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# POLYNOMIALS

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• Teachers: This definition for ‘degree’ has been simplified intentionally to help students understand the concept quickly and easily.
• ### Transcript of "POLYNOMIALS"

1. 1. MADE BY : DEV YADAV SCHOOL : K.V. N.S.G. MANESAR CLASS : IX “B” SUBJECT : MATHS SUBMITTED TO : Ms Madhur Abhispa
2. 2.  A monomial is a real number, a variable, or a product of a real number and one or more variables with whole number exponents. › Examples:  The degree of a monomial in one variable is the exponent of the variable. 2 3 5, , 3 , 4x wy x
3. 3.  A polynomial is a monomial or a sum of monomials. › Example:  The degree of a polynomial in one variable is the greatest degree among its monomial terms. › Example: 2 3 2 5xy x+ − 2 4 7x x− − +
4. 4.  A polynomial function is a polynomial of the variable x. › A polynomial function has distinguishing “behaviors”  The algebraic form tells us about the graph  The graph tells us about the algebraic form
5. 5.  The standard form of a polynomial function arranges the terms by degree in descending order › Example: 3 2 ( ) 4 3 5 2P x x x x= + + −
6. 6. Polynomial s Kinds of Polynomials (according to number of terms) A polynomial with only one term is called a monomial. A polynomial with two terms is called a binomial. A polynomial with three terms is called a trinomial.
7. 7.  Polynomials are classified by degree and number of terms. › Polynomials of degrees zero through five have specific names and polynomials with one through three terms also have specific names. Degree Name 0 Constant 1 Linear 2 Quadratic 3 Cubic 4 Quartic 5 Quintic Number of Terms Name 1 Monomial 2 Binomial 3 Trinomial
8. 8. Monomial : Algebric expression that consists only one term is called monomial. Binomial : Algebric expression that consists two terms is called binomial. Trinomial : Algebric expression that consists three terms is called trinomial. Polynomial : Algebric expression that consists many terms is called polynomial.
9. 9.  A polynomial is a function in the form  is called the leading coefficient  is the constant term  n is the degree of the polynomial f (x) = an xn + ...+ a1x + a0 an a0
10. 10. Polynomials Degree Classify by degree Classify by no. of terms. 5 0 Constant Monomial 2x - 4 1 Linear Binomial 3x2 + x 2 Quadratic Binomial x3 - 4x2 + 1 3 Cubic Trinomial
11. 11. Standard form means that the terms of the polynomial are placed in descending order, from largest degree to smallest degree. Polynomial in standard form: 2 x3 + 5x2 – 4 x + 7 Degree Constant termLeading coefficient Polynomial s
12. 12. Phase 1Phase 1 Phase 2Phase 2 To rewrite a polynomial in standard form, rearrange the terms of the polynomial starting with the largest degree term and ending with the lowest degree term. The leading coefficient, the coefficient of the first term in a polynomial written in standard form, should be positive. How to convert a polynomial into standard form?
13. 13. 745 24 −−+ xxx x5+4 4x 2 x− 7− Write the polynomials in standard form. 243 5572 xxxx ++−− 3 2x+4 x− 7−x5+2 5x+ )7552(1 234 −+++−− xxxx 3 2x−4 x 7+x5−2 5x− Remember: The lead coefficient should be positive in standard form. To do this, multiply the polynomial by –1 using the distributive property.
14. 14. Write the polynomials in standard form and identify the polynomial by degree and number of terms. 23 237 xx −−1. 2. xx 231 2 ++
15. 15. 23 237 xx −− 23 237 xx −− 3 3x− 2 2x− 7+ ( )7231 23 +−−− xx 723 23 −+ xx This is a 3rd degree, or cubic, trinomial.
16. 16. xx 231 2 ++ xx 231 2 ++ 2 3x x2+ 1+ This is a 2nd degree, or quadratic, trinomial.
17. 17. Let f(x) be a polynomial of degree n > 1 and let a be any real number. When f(x) is divided by (x-a) , then the remainder is f(a). PROOF Suppose when f(x) is divided by (x-a), the quotient is g(x) and the remainder is r(x). Then, degree r(x) < degree (x-a) degree r(x) < 1 [ therefore, degree (x-a)=1] degree r(x) = 0 r(x) is constant, equal to r (say) Thus, when f(x) is divided by (x-a), then the quotient is g9x) and the remainder is r. Therefore, f(x) = (x-a)*g(x) + r (i) Putting x=a in (i), we get r = f(a) Thus, when f(x) is divided by (x-a), then the remainder is f(a).
18. 18. Let f(x) be a polynomial of degree n > 1 and let a be any real number. (i) If f(a) = 0 then (x-a) is a factor of f(x). PROOF let f(a) = 0 On dividing f(x) by 9x-a), let g(x) be the quotient. Also, by remainder theorem, when f(x) is divided by (x-a), then the remainder is f(a). therefore f(x) = (x-a)*g(x) + f(a) f(x) = (x-a)*g(x) [therefore f(a)=0(given] (x-a) is a factor of f(x).
19. 19.  Let α, β and γ be the zeroes of the polynomial ax³ + bx² + cx + d  Then, sum of zeroes(α+β+γ) = -b = -(coefficient of x²) a coefficient of x³ αβ + βγ + αγ = c = coefficient of x a coefficient of x³ Product of zeroes (αβγ) = -d = -(constant term) a coefficient of x³
20. 20. I) Find the zeroes of the polynomial x² + 7x + 12and verify the relation between the zeroes and its coefficients. f(x) = x² + 7x + 12 = x² + 4x + 3x + 12 =x(x +4) + 3(x + 4) =(x + 4)(x + 3) Therefore,zeroes of f(x) =x + 4 = 0, x +3 = 0 [ f(x) = 0] x = -4, x = -3 Hence zeroes of f(x) are α = -4 and β = -3.
21. 21. 2) Find a quadratic polynomial whose zeroes are 4, 1. sum of zeroes,α + β = 4 +1 = 5 = -b/a product of zeroes, αβ = 4 x 1 = 4 = c/a therefore, a = 1, b = -4, c =1 as, polynomial = ax² + bx +c = 1(x)² + { -4(x)} + 1 = x² - 4x + 1 The end
22. 22. Some common identities used to factorize polynomials (x+a)(x+b)=x2+(a+b)x+ab(a+b)2 =a2 +b2 +2ab (a-b)2 =a2 +b2 -2ab a2 -b2 =(a+b)(a-b)
23. 23. Advanced identities used to factorize polynomials (x+y+z)2 =x2 +y2 +z2 +2xy+2yz+2zx (x-y)3 =x3 -y3 - 3xy(x-y) (x+y)3 =x3 +y3 + 3xy(x+y) x3 +y3 =(x+y) * (x2 +y2 -xy) x3 -y3 =(x+y) * (x2 +y2 +xy)
24. 24.  A real number ‘a’ is a zero of a polynomial p(x) if p(a)=0. In this case, a is also called a root of the equation p(x)=0.  Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero, and every real number is a zero of the zero polynomial.
25. 25. POLYNOMIAL FUNCTIONS A POLYNOMIAL is a monomial or a sum of monomials. A POLYNOMIAL IN ONE VARIABLE is a polynomial that contains only one variable.Example: 5x2 + 3x - 7
26. 26. A polynomial function is a function of the form f(x) = an xn + an – 1 xn – 1 +· · ·+ a1 x + a0 Where an ≠ 0 and the exponents are all whole numbers. A polynomial function is in standard form if its terms are written in descending order of exponents from left to right. For this polynomial function, an is the leading coefficient, a0 is the constant term, and n is the degree.
27. 27. POLYNOMIAL FUNCTIONS The DEGREE of a polynomial in one variable is the greatest exponent of its variable. A LEADING COEFFICIENT is the coefficient of the term with the highest degree. What is the degree and leading coefficient of 3x5 – 3x + 2 ?
28. 28. POLYNOMIAL FUNCTIONS A polynomial equation used to represent a function is called a POLYNOMIAL FUNCTION. Polynomial functions with a degree of 1 are called LINEAR POLYNOMIAL FUNCTIONS Polynomial functions with a degree of 2 are called QUADRATIC POLYNOMIAL FUNCTIONS Polynomial functions with a degree of 3 are called CUBIC POLYNOMIAL FUNCTIONS
29. 29. Degree Type Standard Form You are already familiar with some types of polynomial functions. Here is a summary of common types of polynomial functions. 4 Quartic f (x) = a4x4 + a3x3 + a2x2 + a1x + a0 0 Constant f (x) = a0 3 Cubic f (x) = a3x3 + a2 x2 + a1x + a0 2 Quadratic f (x) = a2 x2 + a1x + a0 1 Linear f (x) = a1x + a0
30. 30. The largest exponent within the polynomial determines the degree of the polynomial. Polynomial Function in General Form Degree Name of Function 1 Linear 2 Quadratic 3 Cubic 4 Quarticedxcxbxaxy ++++= 234 dcxbxaxy +++= 23 cbxaxy ++= 2 baxy +=
31. 31. What is the degree of the monomial? 24 5 bx The degree of a monomial is the sum of the exponents of the variables in the monomial. The exponents of each variable are 4 and 2. 4+2 = 6. The degree of the monomial is 6. The monomial can be referred to as a sixth degree monomial.
32. 32. 14 +x 83 3 −x 1425 2 −+ xx The degree of a polynomial in one variable is the largest exponent of that variable. 2 A constant has no variable. It is a 0 degree polynomial. This is a 1st degree polynomial. 1st degree polynomials are linear. This is a 2nd degree polynomial. 2nd degree polynomials are quadratic. This is a 3rd degree polynomial. 3rd degree polynomials are cubic.
33. 33. Classify the polynomials by degree and number of terms. Polynomial a. b. c. d. 5 42 −x xx +2 3 14 23 +− xx Degree Classify by degree Classify by number of terms Zero Constant Monomial First Linear Binomial Second Quadratic Binomial Third Cubic Trinomial
34. 34. Find the sum. Write the answer in standard format. (5x3 – x + 2 x2 + 7) + (3x2 + 7 – 4 x) + (4x2 – 8 – x3 ) Adding Polynomials SOLUTION Vertical format: Write each expression in standard form. Align like terms. 5x3 + 2x2 – x + 7 3x2 – 4x + 7 – x3 + 4x2 – 8+ 4x3 + 9x2 – 5x + 6
35. 35. Find the sum. Write the answer in standard format. (2 x2 + x – 5) + (x + x2 + 6) Adding Polynomials SOLUTION Horizontal format: Add like terms. (2x2 + x – 5) + (x + x2 + 6) = (2x2 + x2 ) + (x + x) + (–5 + 6) = 3x2 + 2x + 1
36. 36. Find the difference. (–2x3 + 5x2 – x + 8) – (–2x2 + 3x – 4) Subtracting Polynomials SOLUTION Use a vertical format. To subtract, you add the opposite. This means you multiply each term in the subtracted polynomial by –1 and add. –2x3 + 5x2 – x + 8 –2x3 + 3x – 4– Add the opposite No change –2x3 + 5x2 – x + 8 2x3 – 3x + 4+
37. 37. Find the difference. (–2x3 + 5x2 – x + 8) – (–2x2 + 3x – 4) Subtracting Polynomials Use a vertical format. To subtract, you add the opposite. This means you multiply each term in the subtracted polynomial by –1 and add. –2x3 + 5x2 – x + 8 –2x3 + 3x – 4– 5x2 – 4x + 12 –2x3 + 5x2 – x + 8 2x3 – 3x + 4+ SOLUTION
38. 38. Find the difference. (3x2 – 5x + 3) – (2x2 – x – 4) Subtracting Polynomials SOLUTION Use a horizontal format. (3x2 – 5x + 3) – (2x2 – x – 4) = (3x2 – 5x + 3) + (–1)(2x2 – x – 4) = x2 – 4x + 7 = (3x2 – 5x + 3) – 2x2 + x + 4 = (3x2 – 2x2 ) + (– 5x + x) + (3 + 4)
39. 39. Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. f(x) = x 2 – 3x 4 – 71 2 SOLUTION The function is a polynomial function. It has degree 4, so it is a quartic function. The leading coefficient is – 3. Its standard form is f(x) = –3x 4 + x 2 – 7. 1 2
40. 40. Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. Identifying Polynomial Functions The function is not a polynomial function because the term 3 x does not have a variable base and an exponent that is a whole number. SOLUTION f(x) = x 3 + 3 x
41. 41. Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. SOLUTION f(x) = 6x 2 + 2x –1 + x The function is not a polynomial function because the term 2x–1 has an exponent that is not a whole number.
42. 42. Identifying Polynomial Functions Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type and leading coefficient. SOLUTION The function is a polynomial function. It has degree 2, so it is a quadratic function. The leading coefficient is π. Its standard form is f(x) = πx 2 – 0.5x – 2. f(x) = – 0.5x + πx 2 – 2 2
43. 43. POLYNOMIAL FUNCTIONS EVALUATING A POLYNOMIAL FUNCTION Find f(-2) if f(x) = 3x2 – 2x – 6 f(-2) = 3(-2)2 – 2(-2) – 6 f(-2) = 12 + 4 – 6 f(-2) = 10
44. 44. POLYNOMIAL FUNCTIONS EVALUATING A POLYNOMIAL FUNCTION Find f(2a) if f(x) = 3x2 – 2x – 6 f(2a) = 3(2a)2 – 2(2a) – 6 f(2a) = 12a2 – 4a – 6
45. 45. POLYNOMIAL FUNCTIONS EVALUATING A POLYNOMIAL FUNCTION Find f(m + 2) if f(x) = 3x2 – 2x – 6 f(m + 2) = 3(m + 2)2 – 2(m + 2) – 6 f(m + 2) = 3(m2 + 4m + 4) – 2(m + 2) – 6 f(m + 2) = 3m2 + 12m + 12 – 2m – 4 – 6 f(m + 2) = 3m2 + 10m + 2
46. 46. POLYNOMIAL FUNCTIONS EVALUATING A POLYNOMIAL FUNCTION Find 2g(-2a) if g(x) = 3x2 – 2x – 6 2g(-2a) = 2[3(-2a)2 – 2(-2a) – 6] 2g(-2a) = 2[12a2 + 4a – 6] 2g(-2a) = 24a2 + 8a – 12
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