.5 The G.C.D. and the L.C.M. of Polynomials Find the G.C.D. and the L.C.M. of a2  ab  3a  3b and a2  2ab  b2. Solution: ∵ a2  ab  3a  3b  a(a  b)  3(a  b)  (a  b)(a  3) a2  2ab  b2  (a  b)2 ∴ G.C.D.  (a  b) L.C.M.  (a  b)2(a  3)

Follow-up 10.17 10.6 Algebraic Fractions Solution: A. Multiplication and Division of Algebraic Fractions Follow-up 10.17 Simplify . Solution:

Follow-up 10.18 10.6 Algebraic Fractions Solution: A. Multiplication and Division of Algebraic Fractions Follow-up 10.18 Simplify . Solution:

Follow-up 10.19 10.6 Algebraic Fractions Solution: B. Addition and Subtraction of Algebraic Fractions Follow-up 10.19 Simplify . Solution:

Follow-up 10.20 10.6 Algebraic Fractions Solution: B. Addition and Subtraction of Algebraic Fractions Follow-up 10.20 Simplify . Solution: Expand and simplify. Factorize the numerator.