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Chapter 5: Polynomials

Chapter 5: Polynomials. 5.1 Addition and Subtraction of Polynomials and Polynomial Functions. A polynomial in x is a finite sum of terms of the form ax n , where the coefficient a is a real number and the degree of the term n is a whole number.

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Chapter 5: Polynomials

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  1. Chapter 5: Polynomials

  2. 5.1 Addition and Subtraction of Polynomials and Polynomial Functions

  3. A polynomialin x is a finite sum of terms of the form axn, where the coefficienta is a real number and the degree of the termn is a whole number.

  4. Monomial: a polynomial with exactly one term • Binomial: a polynomial with exactly two terms • Trinomial:a polynomial with exactly three terms Terms of a polynomial are usually written in descending order according to degree. The first term is the highest-degree term and is called the leading term. Its coefficient is the leading coefficient,and its degree is the degree of the polynomial. If a polynomial has more than one variable, the degree of a term is the sum of the exponents of the variables in the term.

  5. Examples

  6. To add polynomials, group like terms then add like terms. • To subtract polynomials, add the opposite of the second polynomial to the first polynomial: A – B = A + (–B)

  7. Polynomials can be added or subtracted vertically. Placeholders may be used to help align like terms. • A polynomial function is a function defined by a finite sum of terms of the form axn, where a is a real number and n is a whole number.

  8. The length of a rectangle is 3 ft more than twice the width. If x represents the width, then 2x + 3 represents the length. The perimeter of the rectangle can be written as a polynomial function P(x) = 2(2x + 3) + 2(x) = 4x + 6 + 2x = 6x + 6

  9. Exercise 1 Write the polynomial in descending order: p(x) = 19x + 9x3 + 11x2 + 2x5 + 1   A) p(x) = 19x + 11x2 + 9x3 + 2x5 + 1 B) p(x) = 2x5 + 9x3 + 11x2 + 19x + 1 C) p(x) = 1 + 19x + 11x2 + 9x3 + 2x5 D) It's already in descending order.

  10. Exercise 2 Subtract the polynomials: (12x2y3 – 3xy2 + 5y) – (2y + 2xy2 – 9x2y3)  A) 3x2y3 – xy2 + 5+7y B) 3x4y6 – x2y4 + 7y2 C) 21x2y3 – 5xy2 + 3y D) 14x2y4 – x2y4 + 14x2y4

  11. Exercise 3 The width of a rectangular plot of land is 17 less than 4 times its length. Write a polynomial that describes the perimeter of the plot. • P(l) = 5l – 17 • P(l) = l(4l – 17) • P(l) = 10l + 34 D) P(l) = 10l – 34

  12. 5.2 Multiplication of Polynomials

  13. Multiplying Monomials To multiply monomials, use the associative and commutative properties to group coefficients and like bases. Then use properties of exponents to simplify the result.

  14. Multiply a Polynomial by a Monomial Use the distributive property to multiply polynomials: a(b + c) = ab + ac. Then simplify each term.

  15. The FOIL Method To multiply polynomials with more than one term, multiply each term of the first polynomial by each term of the second polynomial. For binomials, use the acronym FOIL to help multiply the terms.

  16. Difference of squares:Multiplying the sum and difference of the same two terms results in a product called a difference of squares. (a + b)(a – b) = a2 – b2 • Perfect square trinomials:When squaring a binomial, the result is called a perfect square trinomial.(a + b)2 = a2 + 2ab + b2(a – b)2 = a2 – 2ab + b2

  17. Find and simplify a polynomial that represents the area of the triangle. The area of a triangle is given by the formula The height of the triangle is 5t and the base is 7t + 3. So the area is given by

  18. Exercise 4 Multiply: –5z6(–10z3 – z2 + 7)   A) 50z9 – 5z8 + 35z6 B) 50z18 + 5z12 – 35z6 C) 50z9 – z2 + 7  D) 50z9 + 5z8 – 35z6

  19. Exercise 5 Multiply: (–6t + s)(t – 5s) • –6t2 –5s2 • –6t2 + 24st • –5t – 4s D) –6t2 + 31st – 5s2

  20. Exercise 6 The volume of a rectangular solid is given by V = lwh, where l is the length of the base, w is the width of the base, and h is the height. A rectangular box with a square base has height 14 inches more than the length of the base. Find an expression for the volume of the box in terms of the length of the base. • V = x3 B) V = x3 + 14x2 C) V = x3 + 14 D) V = x2 + 14x

  21. 5.3 Division of Polynomials

  22. Dividing a Polynomial by a Monomial To divide a polynomial by a monomial, divide each term in the polynomial by the divisor and simplify the result. If a, b, and c are polynomials such that c≠ 0 then and

  23. Example Divide each term in the numerator by 7 Simplify each term.

  24. If the divisor has two or more terms, use a long division process similar to the division of real numbers. • Write the divisor and dividend in descending order. • Divide the leading term in the dividend by the leading term in the divisor. • Multiply the result by the divisor and record the new result. • Subtract, bring down the next column, and repeat the process. • Stop when the degree of the remainder is less than the degree of the divisor. • Write the solution in the form quotient + remainder/divisor

  25. Example The result is

  26. The result of long division of polynomials can be checked in the same way as the division of real numbers.Dividend = (divisor)(quotient) + remainder

  27. Synthetic Division When the divisor is first degree and of the form x – r where r is a constant, synthetic division may be used to divide polynomials. This technique uses the coefficients of the divisor and dividend without writing the variables.

  28. Synthetic Division Steps • Write the value of r in a box. • Write the coefficients of the dividend to the right of the box. • Skip a line and draw a horizontal line below the list of coefficients. • Bring down the leading coefficient from the dividend and write it below the line. • Multiply the value of r by the number below the line. Write the result in the next column above the line. • Add the numbers in the column above the line and write the result below the line. • Repeat steps 5 and 6 until all columns have been completed. • Use the numbers below the line as the coefficients of the quotient. The number in the last column is the remainder.

  29. Example Use synthetic division to divide the polynomials. (5n3 – n2 + 2n – 3)  (n + 3) Note that n + 3 = n – (–3). The result is

  30. Before using synthetic division, the terms of the dividend and divisor should be written in descending order. • Missing powers must be accounted for by using placeholders.3a + 7a3 – 4 + a5 = a5 + 0a4 + 7a3 + 0a2 + 3a – 4

  31. Exercise 7 Divide:    A) 2x3 + 9x4 – 7x3 B) 2x3 + 9x2 – 7x C) 2x5 + 9x4 – 7x3 – x2 D) 2x5/2 + 9x2 – 7x3/2

  32. Exercise 8 Divide: (3y3 – 5y2 – 28y – 15)  (y – 4)   A) 3y2 + 7y – 15 B) C) D) –2y2 – 28y – 15

  33. Exercise 9 Divide using synthetic division: (y4 – 81)  (y – 3) A) y3 – 27 B) y3 + 81y – 27 C) y3 + 3y2 + 9y + 27 D) y3 – 3y2 + 9y – 27

  34. 5.4 Greatest Common Factor and Factoring by Grouping

  35. To factor a polynomial means to express the polynomial as a product of two or more polynomials. • The greatest common factor (GCF) of a polynomial is the greatest factor that divideseach term of the polynomial evenly.

  36. Steps to Remove the GCF • Identify the greatest common factor of all terms of the polynomial. • Write each term as the product of the GCF and another factor. • Use the distributive property to factor out the greatest common factor. To check the factorization, multiply the polynomials.

  37. Factor out the greatest common factor. Identify the GCF: 4mn Write each term as the product of the GCF and another factor. Factor out the GCF by using the distributive property.

  38. Sometimes it is helpful to factor out a negative factor, such as the opposite of the GCF. This changes the signs of the remaining terms. • The distributive property may be used to factor out a common factor of more than one term.

  39. Factor by Grouping Given a polynomial with four terms, we can try to factor it by grouping: • Identify and factor out the GCF from all four terms. • Factor out the GCF from the first pair of terms. Factor out the GCF from the second pair of terms. • If the two remaining terms share a common binomial factor, factor it out.

  40. Example Factor by grouping. pq – 3p + 7q – 21 pq – 3p + 7q – 21 The GCF of all four terms is 1. = pq – 3p¦ + 7q – 21 Group the first pair of terms and the second pair of terms. = p(q – 3) + 7(q – 3) Factor out p from the first pair of terms and 7 from the second pair. = (q – 3)(p + 7) Factor out the common binomial factor (q – 3).

  41. The order of factors does not matter because multiplication is commutative. (q – 3)(p + 7) = (p + 7) (q – 3) • It is sometimes necessary to rearrange the original four terms to factor by grouping.

  42. Exercise 10 What is the greatest common factor of 3z4t3 + 15z6t2 + 9z3t4 ? A) 3 B) 3z3t2 C) 3z4t3 D) 15z6t4

  43. Exercise 11 Factor out the greatest common factor: –6t6 – 8t5 – 10t3 A) –2(3t6 + 4t5 + 5t3) C) –6t6(–8t5 – 10t3)  B) 6t3(–t3 – t2 – 2) D) –2t3(3t3 + 4t2 + 5)

  44. Exercise 12 Factor by grouping: 5x3 – 21 + 35x2 – 3x A) (5x2 – 3)(x + 7) B) 5x3 – 21 + x(35x – 3) C) (7x2 – 3)(x + 5)  D) (5x2 – 7)(x + 3)

  45. 5.5 Factoring Trinomials

  46. The product of two binomials results in a four-term expressions that can sometimes be simplified to a trinomial. To factor the trinomial, reverse the process.

  47. Grouping Method To factor trinomials of the form ax2 + bx + c, a≠ 0: • Factor out the GCF. • Find the product ac. • Find two integers whose product is ac and whose sum is b. • Rewrite the middle term bx as the sum of two terms whose coefficients are the numbers found in step 3. • Factor the polynomial by grouping. If no pair of integers can be found in step 3, the trinomial cannot be factored further and is called a prime polynomial.

  48. Example Factor 2x2 – 13x – 7. • Find the product ac = 2(– 7) = –14. • List all the factors of –14 and find the pair whose sum is –13. (–1)(14) (2)(–7) (1)(–14) (–2)(7)The numbers –14 and 1 produce a product of –14 and a sum of –13.

  49. Example (continued) • Write the middle term as two terms whose coefficients are –14 and 1. 2x2 – 13x – 7 = 2x2 – 14x + x – 7 • Factor by grouping. 2x2 – 14x¦ + x – 7= 2x(x – 7) + 1(x – 7)= (2x + 1)(x – 7) Check: (2x + 1)(x – 7) = 2x2 – 14x + x – 7 = 2x2 – 13x – 7 

  50. Trial-and-Error Method To factor trinomials of the form ax2 + bx + c, a≠ 0: • Factor out the GCF. • List the pairs of the factors of a and the pairs of the factors of c. Consider the reverse order in either list. • Construct two binomials of the form • Test each combination of factors until the product of the outer terms and the product of the inner terms add to the middle term. • If no combination of factors works, the polynomial is prime.

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