3º ESO Bilingüe Page 1 UNIT 5 VOCABULARY: POLYNOMIALS 1.1. Monomials A monomial is an algebraic expression consisting of only one term. A monomial can be any of the following: A constant: 2 4-5 A variable: x y z 2 3 are monomials. The product of a constant and one or more variables: 2a 3xy 5x³ -x²y² THE VARIABLES OF A MONOMIAL SHOULD NOT HAVE NEGATIVE OR FRACTIONAL EXPONENTS! Therefore, these algebraic expressions are not monomials: 3x 2 3xy 1/2 3 t 2 The degree of a monomial is defined as the sum of all the exponents of the variables. For example: The degree of 3x is 1 (the exponent of x is 1). The degree of 3x 2 y 2 z is 5 (because 2 + 2 + 1 =5). The degree of 6 is 0 (there is no variable). Exercise. Indicate which of the following expressions are monomials. If it is, indicate the degree and the coefficient of the monomial. If it is not, state why it is not a monomial: a) 3x 3 b) 5x 3 c) 3x + 1 d) 2 x e) 3 4 x2 y f) 2 x 1.2. Polynomials A polynomial is an expression which is made up of monomials that are added or subtracted. Polynomials can have: No variable at all (for example, 21 is a polynomial that has just one term, which is a constant). One variable (for example x 4 2x 2 + x has three terms, but only one variable, x). Two or more variables (for example, xy 4-5x 2 z has two terms, and three variables,x, y and z).
3º ESO Bilingüe Page 2 The degree of a polynomial with only one variable is the largest exponent of that variable. A term that doesn't have a variable in it is called constant term. The coefficient of the term with the highest degree is called the leading coefficient. Exercises. 1. Complete this table: Polynomial Degree Leading coefficient Constant term x 4 3x 5 + 2x 2 + 5 1 x 4 x 2 2 x x 3 x 7 4 2. Write: a) An ordered polynomial without a constant term. b) A complete polynomial but not orderly. c) A polynomial of degree 4, complete with odd leading coefficient. 1.3. Evaluating a polynomial To evaluate a polynomial, you plug in the given value of x, and compute the value. For example, to evaluate P(x) = 2x 3 x 2 4x + 2 at x = 3: P( 3) = 2 ( 3) 3 ( 3) 2 4 ( 3) + 2 = 2 ( 27) (9) + 12 + 2 = 54 9 + 14 = 63 + 14 = 49 ALWAYS REMEMBER TO BE CAREFUL WITH BRACKETS AND THE MINUS SIGNS!
3º ESO Bilingüe Page 3 Exercise. Evaluate x 4 + 3x 3 x 2 + 6 a) for x = 3 b) for x = 3 c) for x = 0 d) for x= 1 2 1.4. Addition of polynomials To add two polynomials, you have to follow these steps: Place like terms together. Add the like terms For example, to add p(x) = 2x 2 + 6x + 5 and q(x) = 3x 2-2x - 1 Start with: 2x 2 + 6x + 5 + 3x 2-2x - 1 = Place like terms together: (2x 2 + 3x 2 ) + (6x - 2x) + (5-1) = Add the like terms: = 5x 2 + 4x + 4 You could also add two polynomials in columns like this: 1.5. Subtraction of polynomials To subtract polynomials, follow these steps: First reverse the sign of each term you are subtracting (in other words, turn "+" into "-", and "-" into "+"). Add as usual. For example, to subtract p(x) = 2x 2 + 6x + 5 and q(x) = 3x 2-2x - 1 Start with: 2x 2 + 6x + 5 - (3x 2-2x - 1) = Reverse the signs of each term: 2x 2 + 6x + 5-3x 2 + 2x + 1 = Add the like terms: = 2x 2 3x 2 + 6x + 2x + 5 + 1 = -x 2 + 8x + 6 And again, you can also do it in columns: Exercise. Given the polynomials: P(x) = 4x 2 1 Q(x) = x 3 3x 2 + 6x 2 R(x) = 6x 2 + x + 1 S(x) = 1 3 2 x2 +4 T(x) = 2 x2 +5 U(x) = x 2 + 2 Calculate: a) P(x) + Q (x) b) P(x) U(x) c) P(x) + R (x) d) 2P(x) R(x) e) S(x) + T(x) + U(x) f) S(x) T(x) + U(x)
3º ESO Bilingüe Page 4 2.1. Multiplication of polynomials Before multiplying polynomials, let us look at the simplest cases first: Monomial times monomial To multiply one monomial by another monomial, first multiply the coefficients, and then multiply each variable together, adding the exponents: Monomial times binomial Multiply the monomial by each of the two terms, like this: Polynomial times polynomial Follow these steps: Multiply each term in the first polynomial by each term in the second polynomial. Always remember to add like terms. For example, to multiply (x + 2y) (3x 4y + 5) We multiply each term in the first polynomial by each term in the second polynomial: 3x 2 4xy + 5x + 6xy 8y 2 + 10y We add like terms: 3x 2 + 2xy + 5x 8y 2 + 10y And once again, you can also do it in columns: Exercise. Multiply: a) (x 4 2x 2 + 2) (x 2 2x + 3) b) (3x 2 5x) (2x 3 + 4x 2 x + 2) 2.2. Special products SQUARE OF A SUM! (a+b) 2 =a 2 +2ab+b 2
3º ESO Bilingüe Page 5 Examples: (x +5) 2 =x 2 +2 x 5+5 2 =x 2 +10x+25 (2x+3) 2 =(2x) 2 +2 2x 3+3 2 =4x 2 +12x+9 SQUARE OF A DIFFERENCE! (a b) 2 =a 2 2ab+b 2 Examples: (x 2 2) 2 =(x 2 ) 2 2 x 2 2+2 2 =x 4 4x 2 +4 (x 3) 2 =x 2 2 x 3+( 3) 2 =x 2 2 3 x +3 SUM TIMES DIFFERENCE! (a+b)(a b)=a 2 b 2 Examples: ( x + 3 2) ( x 3 2) ( =x 2 3 2)2=x 2 9 4 (ab 2 +5)(ab 2 5)=(ab 2 ) 2 5 2 =a 2 b 4 25 Exercises. 1. Expand the following expressions: a) (x + 2) 2 c) (3x + 7)² e) (a 2 b 3 2) 2 b) (2x 5) 2 d) ( x 2 2 ) 2 f) ( x 2 2 + y 3 2 3 ) 2. Expand the following expressions: a) (3x 2) (3x + 2) c) (3x 2) (3x + 2) b) (x + 5) (x 5) d) (3x 5) (3x 5)
3º ESO Bilingüe Page 6 3. Express as a binomial: a) x² + 12x + 36 = (x + )² f) 4x² 9 = b) x² 16 = (x + ) (x - ) g) 64x² - 160x + 100 = c) 4x² 20x + 25 = ( - 5 )² h) 9x² 25 = d) x² + 14x + 49 = i) 25x² + 40x + 16 = e) x² 1 = j) x 2 x + 1 4 2.1. Division of polynomials When the divisor has a single term, it is easy to divide a polynomial by splitting it at the "+" and "-" signs, like this: Or: If the divisor has more than one term, you may need to use the long division (listen to your teacher explanation): Exercise. Find p(x) q(x) if: a) p (x )=x 4 2x 3 11x 2 +30x 20 q (x )=x 2 +3x 2 b) p (x )=x 5 +2 x 3 x +8 q (x )=x 2 2x+1 2.2. Ruffini's Rule (Synthetic Division) Ruffini's rule is a quick method for dividing a polynomial by a binomial of the form x a. Learn the full process watching the following video! http://www.youtube.com/watch?v=3lvlhz3ds-8
3º ESO Bilingüe Page 7 An example: Divide (x 4 3x 2 +2) (x 3) So, the quotient is a polynomial of lower degree and whose coefficients are the ones obtained in the division. q (x )=x 3 +3x 2 +6x+18 The last number obtained, 56, is the remainder. r(x) = 56 Exercise. Find p(x) q(x) if: a) p (x )=x 3 +2 x +70 q (x )=x +4 b) p (x )=x 5 32 q (x )=x 2 c) p (x )=x 4 3x 2 +2 q (x )=x 3 4.1. The Remainder Theorem The Remainder Theorem When you divide a polynomial p(x) by x-a, the remainder of the division will be p(a). For example: Imagine that we want to divide p(x) = 5x 4-3x 3 + 2x 2 7x + 3 (dividend) by q(x) = x - 1 (divisor). To find the remainder, we don't need to divide by x-1... just find out p(1)!!! p(1) = 5 1 4-3 1 3 + 2 1 2 7 1 + 3 = 0 As you can see, if we calculate the division using Ruffini's Rule, we get the same answer. Exercises. 1. Find the remainder of the following divisions: a) (x 5 2x 2 3) : (x 1) b) (2x 4 2x 3 + 3x 2 + 5x + 10) : (x + 2) c) ( x 4 3x 2 + 2) : (x 3) 2. Indicate which of these divisions are exact: a) (x 3 5x 1) : (x 3) b) (x 6 1) : (x + 1) c) (x 4 2x 3 + x 2 + x 1) : (x 1)
3º ESO Bilingüe Page 8 4.2. Roots of a polynomial A root (or "zero") of a polynomial is where the polynomial is equal to zero, all the values of the variables that make the polynomial be 0. A polynomial of degree 2 will have a maximum of 2 real roots (places where the polynomial is equal to zero). A polynomial of degree 3 will have a maximum of 3 real roots. And so on. For example, p(x) = x 2-9 has a degree of 2, so there will be a maximum of 2 roots. Let us solve it. We want it to be equal to zero, so we write x 2 9 = 0 x 2 = 9 x = ± 3 So the roots are +3 and -3. Let's check it: p(3) = 3 2 9 = 9 9 = 0 p(-3) = (-3) 2 9 = 9 9 = 0 4.3. The Factor Theorem The Factor Theorem When p(a)=0 then x-a is a factor of the polynomial p(x) And the other way around, too: When x-a is a factor of the polynomial, then p(a)=0 Why Is This Useful? This theorem means that knowing that x-a is a factor is the same as knowing that "a" is a root (and vice versa). It means that you can quickly check if (x-a) is a factor of the polynomial. Exercise. Check the following statements: a) (x 3 5x 1) has of factor (x 3) c) (x 6 1) has of factor (x + 1) b) (x 4 2x 3 + x 2 + x 1) has of factor (x 1 ) d) (x 10 1024) has of factor (x + 2)