Revision on Polynomials

Do you remember what a monomial is? A monomial is an algebraic expression which can be one of the following:

Do you remember what a monomial is? (a) Number: 3, –6 (b) Variable: x, y

Do you remember what a monomial is? (c) Product of a number and variable(s) : 3x, y, –4xy2

Good. Note that expressions like and are not monomials. 1 x 4 w 2 x 4 w (c) Product of a number and variable(s) : 3x, y, –4xy2

Do you remember what a polynomial is? A polynomial can be a monomial or the sum of two or more monomials.

Do you remember what a polynomial is? For example, 3bc, 4 + 2a , 5 + a + 2c are polynomials.

The following are some key terminologies of polynomials: Degree of the polynomial : The highest degree of all its term(s). Constant term: The term that does not contain a variable. Coefficient of a term: The numerical part of the term.

Degree of the polynomial = 3 For example, Degree of the polynomial = 3 Constant term = –7 4x2 – 6x3 – 7 Coefficient of x3 = –6 Coefficient of x2 = 4 What is the coefficient of x ?

Degree of the polynomial = 3 For example, Degree of the polynomial = 3 Constant term = –7 4x2 – 6x3 – 7 Coefficient of x3 = –6 Coefficient of x2 = 4 The coefficient of x is 0, i.e. 4x2 – 6x3 + 0x – 7.

Degree of the polynomial = 3 For example, Degree of the polynomial = 3 Constant term = –7 4x2 – 6x3 – 7 Coefficient of x3 = –6 Coefficient of x2 = 4 Coefficient of x = 0

Degree of the polynomial = For example, Degree of the polynomial = 3 Constant term = 8 5 x + 8 – 2 x 3 Coefficient of x3 = –2 Coefficient of x2 = Coefficient of x = 5

We usually arrange the terms of a polynomial in descending powers or ascending powers of a variable. e.g. For the polynomial 2x + 5x3 – 4 + 3x2 , we have: (i) Arrange in descending powers of x: 5x3 + 3x2 + 2x – 4 (ii) Arrange in ascending powers of x: –4 + 2x + 3x2 + 5x3

Grouping the like terms. Addition and Subtraction of Polynomials Addition and subtraction of polynomials can be performed by combing like terms after removing the brackets. Addition of polynomials: (4x + 3) + (x – 6) Remove the brackets, .i.e. +(x – 6) = + x – 6. = 4x + 3 + x – 6 = 4x + x + 3 – 6 Grouping the like terms. = 5x – 3

Same result! Subtraction of polynomials: (6x2 + 2x) – (4x2 + x) We can also simplify the above expression as follows: 6x2 + 2x 6x2 – 4x2 2x – x –) 4x2 + x Same result! 2x2 + x

Follow-up question (a) Simplify (x2 – 5x + 1) – (3x2 – 6x + 4). (a) Alternative Solution –) 3x2 – 6x + 4 x2 – 5x + 1 –2x2 + x – 3 Line up the like terms in the same column.

Follow-up question (cont’d) (b) Simplify (1 + x2 – 2x) + (2 – 3x2 + 4x) and arrange the terms in descending powers of x. (b) (1 + x2 – 2x) + (2 – 3x2 + 4x) = 1 + x2 – 2x + 2 – 3x2 + 4x = x2 – 3x2 – 2x + 4x + 1 + 2 = –2x2 + 2x + 3 Alternative Solution +) –3x2 + 4x + 2 x2 – 2x + 1 –2x2 + 2x + 3 Arrange the terms in descending powers of x.

Multiplication of Polynomials We can perform multiplication by applying the distribution law of multiplication: a(x + y) = ax + ay or (x + y)a = xa + ya e.g. (m + 1)(m + 2) = (m + 1)m + (m + 1)2 = m2 + m + 2m + 2 (x + y) a = xa + ya = m2 + 3m + 2

We can also do the expansion by using the method of long multiplication. Line up the like terms in the same column. m + 1 ×) m + 2 m2 + m +) 2m + 2 m2 + 3m + 2

Follow-up question Expand (2 – a2 + 3a)(–2a + 4), and arrange the terms in descending powers of a. (2 – a2 + 3a)(–2a + 4) =(2 – a2 + 3a)(–2a) + (2 – a2 + 3a)(4) = –4a + 2a3 – 6a2 + 8 – 4a2 + 12a –a2 + 3a + 2 ×) –2a + 4 2a3 – 10a2 + 8a + 8 – 4a 2a3 Alternative Solution – 6a2 + 8 +) – 4a2 + 12a = 2a3 – 6a2 – 4a2 – 4a + 12a + 8 = 2a3 – 10a2 + 8a + 8

Division of Polynomials

Divide a polynomial by a monomial We can carry out the division term by term. dividend 2 ) 8 ( ¸ + x divisor 2 8 + = x Rough Work x 2 8 + 2 8 + = x x 1 4  Divide the polynomial term by term. 4 + = x The result (quotient) is x + 4 and the remainder is equal to 0. We say that 2x2 + 8x is divisible by 2x.

You may also use the method of long division. Quotient x 2 x 2 8 x 4 + Divisor 2 x Dividend 2 x + 8 x 2 x 8 x 8 x Remainder = 0

Long division (divisor is not a monomial) Consider the long division of ). 1 2 ( ) 3 6 8 + ¸ x Step 1 3 6 8 1 2 x + 4 x 2 8 Step 2 2 8 x + 1 3 6 4 8 2 x (+ 1)(+ 4x) Step 3 4 8 3 6 1 2 x + 4 x + 3 2 + x  Subtract 8x2 + 4x from 8x2 + 6x + 3.

Step 4 4 8 3 6 1 2 x + + 1 x 2 Step 5 2 x 4 8 3 6 + 1 2 x (+ 1)(+ 1) Step 6 4 8 3 6 1 2 x + 1 + Subtract 2x + 1 from 2x + 3. 2

4 8 3 6 1 2 x + divisor ∵ Degree of 2 = 0, degree of 2x + 1 = 1 ∴ Degree of 2 < degree of 2x + 1 ∴ Stop the division process. remainder ∴ Quotient = 4x + 1, remainder = 2 Since the remainder is not equal to 0, we say that 8x2 + 6x + 3 is not divisible by 2x + 1.

Follow-up question Find the quotient and the remainder of (3x2 + 2x3 – 9)  (2x – 1). 1. Rearrange the terms in descending powers of x. 2. Insert the missing term ‘0x’. 2 x 2 + x 1 + 9 3 2 1 - + x 2 3 - x  x2(2x – 1) = 2x3 – x2 4 2 + x - 9 2 4 - x  2x(2x – 1) = 4x2 – 2x 9 2 - x ∴ Quotient = x2 + 2x + 1 Remainder = -8 1 2 - x  1(2x – 1) = 2x – 1 8 -

What is the relation among the dividend, divisor, quotient and remainder? 4 divisor 29 6 dividend 24 29 = 6 4 + 5 remainder 5 In arithmetic, dividend = divisor quotient + remainder

This is known as the division algorithm. In fact, the mentioned relation is also true for polynomials. Consider f(x)  p(x). Let Q(x) and R(x) be the quotient and the remainder respectively. ) ( x Q R f p M dividend = divisor quotient + remainder f(x) = p(x)  Q(x) + R(x) degree of R(x) < degree of p(x) This is known as the division algorithm.

From the above result, we have 8x2 + 6x + 3  (2x + 1)(4x + 1) + 2 e.g. (8x2 + 6x + 3)  (2x + 1) 4 8 3 6 1 2 x + From the above result, we have 8x2 + 6x + 3  (2x + 1)(4x + 1) + 2 dividend divisor quotient remainder

= (x + 3)(2x – 1) + 5 = (x + 3)(2x) – (x + 3)(1) + 5 When a polynomial is divided by x + 3 , the quotient and the remainder are 2x – 1 and 5 respectively. Find the polynomial. By division algorithm, we have the required polynomial = (x + 3)(2x – 1) + 5 = (x + 3)(2x) – (x + 3)(1) + 5 = 2x2 + 6x – x – 3 + 5 = 2x2 + 5x + 2

Follow-up question When 3x2 + 10x – 5 is divided by a polynomial, the quotient and the remainder are x + 4 and 3. Find the polynomial. Let p(x) be the required polynomial. By division algorithm, we have dividend divisor quotient remainder ∴ The required polynomial is 3x – 2.

Remainder Theorem

When a polynomial f(x) is divided by x – a, we can find the remainder as follows: Let Q(x) and R be the quotient and the remainder respectively. By division algorithm, we have Since the degree of R is less than that of x – a, the degree of R is 0. So, R is a constant.  f(x) = (x – a)  Q(x) + R When x = a, the value of the polynomial is f(a) = (a – a)  Q(a) + R = 0  Q(a) + R = R ∴ Remainder = f(a)

In conclusion, we have the following theorem: Theorem 5.1 Remainder theorem When a polynomial f(x) is divided by x – a, the remainder is equal to f(a). When f(x) = x2 + 2x – 1 is divided by x – 1, remainder = f(1) = (1)2 + 2(1) – 1 = 2

In conclusion, we have the following theorem: Theorem 5.1 Remainder theorem When a polynomial f(x) is divided by x – a, the remainder is equal to f(a). When f(x) = x2 + 2x – 1 is divided by x + 2, x + 2 = x – (–2) remainder = f(–2) = (–2)2 + 2(–2) – 1 = –1

Follow-up question Find the remainder when the polynomial x3 – 2x2 – 3x + 5 is divided by x + 1. Let f(x) = x3 – 2x2 – 3x + 5. By the remainder theorem, x + 1 = x – (–1) remainder = f(–1) = (–1)3 – 2(–1)2 – 3(–1) + 5 = –1 – 2 + 3 + 5 = 5

What is the remainder when a polynomial f(x) is divided by mx – n? The remainder theorem can be extended as follows: Theorem 5.2 When a polynomial f(x) is divided by mx – n, the remainder is equal to . | ø ö ç è æ m n f

Find the remainder when 9x2 + 6x – 7 is divided by 3x + 1. Let f(x) = 9x2 + 6x – 7. By the remainder theorem, remainder = | ø ö ç è æ – 3 1 f 3x + 1 = 3x – (–1) 7 3 1 6 9 2 – | ø ö ç è æ + = = 1 – 2 – 7 = –8

Follow-up question When 4x3 + kx2 – x + 2 is divided by 4x – 1, the remainder is 2. Find the value of k. Let f(x) = 4x3 + kx2 – x + 2. By the remainder theorem, 3 k = 16

Factor Theorem

Consider a polynomial f(x) divided by x – a. By the remainder theorem, we have remainder = f(a) If remainder = 0, then f(x) is divisible by x – a. i.e. If f(a) = 0, then x – a is a factor of f(x).

In conclusion, we have the following theorem: Theorem 5.3 Factor theorem For a polynomial f(x), if f(a) = 0, then x – a is a factor of f(x). f(a) = 0 x – a is a factor of f(x). Consider f(x) = x2 – 5x + 6. f(2) = (2)2 – 5(2) + 6 = 0 ∴ x – 2 is a factor of f(x).

Theorem 5.4 | ø ö ç è æ | ø ö ç è æ How about if a is not an integer? The factor theorem can be generalized as follows: Theorem 5.4 For a polynomial f(x), if f = 0, then mx – n is a factor of f(x). | ø ö ç è æ m n f = 0 | ø ö ç è æ m n mx – n is a factor of f(x).

Is 2x – 1 a factor of 2x2 + x – 1? Let f(x) = 2x2 + x – 1. ∴ 2x – 1 is a factor of 2x2 + x – 1.

∴ x – 2 is not a factor of f(x). Follow-up question Let f(x) = 2x2 – 5x – 3. Use factor theorem to determine whether each of the following is a factor of f(x). (a) x – 2 (b) 2x + 1 (a) f(2) = 2(2)2 – 5(2) – 3 = 8 – 10 – 3 = –5  0 ∴ x – 2 is not a factor of f(x).

Follow-up question Let f(x) = 2x2 – 5x – 3. Use factor theorem to determine whether each of the following is a factor of f(x). (a) x – 2 (b) 2x + 1 (b) ∴ 2x + 1 is a factor of f(x).

In fact, the converse of the factor theorem is also true. Consider a polynomial f(x). If x – a is a factor of f(x), then f(a) = 0. (a) f(a) = 0 x – a is a factor of f(x). (b) If mx – n is a factor of f(x), then f = 0. | ø ö ç è æ m n f = 0 | ø ö ç è æ m n mx – n is a factor of f(x).

If x + 3 is a factor of f(x) = –4x2 + px + 6, what is the value of p?  By the converse of the factor theorem –4(–3)2 + p(–3) + 6 = 0 –36 – 3p + 6 = 0 –3p = 30 p = –10

Follow-up question If 8x3 + 6x2 + qx + 3 is divisible by 4x – 3, find the value of q. Let f(x) = 8x3 + 6x2 + qx + 3. i.e. 4x – 3 is a factor of f(x). ∵ f(x) is divisible by 4x – 3, 4 3 = | ø ö ç è æ f ∴ 3 4 6 8 2 = + | ø ö ç è æ q 4 39 3 - = q 13 - = q

Factorizing Polynomials by Factor Theorem

How can we factorize a cubic polynomial? The key step is to find a linear factor of a polynomial by the factor theorem. Let’s consider f(x) = 2x3 + x2 – 4x – 3.

Find a linear factor of f(x) = 2x3 + x2 – 4x – 3. Suppose f(x) can be factorized as (mx + n)(ax2 + bx + c), i.e. 2x3 + x2 – 4x – 3  (mx + n)(ax2 + bx + c), …… (*) where m, n, a, b and c are integers with m > 0. Consider the coefficient of x3. 2x3 + x2 – 4x – 3  (mx + n)(ax2 + bx + c) 2x3 max3 By comparing the coefficient of x3 on both sides, we have 2 = ma

Find a linear factor of f(x) = 2x3 + x2 – 4x – 3. Suppose f(x) can be factorized as (mx + n)(ax2 + bx + c), i.e. 2x3 + x2 – 4x – 3  (mx + n)(ax2 + bx + c), …… (*) where m, n, a, b and c are integers with m > 0. Consider the constant term. 2x3 + x2 – 4x – 3  (mx + n)(ax2 + bx + c) –3 nc By comparing the constant term on both sides, we have –3 = nc

Find a linear factor of f(x) = 2x3 + x2 – 4x – 3. Suppose f(x) can be factorized as (mx + n)(ax2 + bx + c), i.e. 2x3 + x2 – 4x – 3  (mx + n)(ax2 + bx + c), …… (*) where m, n, a, b and c are integers with m > 0. 2 = ma Possible values of m = 1 and 2 –3 = nc Possible values of n = ± 1 and ± 3 ∴ The possible linear factors of f(x) are x ± 1, x ± 3, 2x ± 1, 2x ± 3.

Then, apply the factor theorem to the possible linear factors. ∴ x + 1 is a factor of f(x).

Consider a cubic polynomial f(x) = px3 + qx2 + rx + s, where p, q, r and s are integers. If mx + n is a linear factor of f(x), then we have: px3 + qx2 + rx + s  (mx + n) Q(x), where Q(x) is a polynomial. We can see that m is a factor of leading coefficient p, and n is a factor of the constant term s. Therefore, a linear factor mx + n can be found as follows: 1. List all the factors of leading coefficient p and constant term s. 2. Apply the factor theorem to possible linear factors until a linear factor mx + n is found.

Follow-up question Consider f(x) = 3x3 – 11x2 – 19x – 5. (a) By considering the coefficient of x3 and the constant term, which of the following is/are possible linear factor(s) of f(x)? x + 1, 2x – 1, 3x + 2, 3x – 5 (b) Among the result(s) obtained in (a), which of them is a factor of f(x)? (a) ∵ Leading coefficient = 3 and constant term = –5 ∴ Possible values of m = 1 and 3, possible values of n = ± 1 and ± 5. ∴ The possible linear factors of f(x) are x + 1 and 3x – 5.

Follow-up question Consider f(x) = 3x3 – 11x2 – 19x – 5. (a) By considering the coefficient of x3 and the constant term, which of the following is/are possible linear factor(s) of f(x)? x + 1, 2x – 1, 3x + 2, 3x – 5 (b) Among the result(s) obtained in (a), which of them is a factor of f(x)? (b) f(–1) = 3(–1)3 – 11(–1)2 – 19(–1) – 5 = 0 ∴ x + 1 is a factor of f(x). ∴ 3x – 5 is not a factor of f(x).

Let consider f(x) = 3x3 – 11x2 + 8x + 4 Let consider f(x) = 3x3 – 11x2 + 8x + 4. We can summarize the steps for factorizing a cubic polynomial.

Factorize f(x) = 3x3 – 11x2 + 8x + 4. Step 1: List all the factors of leading coefficient 3 and constant term +4. Positive factor of the leading coefficient 3: 1, 3 Factors of the constant term +4: ±1, ±2, ±4  Possible linear factors of f(x): x ± 1, x ± 2, x ± 4, 3x ± 1 , 3x ± 2, 3x ± 4.

Factorize f(x) = 3x3 – 11x2 + 8x + 4. Step 2: Use the factor theorem to find a factor of f(x). f(1) = 3(1)3 – 11(1)2 + 8(1) + 4 = 4 0 f(–1) = 3(–1)3 – 11(–1)2 + 8(–1) + 4 = –18 0 f(2) = 3(2)3 – 11(2)2 + 8(2) + 4 = 0  x – 2 is a factor of f(x). After we find a factor of f(x), we can perform long division.

Factorize f(x) = 3x3 – 11x2 + 8x + 4. Step 3: Perform long division. 4 8 11 2 3 + - x 3x 2 4 8 3 - + 5x 3x x 10x -5x 6x the quotient of f(x) ÷ (x – 2). a factor of f(x) 4 +  f(x) = (x – 2)(3x2 – 5x – 2)

Factorize f(x) = 3x3 – 11x2 + 8x + 4. Step 4: Further factorize the quotient if possible. f(x) = (x – 2)(3x2 – 5x – 2) = (x – 2)(x – 2)(3x + 1) Factorize 3x2 – 5x – 2. Rough Work 3x +1 x –2 +x –6x = –5x ∴ f(x) = (x – 2)2(3x + 1)

Follow-up question Factorize . ∵ ∴ x – 3 is a factor of f(x).

Follow-up question Factorize . By long division, ∴

H.C.F. and L.C.M. of Polynomials

Do you know how to find the H.C.F. and L.C.M. of numbers? Consider two numbers 60 and 72. 60 = 22 31 51 Express as products of prime factors. 72 = 23 32 H.C.F.= 22 31 = 12 For each COMMON prime factor, take the one with the smallest exponent.

Do you know how to find the H.C.F. and L.C.M. of numbers? Consider two numbers 60 and 72. 60 = 22 31 51 Express as products of prime factors. 72 = 23 32 L.C.M.= 23 32 51 = 360 For each prime factor, take the one with the largest exponent.

How about finding the H.C.F. and L.C.M. of polynomials? Similar techniques can be applied to find the H.C.F. and L.C.M. of polynomials.

For each COMMON factor, take the one with the lowest degree. In fact, The highest common factor (H.C.F.) of two or more polynomials is the polynomial of the highest degree which is a common factor of all the polynomials. Consider two polynomials x2yz and x3y2. x2yz = x2 y1 z1 x3y2 = x3 y2 H.C.F.= x2 y1 = x2y For each COMMON factor, take the one with the lowest degree.

For each factor, take the one with the highest degree. The lowest common multiple (L.C.M.) of two or more polynomials is the polynomial of the lowest degree which is a common multiple of all the polynomials. Again, consider two polynomials x2yz and x3y2. x2yz = x2 y1 z1 x3y2 = x3 y2 L.C.M.= x3 y2 z1 = x3y2z For each factor, take the one with the highest degree.

Follow-up question Find the H.C.F. and L.C.M. of 8x2y5z3, 6xy2 and 18x3yz2. 8x2y5z3 = 23 x2 y5 z3 6xy2 = 2 3 x y2 18x3yz2 = 2 32 x3 y z2 ∴ H.C.F. = 2 x y = 2xy L.C.M. = 23 32 x3 y5 z3 = 72x3y5z3

How to find the H.C.F. and L.C.M. of x3 – 9x and 2x2 – 4x – 30? We can factorize the polynomials first.

Find the H.C.F. and L.C.M. of x3 – 9x and 2x2 – 4x – 30. x3 – 9x = x(x2 – 9) = x(x2 – 32) = x(x + 3)(x – 3)  a2 – b2 ≡ (a + b)(a – b) 2x2 – 4x – 30 = 2(x2 – 2x – 15) = 2(x + 3)(x – 5)  By the cross method ∴ H.C.F. = x + 3 L.C.M. = 2 x (x + 3) (x – 3) (x – 5) = 2x(x + 3)(x – 3)(x – 5)

Follow-up question Find the H.C.F. and L.C.M. of 2x2 + 5x2 – 3 and 2x2 + 12x + 18. 2x2 + 5x – 3 = (x + 3)(2x – 1) 2x2 + 12x + 18 = 2(x2 + 6x + 9) = 2(x + 3)2 ∴ H.C.F. = 3 + x L.C.M. =

Rational Functions and their Manipulations

Q are polynomials and Q  0, is called a rational function. Rational Functions A function that can be written as ,where P and Q are polynomials and Q  0, is called a rational function. Q P Example: 2 3 – x (i)  3 and x – 2 are polynomials. 5 3 2 – + x y (ii)  x2 – y and x3 + x – 5 are polynomials.

Do you remember the techniques in simplifying a rational function?  Factorize both the numerator and the denominator first.  Cancel out the common factor.

Multiplication and Division of Rational Functions Before doing multiplication and division, reduce each rational function to its simplest form first. 2  Simplify each rational function.  Cancel out the common factors.

Follow-up question Simplify .  Simplify each rational function.  Cancel out the common factors.

Addition and Subtraction of Rational Functions The steps for addition and subtraction of rational functions are shown in the following example. Step 1: Simplify each rational function. Step 2: Find the L.C.M. of the denominators. Step 3: Change each rational function using the L.C.M. as its denominator. ) 7 )( 3 2 ( - + = x

Addition and Subtraction of Rational Functions Step 4: Find the algebraic sum of the new numerators. (For subtraction, find the algebraic difference.) Step 5: Simplify the result.

Follow-up question Simplify .  Simplify each rational function.  L.C.M. of (x + y)(2x – y) and (x + y) = (x + y)(2x – y)