Victoria School AM #4 Polynomials and Partial Fractions 2024

Secondary 4 Additional Mathematics Polynomials and Partial Fractions 1 Teacher’s Name: Class Reg No. Date: 4 Chapter 4: Polynomials and Partial Fractions Reference Book: Additional Maths 360 Textbook, Marshall Cavendish 4.1 Polynomials and Identities You will learn how to, • Add, subtract, multiply and divide polynomials, • Find the unknown constants in a polynomial identity, • Divide one polynomial by another using the division algorithm. What are Polynomials? “Polynomial” is a Greek word which literally means ‘many terms’. (a) A polynomial is a sum of terms, each of the form nax , where a is a constant and the power n is a non-negative integer. Examples of polynomials include x + 1, 3x2 – 4x + 5, x3 – 1 4 x2 + 1 2 , and – x3. (b) In the term nax , a is called the coefficient of nx . (c) The degree or order of a polynomial in x is the highest power of x. Example 1 Determine whether each of the following is a polynomial. Give a brief reason for your answer. If it is a polynomial, state its degree. Polynomial Degree/ Order x + 1 1 3x2 – 4x + 5 2 x3 – 1 4 x2 + 1 2 3 – x3 3 4 0 x2 + 1 2 x 2 (x – 2)(x + 1) 2 322 ++ xx Not a polynomial as it has a term with a fractional power. (d) A polynomial is often denoted by P(x), Q(x), f(x) etc. If Q(x) = 322 ++ xx , then the value of this polynomial at x = 2 is denoted by Q(2) = 22 + 2(2) + 3 = 11. → More generally, the value of the polynomial, Q(x) at x = a, is denoted by Q(a).

Secondary 4 Additional Mathematics Polynomials and Partial Fractions 2 Example 2 Evaluate P(3) and P(−1) if P(x) = x3 – 2x + 1. ( ) ( ) ( ) ( ) ( ) ( ) 3 3 P 3 3 2 3 1 = 22 P 1 1 2 1 1 = 4 = − + − = − − − + Addition and Subtraction and Multiplication of Polynomials When two polynomials are added, subtracted or multiplied, the result is still a polynomial. Addition and subtracting polynomials can be done by combining like terms while multiplication of two polynomials can be done through either using the distributive law. Example 3 Consider the polynomials 2( ) 1P x x x= + + and ( ) 22 3 2.Q x x x= − + Find (i) ( ) ( ),Q x P x− (ii) ( ) ( )2.P x Q x+ ( ) ( ) ( ) 22 22 2 (i) 2 3 2 1 2 3 2 1 41 − = − + − + + = − + − − − = − + Q x P x x x x x x x x x xx ( ) ( ) ( ) 22 22 2 (ii) 2 1 2 2 3 2 1 4 6 4 5 5 5 + = + + + − + = + + + − + = − + P x Q x x x x x x x x x xx Example 4 By observation, find the coefficient of x3 and x2 in the expansion of (i) ( )( ) 323 – 2 5 –1 2 1 ,++x x x x (ii) ( )( ) 222 – 3 1 .+xx ( ) ( ) ( ) ( ) 3 2 (i) Coefficient of 3 1 2 2 1 Coefficient of 2 1 5 2 8 =− =− =− + = x x ( ) ( ) 3 2 (ii) Coefficient of 0 Coefficient of 2 1 3 1 1 = =− =− x x

Secondary 4 Additional Mathematics Polynomials and Partial Fractions 3 Finding Unknowns in Identities Consider the polynomials x + 3 and x2 – x. Now, x + 3 = x2 – x holds only for x = −1 and x = 3. We call x + 3 = x 2 – x an equation with solutions x = −1 and x = 3. Now, consider the polynomials x 2 – 9 and (x – 3)(x + 3). These two polynomials are identical, i.e. they are the same except that they are written differently. Since x 2 – 9 = (x – 3)(x + 3) is true for all real values of x, we call such an equation an identity. An identity in x is an equation that is true for all values of the variable x. Identities can be written using the symbol “ ≡ ”. Hence, the equation can be rewritten as x 2 – 9 ≡ (x – 3)(x + 3). Mathematical formulae are actually identities. For example, ( ) 2 22 2a b a ab b+ = + + is an identity. One non-example is 2 2 4 3.x x x+ = + For solving for unknown(s) in an identity, you can 1. Substitute suitable values of x, 2. Equate the coefficients of the corresponding terms and/or constants on both sides of the identity. Example 5 Given that cxbxaxxx +−+−+− )1()1(52 223 , find the values of a, b and c. Method 1: Substituting suitable values of x 32 3 2 2 When 1, 1 2(1) 5 4 When 0, 5 ( 1) 4 1 When 1, ( 1) 2( 1) 5 ( 1)( 1 1) ( 1)( 1 1) 4 1 2 5 4 2 4 1 1, 1, 4 = − + = = = = − +  =− =− − − − + = − − − + − − − + − − + =− + + =  = =− = xc c xb b x a a a a b c Method 2: Comparing the coefficients of like powers of x 3 2 2 32 32 2 5 ( 2 1) 2 2 ( ) By it is n c ot n o ec f es c s comparing the ef i ien ary to expan y ts o d; we can al f so do b observation − +  − + + − +  − + + − +   − + + − +  x x ax x x bx b c ax ax ax bx b c ax ax a b x b c like-terms, 1, 1, 4 = =− =a b c How do you know what values to substitute into the polynomial?

Secondary 4 Additional Mathematics Polynomials and Partial Fractions 4 Example 6 Given that ))(2)(12(64 3 baxxxcxx ++−++ , find the values of a, b and c. 3 3 0 Compare the coefficients of term and con stant term, : 4 2(1) 2 : 6 1(2) 3 When 1, 4 6 (1)(3)( 1) 13 2, 3, 13 − = = =− =− = + + = − =−  = =− =− x xa a xb b xc c a b c Division of Polynomials Recall long division for integers: We can express 123 = 5 × 24 + 3. Note that in any division, dividend= divisor × quotient + remainder A similar process works for division of polynomials. When a polynomial P(x) is divided by another polynomial D(x), we can express the division algorithm as P(x) = D(x) × Q(x) + R(x), where the quotient Q(x) and the remainder R(x) are polynomials of x. In addition, if the degree of D(x) is not zero, then the degree of the remainder R(x) is always less than the degree of the divisor D(x). You can also use a combination of substitution and comparing the coefficients to solve a question! quotient remainder divisor dividend 24 5 1 2 3 10 23 20 3

Secondary 4 Additional Mathematics Polynomials and Partial Fractions 5 Example 7 Complete the following long division. ( ) ( ) ( ) 2 32 2 2 3 32 22 21 24 3 1 2 1 2 3 2 5 3 1 ( 2)(2 1) 3 2 2 5 3 1 −+ −− − + + − − + + −− − + +  − − − + −+ + +x xx xx xx xx x x x x x x x x x x x ( ) ( ) 2 4 4 24 2 2 22 2 2 2 63 66 3 4 7 3 3 4 10 6 3 4 7 ( 1)(6 3) 1 4 0 7 4 1 6 3 − −+ − − + − − − − + ++ + + − + − + − − + x xx xx x x x x x x x x x x x x Class Practice 1 1 Use long division to find the remainder when 4 3 23 5 2 10− + +x x x is divided by (a) 2,− x (b) 2 1.−x Write the expression in the form “dividend = divisor × quotient + remainder” for each part. (a) ( ) ( ) ( ) 32 4 3 2 43 32 32 2 2 3 4 8 2 3 5 2 10 36 2 2 4 4 8 8 0 0 − − − − − + − + + −− + −− −− + + + x x x x x x x xx xx xx x xx x x x ( ) ( )( ) 4 3 2 3 2 10 8 16 26 Remainder 26 3 5 2 10 2 3 4 8 26 −− = − + + = − − − − − + x x x x x x x x (b) ( ) ( ) ( ) 2 2 4 3 2 42 32 3 2 2 3 5 5 1 3 5 2 10 3 3 5 5 5 5 5 5 10 5 5 0 −+ − − + + −− −+ − − + −+ −− + xx x x x x xx xx xx xx x x ( )( ) 4 3 2 2 2 5 15 Remainder 15 5 3 5 2 10 1 3 5 5 15 5 −+ =− − + + = − − + + − x x x x x x x x x

Secondary 4 Additional Mathematics Polynomials and Partial Fractions 6 2 The function P is defined by 32( ) 3 5 4 9P x x x x= − + − and the function Q is defined by ( ) 432 4 6 5.Q x x x x= + − − (a) Find the remainders when ()Px and ( )Qx are each divided by ( )2.x− (b) Deduce the remainder when ( ) ( )P x Q x+ and 2 ( ) ( )P x Q x− are each divide