Excerpt from "Intermediate Algebra" 2014 AoPS Inc.

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1 Ecert from "Intermediate Algebra" 04 AoPS Inc. for which our grah is below the -ais with the oints where the grah intersects the -ais (because the ineuality is nonstrict), we have the same answer as before, (, 5] [ [ 4, 4]. Our consideration of the grah of y = f () is essentially the same aroach to the roblem as the table. In both, we determine the sign of f () for di erent intervals of. We use the roots of f () to determine the intervals we must consider. Eercises 7.. Find all roots of each of the following olynomials. (a) f () = (c) f () = (b) g(t) = t 4 + 5t 9t 65t + 50 (d) h(y) = 6y 5y y Find all solutions to the ineualities below. (a) t + 0t + 7t > 8 (c) r 4 + r + 5r r 8 ale 0 (b) r (6r + r ) ale 5(4r + 5) 7.. There are four roots of f () = We can easily test to find that f () = 0. We can then check all the other divisors of 6, both ositive and negative, and find that no divisors of 6 besides are roots of the olynomial. Is it correct to deduce that the other three roots of f () are not integers? 7..4? Suose that f () is a olynomial with integer coe cients such that f () = and f (7) = 7. Show that f () has no integer roots. Hints:, 9 7. Rational Roots A rational number is a number that can be eressed in the form /, where and are integers. A number that is not rational is called an irrational number. Just as we can use the coe cients of a olynomial with integer coe cients to narrow the search for integer roots, we can also use these coe cients to narrow the search for non-integer rational roots. Problems Problem 7.: In this roblem we find the roots of the olynomial g() = (a) Can you find any integers n such that g(n) = 0? (b) Suose g(/) = 0, where and are integers and / is in reduced form. Rewrite g(/) = 0 using the definition of g. (c) Get rid of the fractions in your euation from (b) by multilying by the aroriate ower of. (d) What terms in your euation from (c) have? Why must divide 0? (e) What terms in your euation from (c) have? Why must divide? (f) 06 Find all the roots of g(). Coyrighted Material

2 Ecert from "Intermediate Algebra" 04 AoPS Inc RATIONAL ROOTS Problem 7.: Let f () be the olynomial f () = a n n + a n n + a n n + + a + a 0, where all the coe cients are integers and both a n and a 0 are nonzero. Let and be integers such that / is a fraction in simlest terms, and f (/) = 0. In this roblem, we show that divides a 0 and divides a n. (a) Suose f (/) = 0. Why must we have a n n + a n n + a n n + a n n + + a n + a 0 n = 0? (b) (c) Which terms in the euation in (a) must be divisible by? Use this to show that a 0 must be divisible by. Which terms in the euation in (a) must be divisible by? Use this to show that a n must be divisible by. Problem 7.4: Find all the roots of each of the following olynomials: (a) f () = (b) g() = Problem 7.5: Find all r such that r 4 6r > 4r 69r + 8. We ve found a way to narrow our search for integer roots. How about rational roots? Problem 7.: Find all such that = 0. Solution for Problem 7.: We let f () = and begin our search for roots of f () by seeing if we can find any integer roots. An integer root of f () must be a divisor of its constant term, which is 0 = 5. We try each divisor (ositive and negative): f () = = 7, f ( ) = ( ) +6 ( ) ( ) +0 = 45, f () = = 08, f ( ) = ( ) +6 ( ) ( ) +0 = 40, f (5) = = 755, f ( 5) = ( 5) +6 ( ) ( 5) +0 = 95, f (0) = = 00, f ( 0) = ( 0) +6 ( 0) ( 0) +0 = (We might also have checked some of these a little faster with synthetic division.) We know we don t have to check any other integers, since any integer root must divide the constant term of f (). Since all of these fail, we know there are no integer roots of f (). Back to the drawing board. A little number theory heled us find integer roots, so we hoe that a similar method might hel us find rational roots. Letting and be relatively rime integers (meaning they have no common ositive divisor besides ), we eamine the euation f (/) = 0: f! =! + 6!! Coyrighted Material + 0 = 0. 07

3 Ecert from "Intermediate Algebra" 04 AoPS Inc. Multilying by to get rid of all the denominators, we have the euation = 0. (7.) Previously, we isolated the last term of such an eansion in order to find information about integer roots. Doing so here gives us 0 = ( 6 + ). Dividing this euation by gives us 0 = 6 +. The right side of this euation must be an integer, so 0 / must also be an integer. Therefore, divides 0. But, since and are relatively rime, must divide 0. If we isolate the first term of the left side of Euation (7.) instead of the last term, we have Dividing this euation by gives us = ( ). = As before, the right side is an integer, so the left side must be as well. Thus divides. Because and are relatively rime, we know that must divide. Combining what we have learned about and, we see that if / is a root of f (), then is a divisor of 0 and is a divisor of. Now, we make a list of ossible rational roots /: ±, ±, ±5, ±0, ±, ±5, ±, ±, ±5, ±0, ± 4, ±5 4, ± 6, ±5 6, ±, ± 5. While this is not a terribly small list, it is a comlete list of ossible rational roots of a olynomial with leading coe cient and constant term 0. We already know that f () has no integer roots, so we continue our search for roots with the fractions in the list above. We start with the fractions with small denominators, since these will be easiest to check. After a little eerimentation, we find that f (/) = 0 via the synthetic division at right. (You may have found one of the other roots first.) We therefore have f () = ( + 0) We continue our search by finding roots of + 0. We find + 0 = (6 + 0) = ( )( + 5), so we have f () = ( )( + 5). 08 Coyrighted Material

4 Ecert from "Intermediate Algebra" 04 AoPS Inc RATIONAL ROOTS When factoring olynomials, we often factor constants out of each linear factor so that the coe in each factor is. This makes identifying the roots articularly easy: f () = () () + 5 = + 5. Now we can easily see that the roots of f () are,, and 5. cient of In our solution, we found a way to narrow our search for rational roots of f () in much the same way we can narrow our search for integer roots. Let s see if we can aly this method to any olynomial with integer coe cients. Problem 7.: Let f () be the olynomial f () = a n n + a n n + a n n + + a + a 0, where all the coe cients are integers and both a n and a 0 are nonzero. Let and be integers such that / is a fraction in simlest terms, and f (/) = 0. Show that divides a 0 and divides a n. Solution for Problem 7.: We use our solution to Problem 7. as a guide. Since f (/) = 0, we have a n! n + a n! n + a n! n! + + a + a 0 = 0. We get rid of the fractions by multilying both sides by n, which gives us a n n + a n n + a n n + + a n + a 0 n = 0. (7.) To see why must divide a n, we divide Euation (7.) by to roduce a n / in the first term: a n n + a n n + a n n + + a n + a 0 n = 0. Isolating this first term gives us a n n = a n n a n n a n a 0 n. Each term on the right side of this euation is an integer, so the entire right side is an integer. Therefore, a n n / must be an integer. Since / is in lowest terms and a n n / is an integer, we know that divides a n. We take essentially the same aroach to show that a 0. We divide Euation (7.) by and isolate the term with a 0 in it, which gives us a 0 n = a n n a n n a n n a n. Each term on the right is an integer, so the entire right side euals an integer. So, the eression a 0 n / must eual an integer, which means that must divide a 0. Etra! If f (n) = n + n + 4, then f (), f (), f (), and f (4) are all rime. Maybe f (n) is rime for all integers n. Is it? (Continued on age 0) Coyrighted Material 09

5 Ecert from "Intermediate Algebra" 04 AoPS Inc. In Problem 7., we have roved the Rational Root Theorem: Imortant: Let f () be the olynomial f () = a n n + a n n + a n n + + a + a 0, where all the a i are integers, and both a n and a 0 are nonzero. If and are relatively rime integers and f (/) = 0, then a 0 and a n. Problem 7.4: Find all the roots of each of the following olynomials: (a) f () = (b) g() = Solution for Problem 7.4: (a) We first look for easy roots by evaluating f () and f ( ). We find that f () = = 56 and f ( ) = = 50. Since f (0) = 54, we see that there is a root between and 0 (because f ( ) < 0 and f (0) > 0) and a root between 0 and (because f (0) > 0 and f () < 0). Concet: We often start our hunt for rational roots by evaluating f ( ), f (0), and f (), because these are easy to comute and the results may tell us where to continue our search. We continue our hunt for roots with, which may be a root because divides the leading coe cient of f (). Using synthetic division, we find f ( ) = 4, so there is a root between and. We try and it works, giving us f () = ( 99 8) = (4 7). Factoring the uadratic then gives us f () = (4 + )( 9). So, the roots of f () are 4,, and 9. (b) We start our hunt for roots by finding g(0) = 45, g() = 80, and g( ) = 9. Unfortunately, that doesn t hel too much; we can t immediately tell if there are any roots between and 0, or between 0 and. So, we test and, knowing that and cannot be roots because the constant term of g() is odd. By synthetic division, we find that g() = 78. This has the same sign as g(), so we continue with g( ). We find that g( ) = 40, so because g( ) > 0 and g( ) < 0, we know there is a root between and. (Note that we didn t need to find the actual value of g( ). Once it is clear that g( ) is a large ositive number, we know that there is a root between and.) With synthetic division, we find that g( ) is also a large ositive number, so there is a root between and. We try /. We know that / might be a root of g() because divides the constant term and divides the leading coe cient. If / doesn t work, we will at least narrow our search to between and 0 Coyrighted Material

6 Ecert from "Intermediate Algebra" 04 AoPS Inc RATIONAL ROOTS / or between / and. Fortunately, the synthetic division shows us that g() = + ( ) = + ( ). We now note that has no negative roots because if is negative, each term of this olynomial is negative. So, we only have to search for ositive roots. We already know that and don t work, and the only two other ositive integers that divide 5 are 5 and 5. We find that 5 works, and we have g() = + ( 5)(5 4 + ). Now, we can factor the uadratic or use the uadratic formula to find g() = + ( 5)( )(5 ) = 0 + ( 5) The roots of g() are /, /, /5, and 5. Notice that there are roots between 0 and. Why doesn t testing g(0) and g() reveal the fact that these roots eist? (You ll have a chance to answer this uestion as an Eercise.). 5 Problem 7.5: Find all r such that r 4 6r > 4r 69r + 8. Solution for Problem 7.5: First, we move all the terms to the left, which gives us r 4 6r 4r + 69r 8 > 0. Now, we must factor the olynomial on the left. Let this olynomial be f (r). Because f (0) = 8 and f () = 6, there is a root between 0 and. We find that f (/) = 5, so the root is between 0 and /. Trying / gives us f (r) = r (r r 45r + 54) = r (4r 4r 5r + 8). Continuing with 4r 4r 5r + 8, we find that and are not roots, but is, and we have f () = r (r + )(4r r + 9) = r So, our ineuality is r (r + )(r ) > 0. (r + )(r ). The eression on the left side of the ineuality euals 0 when r =, /, or /. We consider the intervals between (and beyond) these values to build the table at right. We see that f (r) > 0 for r (, )[(, )[(, +). The ineuality is strict, so the roots of f (r) do not satisfy the ineuality. r + r (r ) f (r) r < + + < r < + + < r < r > Coyrighted Material

7 Ecert from "Intermediate Algebra" 04 AoPS Inc. Eercises 7.. Find all roots of the following olynomials: (a) g(y) = y 8y 9y + 0 (b) f () = In art (b) of Problem 7.4, we saw that g(0) and g() are both negative, and yet we still found roots between 0 and. How can that be? Shouldn t g(0) and g() have di erent signs if g() = 0 for some between 0 and? 7.. Suose f () = Notice that f () = 0, so is a root of f (). But doesn t evenly divide the constant term of f (). This seems to violate the Rational Root Theorem! Does it really? 7..4 In Problem 7.5, we built a table to find the values of r for which (r )(r + )(r ) > 0. How could we have solved this ineuality by thinking about the grah of y = ( )( + )( )? 7..5 Solve the two ineualities below: (a) (b)? 6s 4 + s s + 5s < Bounds In this section, we learn how to use synthetic division to determine bounds on the roots of a olynomial. Problems Problem 7.6: (a) Let f () = Without even finding the roots, how can we tell that there are no ositive values of for which f () = 0? (b) Let f () = Without finding the roots, how can we tell that there are no negative values of for which f () = 0? Problem 7.7: Let f () = a n n + a n n + + a + a 0. (a) Show that if all the coe cients of f () have the same sign (ositive or negative), then f () has no ositive roots. (b) (c) Show that if a i > 0 for all odd i and a i < 0 for all even i, then f () has no negative roots. Is it ossible for f () to have negative roots if a i is negative for all odd i and ositive for all even i? Problem 7.8: Let g() = (a) Use synthetic division to divide g() by 6. (b) What do you notice about the coe cients in the uotient and the remainder from art (a)? (c) How can you use your answer to art (b) to deduce that there are no roots of g() greater than 6? (In other words, how do you know you don t have to test 8,, 6, etc.?) Coyrighted Material