P.3 Division of Polynomials

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1 00 section P3 P.3 Division of Polynomials In this section we will discuss dividing polynomials. The result of division of polynomials is not always a polynomial. For example, xx + 1 divided by xx becomes xx + 1 xx = xx xx + 1 xx = xx, which is not a polynomial. Thus, the set of polynomials is not closed under the operation of division. However, we can perform division with remainders, mirroring the algorithm of division of natural numbers. We begin with dividing a polynomial by a monomial and then by another polynomial. Division of Polynomials by Monomials To divide a polynomial by a monomial, we divide each term of the polynomial by the monomial, and then simplify each quotient. In other words, we use the reverse process of addition of fractions, as illustrated below. aa + bb dd = aa dd + bb dd Dividing Polynomials by Monomials Divide and simplify. a. (6xx xx xx) (3xx) b. xxyy 8xx yy+6xx 3 yy xxyy Solution a. (6xx xx xx) (3xx) = 6xx3 +15xx xx 3xx 5 = 6xx3 + 15xx 3xx 3xx xx 3xx = xx b. xxyy 8xx yy+6xx 3 yy xxyy 4 3 = xxyy + 8xx yy 6xx3 yy xxyy xxyy xxyy = yy 33xx Division of Polynomials by Polynomials Polynomials and Polynomial Functions To divide a polynomial by another polynomial, we follow an algorithm similar to the long division algorithm used in arithmetic. For example, observe the steps taken in the long division algorithm when dividing 158 by 13 and the corresponding steps when dividing xx + 5xx + 8 by xx + 3. Step 1: Place the dividend under the long division symbol and the divisor in front of this symbol. 13 ) 158 xx + 3 dddddddddddddd ) xx + 5xx + 8 dddddddddddddddd

2 section P3 01 Remember: Both polynomials should be written in decreasing order of powers. Also, any missing terms after the leading term should be displayed with a zero coefficient. This will ensure that the terms in each column are of the same degree. Step : Divide the first term of the dividend by the first term of the divisor and record the quotient above the division symbol. qqqqqqqqqqqqqqtt 1 xx 13 ) 15 8 xx + 3 ) xx + 5xx + 8 Step 3: Multiply the quotient from Step by the divisor and write the product under the dividend, lining up the columns with the same degree terms ) xx xx + 3 ) xx + 5xx + 8 xx + 3xx Step 4: Underline and subtract by adding opposite terms in each column. We suest to record the new sign in a circle, so that it is clear what is being added ) xx xx + 3 ) xx + 5xx + 8 (xx + 3xx) xx Step 5: Drop the next term (or digit) and repeat the algorithm until the degree of the remainder is lower than the degree of the divisor ) xx + xx + 3 ) xx + 5xx + 8 (xx + 3xx) xx + 8 (xx + 6) rrrrrrrrrrrrrrrrrr In the example of long division of numbers, we have 158 = So, the quotient can be written as = In the example of long division of polynomials, we have So, the quotient can be written as xx +5xx+8 xx + 5xx + 8 = (xx + 3) (xx + ) +. xx+3 = xx + + xx+33. Generally, if PP, DD,, and RR are polynomials, such that PP(xx) = DD(xx) (xx) + RR(xx), then the ratio of polynomials PP and DD can be written as PP(xx) RR(xx) = (xx) + DD(xx) DD(xx), Division of Polynomials

3 0 section P3 where (xx) is the quotient polynomial, and RR(xx) is the remainder from the division of PP(xx) by the divisor DD(xx). Observe: The degree of the remainder must be lower than the degree of the divisor, as otherwise, we could apply the division algorithm one more time. Dividing Polynomials by Polynomials Divide. a. (3xx 3 xx + 5) (xx 3) b. pp 3 +pp+5pp 1+pp Solution a. When writing the polynomials in the long division format, we use a zero placeholder term in place of the missing linear terms in both, the dividend and the divisor. So, we have 3333 xx + 0xx 3 ) 3xx 3 xx + 0xx + 5 ( 3xx 3 + 0xx + 9xx) xx 9xx + 5 ( xx 0xx + 6) Thus, (3xx 3 xx + 5) (xx 3) = 3xx + 9xx 1 xx 3 = xx 33. b. To perform this division, we arrange both polynomials in decreasing order of powers, and replace the constant term in the dividend with a zero. So, we have pp pp + 77 pp + 5 ) pp 3 + 3pp + pp + 0 ( pp 3 + 5pp ) pp + pp ( pp 5pp) 7pp + 0 7pp Thus, pp3 +pp+5pp 5+pp = pp pp = pp+5 pp pp Observe in the above answer that is written in a simpler form,. This is pp+5 4pp+10 because 35 = 35 1 = 35. pp+5 pp+5 4pp+10 Polynomials and Polynomial Functions

4 section P3 03 Quotient Functions Similarly as in the case of polynomials, we can define quotients of functions. Definition 3.1 Suppose ff and are functions of xx with the corresponding domains DD ff and DD. Then the quotient function, denoted ff, is defined as ff ff(xx) (xx) = (xx). The domain of the quotient function is the intersection of the domains of the two functions, DD ff and DD, excluding the xx-values for which (xx) = 0. So, DD ff = DD ff DD \{xx (xx) = 00} Dividing Polynomial Functions Suppose PP(xx) = xx xx 6 and (xx) = xx. Find the following: a. PP (xx) and PP (aa), Solution b. PP ( 3) and PP (), c. domain of PP. Notice that this equation holds only for xx. a. By Definition 3.1, PP PP(xx) (xx) = = xx xx 6 = (xx+3)(xx ) = + 33 (xx) xx xx So, PP ( 3) = ( 3) + 3 = 33. One can verify that the same value is found by evaluating PP( 3) ( 3). PP () is undefined, so is not in the domain of PP b. Since the equation (xx+3)(xx ) = xx + 3 is true only for xx, the simplified formula xx PP (xx) = xx + 3 cannot be used to evaluate PP (). However, by Definition 3.1, we have PP PP() () = () = () () 6 = 8 6 = 0 () 0 0 = uuuuuuuuuuuuuuuuuu To evaluate PP (aa), we first notice that if aa 1, then aa. So, we can use the simplified formula PP (xx) = xx + 3 and evaluate PP (aa) = (aa) + 3 = c. The domain of any polynomial is the set of all real numbers. So, the domain of PP is the set of all real numbers except for the xx-values for which the denominator (xx) = Division of Polynomials

5 04 section P3 xx is equal to zero. Since the solution to the equation xx = 0 is xx =, then the value must be excluded from the set of all real numbers. Therefore, DDPP = R \ {}. P.3 Exercises Vocabulary Check Complete each blank with the most appropriate term from the given list: divisor, remainder, decreasing, zero, monomial, lower, domain, factor. 1. The algorithm of long division of polynomials requires that both polynomials are written in order of powers and any missing term is replaced with a term which plays the role of a placeholder.. To check a division problem, multiply the by the quotient and then add the. 3. When dividing a polynomial by a, there is no need to use the long division algorithm. 4. In polynomial division, when the degree of the remainder is than the degree of the divisor, the division process ends. 5. The of a quotient function ff does not contain the zeros of the devisor function. 6. If the remainder in the division of PP(xx) by (xx) is zero, then (xx) is a of PP(xx). Concept Check 7. True or False? When a tenth-degree polynomial is divided by a second-degree polynomial, the quotient is a fifth-degree polynomial. 8. True or False? When a polynomial is divided by a second-degree polynomial, the remainder is a first-degree polynomial. Divide. 9. 0xx 3 15xx +5xx 5xx 10. 7yy4 +18yy 9yy 9yy 11. 8xx yy 4xxxx 4xxxx 1. 5cc3 dd+10cc dd 15ccdd 3 5cccc 13. 9aa5 15aa 4 +1aa 3 3aa 14. 0xx3 yy +44xx yy 3 4xx yy 4xx yy xx3 7xx +1xx 16. 4mm nn 1mmnn 3 +18mmnn 17. 1aabb cc+10aa bbbb+18aaaacc 8xx 3 14mm nn 3 6aa bbbb Divide. 18. (xx + 3xx 18) (xx + 6) 19. (3yy + 17yy + 10) (3yy + ) 0. (xx 11xx + 16) (xx + 8) 1. (tt 7tt 9) (tt 3) Polynomials and Polynomial Functions

6 section P yy3 yy 10 3yy zz3 6zz +17zz 13 5zz kk4 +6kk 3 +3kk 1 kk tt +19tt+7 4tt xx3 +1 xx aa3 +6aa +14 aa+4 6. xx3 +4xx xx+ xx +xx kk4 +1kk 3 4kk 1 3kk 1 3. xx4 4xx 3 +5xx 3xx+ xx yy4 +16 yy+ 4. 4xx3 +8xx 11xx+3 4xx xx3 xx +5xx 4 xx xx pp3 +7pp +9pp+3 pp+ 33. pp3 1 pp xx5 3 xx Concept Check For each pair of polynomials, PP(xx) and DD(xx), find such polynomials (xx) and RR(xx) that PP(xx) = (xx) DD(xx) + RR(xx). 37. PP(xx) = 4xx 3 4xx + 13xx and DD(xx) = xx PP(xx) = 3xx 3 xx + 3xx 5 and DD(xx) = 3xx Concept Check For each pair of functions, ff and, find the quotient function ff (xx) and state its domain. 39. ff(xx) = 6xx 4xx, (xx) = xx 40. ff(xx) = 6xx + 9xx, (xx) = 3xx 41. ff(xx) = xx 36, (xx) = xx ff(xx) = xx 5, (xx) = xx ff(xx) = xx xx 3, (xx) = xx ff(xx) = 3xx + xx 4, (xx) = 3xx ff(xx) = 8xx , (xx) = xx ff(xx) = 64xx 3 7, (xx) = 4xx 3 Let PP(xx) = xx 44, (xx) =, and RR(xx) = xx. Find each of the following. If the value can t be evaluated, say DNE (does not exist). 47. RR (xx) 48. PP (xx) 49. RR RR (xx) PP 50. RR () 51. RR (0) 5. PP (3) RR 53. RR ( ) 54. PP RR () 55. PP PP (aa), for aa RR 56. RR (xx) 58. RR (aa 1) RR Analytic Skills Solve each problem. 59. The area AA of a rectangle is 5xx + 1xx + 4 and its width WW is xx +. a. Find the length LL of the rectangle. b. Find the length if the width is 8 meters. LL xx The area AA of a triangle is 6xx xx 15. Find its height h, if the base of the triangle is 3xx 5. Then, find the height h, if the base is 7 centimeters. h 3xx 5 Division of Polynomials

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