POLYNOMIALS Maths 4 th ESO José Jaime Noguera 1
Algebraic expressions Book, page 26 YOUR TURN: exercises 1, 2, 3. Exercise: Find the numerical value of the algebraic expression xy 2 8x + y, knowing that x=-2 and y=3. 2
Definitions A monomial is an algebraic expression as ax n, where a is a real number, and n is a non negative integer. A polynomial is an algebraic expression that consists of a sum of several monomials. 3
Standard form of a polynomial Example: P x = 3x 2 5 + 5x 3 8x 5 a) Write in standard form: P x = 8x 5 + 5x 3 + 3x 2 5 b) Degree: 5 c) Constant term: -5 d) Leading coefficient: -8 4
Exercise Find the standard form, the leading term, the leading coefficient, the linear term, the quadratic term, the constant term and the degree of: a) P x = 5x 3 7x 2 + 5x 4 8 + 9x b) P x = 2x 7 3x 5 + 7x 4 c) P x = 4x 3 7x 5 + x 4 5x 5
Reviewing operations with monomials ADDITION/SUBTRACTIONS. You can sum or subtract monomials WITH THE SAME LITERAL PART (called similar monomials or like terms in a polynomial): ax n + bx n = a + b x n 3x 5 5x 5 = 2x 5 2x 3 5x 2 you can t do anything PRODUCT: ax n bx m = a b x n+m 3x 2 5x 7 = 15x 9 DIVISION: ax n : bx m = a: b x n m 8x 5 : 5x 7 = 8 5 x 2 POWER: ax n m = a m x n m 3x 2 3 = 3 3 x 6 = 27x 6 6
Exercises a) 5x 2 +3x 2 = b) 3x 5 9x2 = c) 8x-3x = d) 21x 6 :3x 3 = e) -3x 3 5x7 = f) 30x 5 : (-5x 3 ) = g) 5x 3-3x 2-7x 3 +3x 2 +5x = h) 5x 2-3x+7x 2 +6x-2x 2 = 7
Extracting common factors 1. Factorize all the coefficients 2. Find the greatest common factor of the terms (considering both coefficients and variables). 3. Divide each term of the expression in parentheses by the greatest common factor, and write the variable factor outside the parentheses. Example: 21x 3 y 2 15x 2 y 4 + 3x 2 y 2 1. 3 7x 3 y 2 3 5x 2 y 4 + 3x 2 y 2 2. 3 7xxxyy 3 5xxyyyy + 3xxyy 3. 3x 2 y 2 (7x 5y 2 + 1) 8
Exercise Extract common factors: a) 7x 2 + 49x 14x 3 b) 28a 3 b 2 c 12a 2 bc 3 8 a 4 b 2 c) 5xy 3 35x 2 y 3 + 15xz d) (x-2)(x+7)-(x-2)(2x-6) e) 3(x+2)-(2x+4) 9
Binomial theorem Where the binomial coefficients can be calculated using the Pascal s triangle (or Tartaglia s triangle): 10
Click on the image 11
What happens with a b n? It s the same but you have to change the signs: a b 5 = +1a 5 b 0 5a 4 b 1 + 10a 3 b 2 10a 2 b 3 + 5a 1 b 4 1a 0 b 5 Exercise: Find the following powers: 1. x 2 2 2. x + 5 3 3. 3x 1 2 4. x + 2y 3 5. 3x 5 4 6. 2x + 1 5 12
Now, we are going to follow the book: Page 27. 3.1 Important equalities Exercises 5, and Exercise: Write as a power: a) 4x 2 + 20x + 25 b) 25x 2 1 c) 4x 2 12xy + 9y 2 d) 9x 4 49 13
Operations with polynomials Page 27, 28, 29 3.2 Operations with polynomials: +, -,, :, Ruffini s rule or synthetic division. Exercises: Pag 27: 6, 7 Pag 28: 8, 9 Pag 29: 10 14
Remainder theorem: P(x) Remainder x-a quotient Remainder = P(a) Example: P x = 2x 3 x 2 7x + 5 If we calculate P(x) : x-2 we obtain Quocient: 2x 2 + 3x 1 Remainder: 3 If we calculate P 2 = 2 2 3 2 2 72 + 5 = 3 15
Roots of a polynomial We say that a is a root of P(x) if P(a)=0 The roots of a polynomial fulfills: If P(x):(x-a) is exact, then a is a root of P(x) If a is a root of P(x), then P(x):(x-a) is exact The number of roots of a polynomial is less than or equal to the degree of P(x) If a is a root of P(x), then a has to be a divisor of the constant term of P(x) 16
Examples Is 3 a root of P x = 3x 2 5x + 1? No, because P 3 = 3 3 2 5 3 + 1 = 13 0 Is 3 a root of P x = 3x 2 8x 3? Yes, because P 3 = 3 3 2 8 3 3 = 0 What is the remainder of (3x 2 8x 3):(x-3)? The remainder is 0 because 3 is a root of 3x 2 8x 3 17
EXERCISES Pag 29: 11, 12, 13 EXERCISE: True of false? a) The division 3x 3 2x 2 + 1: x 1 is exact. b) The polynomial P x = 3x 3 2x 2 + 1 has 4 roots. c) 3 is a possible root of 3x 4 2x 2 + 12 d) If P(-2)=0, then the remainder of P(x):(x+2) is 0. 18
Factorization of a polynomial The goal is to write the polynomial as a product of polynomials with the lowest possible degree. Examples: a) x 2 6x + 9 = x 3 x 3 b) 2x 4 6x 3 26x 2 + 30x = =2x(x 1)(x + 3)(x 5) 19
How can we do that? There are several ways but you can follow the steps: STEP 1: Extract common factors. Now we are going to work with the polynomial inside the parentheses. (If x is a common factor, then 0 is a root) STEP 2: The degree of the polynomial inside the parentheses is 1. Great!, you have finished. GO TO STEP 5. STEP 3: The degree of the polynomial inside the parentheses is 2. Solve the equation polynomial=0. You will obtain 2 solutions=two roots of the polynomial. Save them. You have finished. GO TO STEP 5. If the equation has no solution, you have finished, GO TO STEP 5. (In this step you can also use the important equalities) 20
STEP 4: The degree of the polynomial inside the parentheses is greater than or equal to 3. You have to use the Ruffini s rule: The possible roots are the divisors of the constant term. When you obtain REMAINDER=0 you have found a root. Save it. Dividend = Divisor Quotient (+remainder=0). Working with the quotient GO TO STEP 2. If you don t find any root, you have finished. GO TO STEP 5. STEP 5: The factorized polynomial is: a(x-root_1)(x-root_2).(x-root_n)(maybe a polynomial that you cannot factorize) VERY IMPORTANT. YOU HAVE TO CHECK WHETER THE LEADING COEFFICIENT IS THE SAME THAT THE ORIGINAL ONE. If THIS IS NOT THE CASE, YOU JUST HAVE TO MULTIPLY THE FACTORIZED EXPRESSION BY THE ORIGINAL LEADING COEFFICIENT. 21
EXAMPLES 3x 2 15x STEP 1: 3x (x-5) STEP 2: The degree of (x-5) is 1, so you have finished. STEP 5 The leading term of 3x 2 15x is the same as the leading term of 3x (x-5), so this is the correct result. 22
4x 3 12x 2 + 9x STEP 1: x 4x 2 12x + 9 (0 is a root) STEP 2: 4x 2 12x + 9 don t have degree 1 STEP 3: We have to solve 4x 2 12x + 9 = 0 STEP 5 x = 12 ± 12 2 4 4 9 2 4 = x 1 = 3 2 x 2 = 3 2 2 4x 3 12x 2 + 9x = 4 x x 3 2 x 3 2 = 4x x 3 2 In step 3 is also possible 4x 2 12x + 9 = 2x 3 2 In this case 4x 3 12x 2 + 9x = x 2x 3 2 (pay attention with the leading coefficient). 23
x 3 3x 2 x + 3 STEP 1: There is no common factor STEP 2: The degree isn t 1 STEP 3: The degree isn t 2 STEP 4: The degree is 3. Ruffini.The divisors of the constant term are: +1,-1,+3,-3 Let s begin with 1: QUOTIENT=x 2 2x 3 REMAINDER=0 Then, 1 is a root of x 3 3x 2 x + 3. We go to step 2 using the quotient. 24
We are working with x 2 2x 3 STEP 2: The degree isn t 1 STEP 3: The degree is 2. We have to solve x 2 2x 3=0 x = 2 ± 2 2 4 1 ( 3) 2 1 The roots are -1 and 3. = x 1 = 1 x 2 = 3 STEP 5: We have the roots: 1, -1 and 3, so: x 3 3x 2 x + 3=(x-1)(x+1)(x-3) 25
2x 4 + 14x 3 + 16x 2 32x STEP 1: 2x x 3 + 7x 2 + 8x 16 It means that 0 is also a root STEP 2: NO STEP 3: NO STEP 4: Ruffini. Possible roots: +1,-1,2,-2,4,-4,8,-8,16,-16 QUOTIENT: x 2 + 8x + 16 REMAINDER: 0 1 is a root of x 3 + 7x 2 + 8x 16 26
Working with x 2 + 8x + 16 Step 2: No Step 3: We have to solve x 2 + 8x + 16=0 x = 8 ± 8 2 4 1 (16) = x 1 = 4 2 1 x 2 = 4 Step 5: The roots are 0, 1, 4 (double): 2 x 0 x 1 (x 4)(x 4)= 2x(x 1)(x 4) 2 Sometimes it s clearer to write all the steps: 2x 4 + 14x 3 + 16x 2 32x x x 3 + 7x 2 + 8x 16 x(x-1)(x 2 + 8x + 16 ) 2x(x 1)(x 4) 2 27
Exercises Pag 30: 14 Exercise: Factorize: 28
Algebraic fractions Book. Page 30 and 31: 3.6 Algebraic fractions. Exercises: PAG 30: 15 PAG 31: 16 29