0.6 Factoring 73. As always, the reader is encouraged to multiply out (3

Transcription

1 0.6 Factoring 7 5. The G.C.F. of the terms in 81 16t is just 1 so there is nothing of substance to factor out from both terms. With just a difference of two terms, we are limited to fitting this olynomial into either the Difference of Two Squares or Difference of Two Cubes formula. Since the variable here is t, and is a multile of, we can think of t (t ). This means that we can write 16t (t ) which is a erfect square. (Since is not a multile of, we cannot write t as a erfect cube of a olynomial.) Identifying 81 9 and 16t (t ), we aly the Difference of Squares Formula to get: 81 16t 9 (t ) (9 t )(9 + t ) Difference of Squares, a 9, b t At this oint, we have an oortunity to roceed further. Identifying 9 and t (t), we see that we have another difference of squares in the first quantity, which we can reduce. (The sum of two squares in the second quantity cannot be factored over the rationals.) 81 16t (9 t )(9 + t ) ( (t) )(9 + t ) ( t)( + t)(9 + t ) Difference of Squares, a, b t As always, the reader is encouraged to multily out ( t)(+t)(9+t ) to check the result. 6. With a G.C.F. of 1 and just two terms, x 6 6 is a candidate for both the Difference of Squares and the Difference of Cubes formulas. Notice that we can identify x 6 (x ) and 6 8 (both erfect squares), but also x 6 (x ) and 6 (both erfect cubes). If we follow the Difference of Squares aroach, we get: x 6 6 (x ) 8 (x 8)(x + 8) Difference of Squares, a x and b 8 At this oint, we have an oortunity to use both the Difference and Sum of Cubes formulas: x 6 6 (x )(x + ) (x )(x +x + )(x + )(x x + ) Sum / Difference of Cubes, a x, b (x )(x + )(x x + )(x +x + ) Rearrange factors From this aroach, our final answer is (x )(x + )(x x + )(x +x + ). Following the Difference of Cubes Formula aroach, we get x 6 6 (x ) (x )((x ) +x + ) Difference of Cubes, a x, b (x )(x +x + 16) At this oint, we recognize x as a difference of two squares: x 6 6 (x )(x +x + 16) (x )(x + )(x +x + 16) Difference of Squares, a x, b

2 7 Prerequisites Unfortunately, the remaining factor x +x + 16 is not a erfect square trinomial - the middle term would have to be 8x for this to work - so our final answer using this aroach is (x )(x + )(x +x + 16). This isn t as factored as our result from the Difference of Squares aroach which was (x )(x + )(x x + )(x +x + ). While it is true that x +x +16 (x x +)(x +x +), there is no intuitive way to motivate this factorization at this oint. 5 The moral of the story? When given the otion between using the Difference of Squares and Difference of Cubes, start with the Difference of Squares. Our final answer to this roblem is (x )(x + )(x x + )(x +x + ). The reader is strongly encouraged to show that this reduces down to x 6 6 after erforming all of the multilication. The formulas on age 71, while useful, can only take us so far, so we need to review some more advanced factoring strategies. Advanced Factoring Formulas un-f.o.i.l.ing : Given a trinomial Ax + Bx + C, try to reverse the F.O.I.L. rocess. That is, find a, b, c and d such that Ax + Bx + C (ax + b)(cx + d). NOTE: This means ac A, bd C and B ad + bc. Factor by Grouing: If the exression contains four terms with no common factors among the four terms, try factor by grouing : ac + bc + ad + bd (a + b)c +(a + b)d (a + b)(c + d) The techniques of un-f.o.i.l.ing and factoring by grouing are difficult to describe in general but should make sense to you with enough ractice. Be forewarned - like all Rules of Thumb, these strategies work just often enough to be useful, but you can be sure there are excetions which will defy any advice given here and will require some insiration to solve. 6 Even though Chater will give us more owerful factoring methods, we ll find that, in the end, there is no single algorithm for factoring which works for every olynomial. In other words, there will be times when you just have to try something and see what haens. Examle Factor the following olynomials comletely over the integers x x 6. t 11t y 1y. 18xy 5xy 180x 5. t 10t +t x +x Of course, this begs the question, How do we know x x + and x +x + are irreducible? (We were told so on age 71, but no reason was given.) Stay tuned! We ll get back to this in due course. 6 Jeff will be sure to eer the Exercises with these. 7 This means that all of the coefficients in the factors will be integers. In a rare dearture from form, Carl decided to avoid fractions in this set of examles. Don t get comlacent, though, because fractions will return with a vengeance soon enough.

3 0.6 Factoring 75 Solution. 1. The G.C.F. of the terms x x 6 is 1 and x x 6 isn t a erfect square trinomial (Think about why not.) so we try to reverse the F.O.I.L. rocess and look for integers a, b, c and d such that (ax + b)(cx + d) x x 6. To get started, we note that ac 1. Since a and c are meant to be integers, that leaves us with either a and c both being 1, or a and c both being 1. We ll go with a c 1, since we can factor 8 the negatives into our choices for b and d. This yields (x + b)(x + d) x x 6. Next, we use the fact that bd 6. The roduct is negative so we know that one of b or d is ositive and the other is negative. Since b and d are integers, one of b or d is ±1 and the other is 6 OR one of b or d is ± and the other is. After some guessing and checking, 9 we find that x x 6(x + )(x ).. As with the revious examle, we check the G.C.F. of the terms in t 11t + 5, determine it to be 1 and see that the olynomial doesn t fit the attern for a erfect square trinomial. We now try to find integers a, b, c and d such that (at + b)(ct + d) t 11t + 5. Since ac, we have that one of a or c is, and the other is 1. (Once again, we ignore the negative otions.) At this stage, there is nothing really distinguishing a from c so we choose a and c 1. Now we look for b and d so that (t + b)(t + d) t 11t + 5. We know bd 5 so one of b or d is ±1 and the other ±5. Given that bd is ositive, b and d must have the same sign. The negative middle term 11t guides us to guess b 1 and d 5 so that we get (t 1)(t 5) t 11t + 5. We verify our answer by multilying. 10. Once again, we check for a nontrivial G.C.F. and see if 6 11y 1y fits the attern of a erfect square. Twice disaointed, we rewrite 6 11y 1y 1y 11y + 6 for notational convenience. We now look for integers a, b, c and d such that 1y 11y +6 (ay + b)(cy + d). Since ac 1, we know that one of a or c is ±1 and the other ±1 OR one of them is ± and the other is ±6 OR one of them is ± while the other is ±. Since their roduct is 1, however, we know one of them is ositive, while the other is negative. To make matters worse, the constant term 6 has its fair share of factors, too. Our answers for b and d lie among the airs ±1 and ±6, ± and ±18, ± and ±9, or ±6. Since we know one of a or c will be negative, we can simlify our choices for b and d and just look at the ositive ossibilities. After some guessing and checking, 11 we find ( y + )(y + 9) 1y 11y Since the G.C.F. of the terms in 18xy 5xy 180x is 18x, we begin the roblem by factoring it out first: 18xy 5xy 180x 18x(y y 10). We now focus our attention on y y 10. We can take a and c to both be 1 which yields (y + b)(y + d) y y 10. Our choices for b and d are among the factor airs of 10: ±1 and ±10 or ± and ±5, where 8 Pun intended! 9 The authors have seen some strange gimmicks that allegedly hel students with this ste. We don t like them so we re sticking with good old-fashioned guessing and checking. 10 That s the checking art of guessing and checking. 11 Some of these guesses can be more educated than others. Since the middle term is relatively small, we don t exect the extreme factors of 6 and 1 to aear, for instance.

4 76 Prerequisites one of b or d is ositive and the other is negative. We find (y 5)(y + ) y y 10. Our final answer is 18xy 5xy 180x 18x(y 5)(y + ). 5. Since t 10t t + 15 has four terms, we are retty much resigned to factoring by grouing. The strategy here is to factor out the G.C.F. from two airs of terms, and see if this reveals a common factor. If we grou the first two terms, we can factor out a t to get t 10t t (t 5). We now try to factor something out of the last two terms that will leave us with a factor of (t 5). Sure enough, we can factor out a from both: t+15 (t 5). Hence, we get t 10t t + 15 t (t 5) (t 5) (t )(t 5) Now the question becomes can we factor t over the integers? This would require integers a, b, c and d such that (at +b)(ct +d) t. Since ab and cd, we aren t left with many otions - in fact, we really have only four choices: (t 1)(t + ), (t + 1)(t ), (t )(t +1) and (t +)(t 1). None of these roduces t - which means it s irreducible over the integers - thus our final answer is (t )(t 5). 6. Our last examle, x +x +16, is our old friend from Examle As noted there, it is not a erfect square trinomial, so we could try to reverse the F.O.I.L. rocess. This is comlicated by the fact that our highest degree term is x, so we would have to look at factorizations of the form (x + b)(x + d) as well as (x + b)(x + d). We leave it to the reader to show that neither of those work. This is an examle of where trying something ays off. Even though we ve stated that it is not a erfect square trinomial, it s retty close. Identifying x (x ) and 16, we d have (x + ) x +8x + 16, but instead of 8x as our middle term, we only have x. We could add in the extra x we need, but to kee the balance, we d have to subtract it off. Doing so roduces and unexected oortunity: x +x + 16 x +x (x x ) Adding and subtracting the same term x +8x + 16 x Rearranging terms (x + ) (x) Factoring erfect square trinomial [(x + ) x][(x + ) + x] Difference of Squares: a (x + ), b x (x x + )(x +x + ) Rearraging terms We leave it to the reader to check that neither x integers, so we are done. x + nor x +x + factor over the Solving Equations by Factoring Many students wonder why they are forced to learn how to factor. Simly ut, factoring is our main tool for solving the non-linear equations which arise in many of the alications of Mathematics. 1 We use factoring in conjunction with the Zero Product Proerty of Real Numbers which was first stated on age 19 and is given here again for reference. 1 Also known as story roblems or real-world examles.

5 0.6 Factoring 77 The Zero Product Proerty of Real Numbers: If a and b are real numbers with ab 0 then either a 0 or b 0 or both. For examle, consider the equation 6x + 11x 10. To see how the Zero Product Proerty is used to hel us solve this equation, we first set the equation equal to zero and then aly the techniques from Examle 0.6.: 6x + 11x 10 6x + 11x 10 0 Subtract 10 from both sides (x + 5)(x ) 0 Factor x or x 0 Zero Product Proerty x 5 or x a x + 5, b x The reader should check that both of these solutions satisfy the original equation. It is critical that you see the imortance of setting the exression equal to 0 before factoring. Otherwise, we d get: 6x + 11x 10 x(6x + 11) 10 Factor What we cannot deduce from this equation is that x 10 or 6x or that x and 6x , etc.. (It s wrong and you should feel bad if you do it.) It is recisely because 0 lays such a secial role in the arithmetic of real numbers (as the Additive Identity) that we can assume a factor is 0 when the roduct is 0. No other real number has that ability. We summarize the correct equation solving strategy below. Strategy for Solving Non-linear Equations 1. Put all of the nonzero terms on one side of the equation so that the other side is 0.. Factor.. Use the Zero Product Proerty of Real Numbers and set each factor equal to 0.. Solve each of the resulting equations. Let s finish the section with a collection of examles in which we use this strategy. Examle Solve the following equations. 1. x 5 16x. t 1+t. (y 1) (y 1). w 8w 1 1 w 5. z(z(18z + 9) 50) 5 6. x 8x 90

6 78 Prerequisites Solution. 1. We begin by gathering all of the nonzero terms to one side getting 0 on the other and then we roceed to factor and aly the Zero Product Proerty. x 5 16x x + 16x 5 0 Add 16x, subtract 5 (x 5)(x + 7) 0 Factor x 5 0 or x Zero Product Proerty x 5 or x 7 We check our answers by substituting each of them into the original equation. Plugging in x 5 5 yields on both sides while x 7 gives 17 on both sides.. To solve t 1+t, we first clear fractions then move all of the nonzero terms to one side of the equation, factor and aly the Zero Product Proerty. t 1+t t 1+t Clear fractions (multily by ) 0 1+t t Subtract 0 t t + 1 Rearrange terms 0 (t 1) Factor (Perfect Square Trinomial) At this oint, we get (t 1) (t 1)(t 1) 0, so, the Zero Product Proerty gives us t 1 0 in both cases. 1 Our final answer is t 1, which we invite the reader to check.. Following the strategy outlined above, the first ste to solving (y 1) (y 1) is to gather the nonzero terms on one side of the equation with 0 on the other side and factor. (y 1) (y 1) (y 1) (y 1) 0 Subtract (y 1) (y 1)[(y 1) ] 0 Factor out G.C.F. (y 1)(y ) 0 Simlify y 1 0 or y 0 y 1 or y Both of these answers are easily checked by substituting them into the original equation. An alternative method to solving this equation is to begin by dividing both sides by (y 1) to simlify things outright. As we saw in Examle 0..1, however, whenever we divide by 1 More generally, given a ositive ower, the only solution to X 0 is X 0.

7 0.6 Factoring 79 a variable quantity, we make the exlicit assumtion that this quantity is nonzero. Thus we must stiulate that y (y 1) (y 1) (y 1) (y 1) y 1 y Divide by (y 1) - this assumes (y 1) 6 0 Note that in this aroach, we obtain the y solution, but we lose the y 1 solution. How did that haen? Assuming y is equivalent to assuming y 6 1. This is an issue because y 1 is a solution to the original equation and it was divided out too early. The moral of the story? If you decide to divide by a variable exression, double check that you aren t excluding any solutions. 1. Proceeding as before, we clear fractions, gather the nonzero terms on one side of the equation, have 0 on the other and factor. 8w 1 w 1 w 8w 1 w 1 1 Multily by 1 1 w (8w 1) (w ) Distribute w 8w 1 w + 1 Distribute 0 8w 1 w + 1 w Subtract w 0 8w w w Gather like terms 0 w (8w w ) Factor out G.C.F. w At this oint, we aly the Zero Product Proerty to deduce that w 0 or 8w w 0. From w 0, we get w 0. To solve 8w w 0, we rearrange terms and factor: w +8w (w 1)( w + ) 0. Alying the Zero Product Proerty again, we get w 1 0 (which gives w 1 ), or w + 0 (which gives w ). Our final answers are w 0, w 1 and w. The reader is encouraged to check each of these answers in the original equation. (You need the ractice with fractions!) 5. For our next examle, we begin by subtracting the 5 from both sides then work out the indicated oerations before factoring by grouing. z(z(18z + 9) 50) 5 z(z(18z + 9) 50) 5 0 Subtract 5 z(18z +9z 50) 5 0 Distribute 18z +9z 50z 5 0 Distribute 9z (z + 1) 5(z + 1) 0 Factor (9z 5)(z + 1) 0 Factor 1 You will see other examles throughout this text where dividing by a variable quantity does more harm than good. Kee this basic one in mind as you move on in your studies - it s a good cautionary tale.

8 80 Prerequisites At this oint, we use the Zero Product Proerty and get 9z 5 0 or z The latter 1 gives z whereas the former factors as (z 5)(z + 5) 0. Alying the Zero Product Proerty again gives z 5 0 (so z 5 ) or z (so z 5.) Our final answers are z 1, z 5 and z 5, each of which good fun to check. 6. The nonzero terms of the equation x 8x 9 0 are already on one side of the equation so we roceed to factor. This trinomial doesn t fit the attern of a erfect square so we attemt to reverse the F.O.I.L.ing rocess. With an x term, we have two ossible forms to try: (ax + b)(cx + d) and (ax + b)(cx + d). We leave it to you to show that (ax + b)(cx + d) does not work and we show that (ax + b)(cx + d) does. Since the coefficient of x is 1, we take a c 1. The constant term is 9 so we know b and d have oosite signs and our choices are limited to two otions: either b and d come from ±1 and ±9 OR one is while the other is. After some trial and error, we get x 8x 9(x 9)(x + 1). Hence x 8x 9 0 reduces to (x 9)(x + 1) 0. The Zero Product Proerty tells us that either x 9 0 or x To solve the former, we factor: (x )(x + ) 0, so x 0 (hence, x ) or x + 0 (hence, x ). The equation x has no (real) solution, since for any real number x, x is always 0 or greater. Thus x + 1 is always ositive. Our final answers are x and x. As always, the reader is invited to check both answers in the original equation.

9 0.6 Factoring Exercises In Exercises 1-0, factor comletely over the integers. Check your answer by multilication. 1. x 10x. 1t 5 8t. 16xy 1x y. 5(m + ) (m + ) 5. (x 1)(x + ) (x 1) 6. t (t 5) + t 5 7. w t 9. 81t z 6y 11. (y + ) y 1. (x + h) (x + h) 1. y y t + 10t x 6x + 7x 16. m + 10m x 18. t 6 + t 19. x 5x 1 0. y 1y t + 16t +5. 6x x m m. 7w w 5. m +9m 1m 6. x + x 0 7. (t 1) + (t 1) x 5x 9x t + t t y +5y +9 In Exercises 1-5, find all rational number solutions. Check your answers. 1. (7x + )(x 5) 0. (t 1) (t + ) 0. (y + )(y + y 10) 0. t t 5. y +y 6. 6x 8x x 9x 8. w(6w + 11) w +5w + (w +1) 0. x (x ) 16(x ) 1. (t + 1) (t + 1). a +6 a. 8t t +. x + x x y y (y + ) 6. With hel from your classmates, factor x +8x With hel from your classmates, find an equation which has,, and 117 as solutions. 15 y +5y +9(y +6y + 9) y

10 8 Prerequisites 0.6. Answers 1. x(1 5x). t (t ). xy(y x). (m + ) (m + 7) 5. (x 1)(x 1) 6. (t 5)(t + 1) 7. (w 11)(w + 11) 8. (7 t)(7 + t) 9. (t )(t + )(9t + ) 10. (z 8y )(z +8y ) 11. (y )(y + 1) 1. (x + h)(x + h 1)(x + h + 1) 1. (y 1) 1. (5t + 1) 15. x(x ) 16. (m + 5) 17. ( x)(9 + 6x +x ) 18. t (t + 1)(t t + 1) 19. (x 7)(x + ) 0. (y 9)(y ) 1. (t + 1)(t + 5). (x 5)(x ). (7 m)(5 + m). ( w + 1)(w ) 5. m(m 1)(m + ) 6. (x )(x + )(x + 5) 7. (t )(t + )(t + 1) 8. (x )(x + )(x 5) 9. (t )(1 t)(1 + t) 0. (y y + )(y + y + ) 1. x 7 or x 5. t 1 or t. y 5 or y. t 0 or t 5. y 1 or y 6. x or x 7 7. x 0 or x ± 8. w 5 or w 9. w 5 or w 1 0. x or x ± 1. t 1, t 1, or t 0. a ±1. t or t. x 5. y ±

11 0.7 Quadratic Equations Quadratic Equations In Section 0.6.1, we reviewed how to solve basic non-linear equations by factoring. The astute reader should have noticed that all of the equations in that section were carefully constructed so that the olynomials could be factored using the integers. To demonstrate just how contrived the equations had to be, we can solve x +5x 0 by factoring, (x 1)(x + ) 0, from which we obtain x 1 and x. If we change the 5 to a 6 and try to solve x +6x 0, however, we find that this olynomial doesn t factor over the integers and we are stuck. It turns out that there are two real number solutions to this equation, but they are irrational numbers, and our aim in this section is to review the techniques which allow us to find these solutions. 1 In this section, we focus our attention on quadratic equations. Definition An equation is said to be quadratic in a variable X if it can be written in the form AX + BX + C 0 where A, B and C are exressions which do not involve X and A 6 0. Think of quadratic equations as equations that are one degree u from linear equations - instead of the highest ower of X being just X X 1, it s X. The simlest class of quadratic equations to solve are the ones in which B 0. In that case, we have the following. Solving Quadratic Equations by Extracting Square Roots If c is a real number with c 0, the solutions to X c are X ± c. Note: If c < 0, X c has no real number solutions. There are a coule different ways to see why Extracting Square Roots works, both of which are demonstrated by solving the equation x. If we follow the rocedure outlined in the revious section, we subtract from both sides to get x 0 and we now try to factor x. As mentioned in the remarks following Definition 0.1, we could think of x x ( ) and aly the Difference of Squares formula to factor x (x )(x + ). We solve (x )(x + ) 0 by using the Zero Product Proerty as before by setting each factor equal to zero: x 0 and x + 0. We get the answers x ±. In general, if c 0, then c is a real number, so x c x ( c) (x c)(x + c). Relacing the with c in the above discussion gives the general result. Another way to view this result is to visualize taking the square root of both sides: since x c, x c. How do we simlify x? We have to exercise a bit of caution here. Note that (5) and ( 5) both simlify to 5 5. In both cases, x returned a ositive number, since the negative in 5 was squared away before we took the square root. In other words, x is x if x is ositive, or, if x is negative, we make x ositive - that is, x x, the absolute value of x. So from x, we take the square root of both sides of the equation to get x. This simlifies to x, which by Theorem 0. is equivalent to x or x. Relacing the in the revious argument with c, gives the general result. 1 While our discussion in this section dearts from factoring, we ll see in Chater that the same corresondence between factoring and solving equations holds whether or not the olynomial factors over the integers.

12 8 Prerequisites As you might exect, Extracting Square Roots can be alied to more comlicated equations. Consider the equation below. We can solve it by Extracting Square Roots rovided we first isolate the erfect square quantity: x x + x + 15 x + ± r 15 Add 15 Divide by Extract Square Roots x + 15 ± 15 x ± ± 15 x Proerty of Radicals Subtract Add fractions Let s return to the equation x +6x 0 from the beginning of the section. We leave it to the reader to show that x + 15 x +6x. (Hint: Exand the left side.) In other words, we can solve x +6x 0bytransforming into an equivalent equation. This rocess, you may recall, is called Comleting the Square. We ll revisit Comleting the Square in Section. in more generality and for a different urose but for now we revisit the stes needed to comlete the square to solve a quadratic equation. Solving Quadratic Equations: Comleting the Square To solve a quadratic equation AX + BX + C 0 by Comleting the Square: 1. Subtract the constant C from both sides.. Divide both sides by A, the coefficient of X. (Remember: A 6 0.). Add B A to both sides of the equation. (That s half the coefficient of X, squared.).. Factor the left hand side of the equation as X + A B 5. Extract Square Roots. B 6. Subtract A from both sides.

13 0.7 Quadratic Equations 85 To refresh our memories, we aly this method to solve x x + 5 0: x x +5 0 x x 5 Subtract C 5 x 5 8x Divide by A x 5 8x Add B A ( ) 16 (x ) r x ± r x ± Factor: Perfect Square Trinomial Extract Square Roots Add At this oint, we use roerties of fractions and radicals to rationalize the denominator: r r We can now get a common (integer) denominator which yields: r 19 x ± ± 1 ± 19 The key to Comleting the Square is that the rocedure always roduces a erfect square trinomial. To see why this works every single time, we start with AX + BX + C 0 and follow the rocedure: AX + BX + C 0 AX + BX C Subtract C X + BX A X + BX A + B A C A C A + B A Divide by A 6 0 B Add A (Hold onto the line above for a moment.) Here s the heart of the method - we need to show that X + BX B A A + X + B A To show this, we start with the right side of the equation and aly the Perfect Square Formula from Theorem 0.7 X + A B B B X + X + X + BX B A A A + X A Recall that this means we want to get a denominator with rational (more secifically, integer) numbers.

14 86 Prerequisites With just a few more stes we can solve the general equation AX + BX + C 0 so let s ick u the story where we left off. (The line on the revious age we told you to hold on to.) X + BX B A A + C B A + A X + A B C A + B A X + A B AC A + B A X + B B AC A A X + B r B A ± AC A X + B B A ± AC A B B X A ± AC A B ± B X AC A Lo and behold, we have derived the legendary Quadratic Formula! Factor: Perfect Square Trinomial Get a common denominator Add fractions Extract Square Roots Proerties of Radicals Subtract B A Add fractions. Theorem 0.9. Quadratic Formula: The solution to AX + BX + C 0 with A 6 0 is: X B ± B AC A We can check our earlier solutions to x +6x 0 and x x using the Quadratic Formula. For x +6x 0, we identify A, B 6 and C. The quadratic formula gives: x 6 ± 6 ()( ) () 6 ± ± 60 Using roerties of radicals ( 60 15), this reduces to ( ± 15) ± 15. We leave it to the reader to show these two answers are the same as ± 15, as required. For x x + 5 0, we identify A, B and C 5. Here, we get: x ( ) ± ( ) ()(5) () ± Since , this reduces to x 1± 19. Think about what ( ± 15) is really telling you.

15 0.7 Quadratic Equations 87 It is worth noting that the Quadratic Formula alies to all quadratic equations - even ones we could solve using other techniques. For examle, to solve x +5x 0 we identify A, B 5 and C. This yields: x 5 ± 5 ()( ) () 5 ± 9 5 ± 7 At this oint, we have x and x the same two answers we obtained factoring. We can also use it to solve x, if we wanted to. From x 0, we have A 1, B 0 and C. The Quadratic Formula roduces x 0 ± 0 (1)() (1) ± 1 ± ± As this last examle illustrates, while the Quadratic Formula can be used to solve every quadratic equation, that doesn t mean it should be used. Many times other methods are more efficient. We now rovide a more comrehensive aroach to solving Quadratic Equations. Strategies for Solving Quadratic Equations If the variable aears in the squared term only, isolate it and Extract Square Roots. Otherwise, ut the nonzero terms on one side of the equation so that the other side is 0. Try factoring. If the exression doesn t factor easily, use the Quadratic Formula. The reader is encouraged to ause for a moment to think about why Comleting the Square doesn t aear in our list of strategies desite the fact that we ve sent the majority of the section so far talking about it. Let s get some ractice solving quadratic equations, shall we? Examle Find all real number solutions to the following equations. 1. (w 1) 0. 5x x(x ) 7. (y 1) y +. 5(5 1x) 59 5x 5..9t + 10t x x 6 Solution. 1. Since (w 1) 0 contains a erfect square, we isolate it first then extract square roots: (w 1) 0 (w 1) Add (w 1) ± w 1 Extract Square Roots 1 ± w Add 1 1 ± w Divide by Unaccetable answers include Jeff and Carl are mean and It was one of Carl s Pedantic Rants.

16 88 Prerequisites We find our two answers w 1±. The reader is encouraged to check both answers by substituting each into the original equation. 5. To solve 5x x(x ) 7, we begin erforming the indicated oerations and getting one side equal to 0. 5x x(x ) 7 5x x +x 7 Distribute x +8x 7 Gather like terms x +8x 7 0 Subtract 7 At this oint, we attemt to factor and find x +8x 7(x 1)( x + 7). Using the Zero Product Proerty, we get x 1 0 or x Our answers are x 1 or x 7, both of which are easy to check.. Even though we have a erfect square in (y 1), Extracting Square Roots won t hel matters since we have a y on the other side of the equation. Our strategy here is to erform the indicated oerations (and clear the fraction for good measure) and get 0 on one side of the equation. (y 1) y + y y + y +1 Perfect Square Trinomial (y y + y + 1) Multily by y + y 6y + 6 Distribute y 6y + 6 (y + ) y 6y + 6+(y + ) 0 Subtract 6, Add (y + ) y 5y 1 0 A cursory attemt at factoring bears no fruit, so we run this through the Quadratic Formula with A, B 5 and C 1. y+ ( 5) ± ( 5) y ()( 1) () y 5 ± y 5 ± 7 6 Since 7 is rime, we have no way to reduce 7. Thus, our final answers are y 5± 7 6. The reader is encouraged to suly the details of the challenging verification of the answers. 5 It s excellent ractice working with radicals fractions so we really, really want you to take the time to do it.

17 0.7 Quadratic Equations 89. We roceed as before; our aim is to gather the nonzero terms on one side of the equation. 5(5 1x) 59 5x x 59 5x 59 (15 105x) 5x Distribute Multily by 500 0x x Distribute 500 0x x 0 Subtract 59, Add 100x 100x 0x Gather like terms With highly comosite numbers like 100 and 1, factoring seems inefficient at best, 6 so we aly the Quadratic Formula with A 100, B 0 and C 1: x ( 0) ± ( 0) (100)(1) (100) 0 ± ± ± To our surrise and delight we obtain just one answer, x Our next equation.9t + 10t + 0, already has 0 on one side of the equation, but with coefficients like.9 and 10, factoring with integers is not an otion. We could make things a bit easier on the eyes by clearing the decimal (by multilying through by 10) to get 9t + 100t but we simly cannot rid ourselves of the irrational number. The Quadratic Formula is our only recourse. With A 9, B 100 and C 0 we get: 6 This is actually the Perfect Square Trinomial (10x 1).

18 90 Prerequisites t 100 q ± (100 ) ( 9)(0) ( 9) 100 ± ± ± ( 50 ± 50) ( 9) 50 ± 50 9 ( 50 ± 50) Reduce Proerties of Negatives Distribute You ll note that when we distributed the negative in the last ste, we changed the ± to a. While this is technically correct, at the end of the day both symbols mean lus or minus, 7 so we can write our answers as t 50 ± Checking these answers are a true test of arithmetic mettle. 6. At first glance, the equation x x 6 seems mislaced. The highest ower of the variable x here is, not, so this equation isn t a quadratic equation - at least not in terms of the variable x. It is, however, an examle of an equation that is quadratic in disguise. 8 We introduce a new variable u to hel us see the attern - secifically we let u x. Thus u (x ) x. So in terms of the variable u, the equation x x 6 is u u 6. The latter is a quadratic equation, which we can solve using the usual techniques: u u 6 0 u u 6 Subtract u After a few attemts at factoring, we resort to the Quadratic Formula with A, B, 7 There are instances where we need both symbols, however. For examle, the Sum and Difference of Cubes Formulas (age 71) can be written as a single formula: a ± b (a ± b)(a ab + b ). In this case, all of the to symbols are read to give the sum formula; the bottom symbols give the difference formula. 8 More formally, quadratic in form. Carl likes Quadratics in Disguise since it reminds him of the tagline of one of his beloved childhood cartoons and toy lines.

19 0.7 Quadratic Equations 91 C 6 and get: u ( ) ± ( ) ()( 6) () ± ± 76 6 ± 19 6 ± 19 6 (1 ± 19) () 1 ± 19 Proerties of Radicals Factor Reduce We ve solved the equation for u, but what we still need to solve the original equation 9 - which means we need to find the corresonding values of x. Since u x, we have two equations: x or x 1 19 q We can solve the first equation by extracting square roots to get x ±. The second equation, however, has no real number solutions because 1 19 is a negative number. For our final answers we can rationalize the denominator 10 to get: x ± s ± s ± + 19 As with the revious exercise, the very challenging check is left to the reader. Our last examle above, the Quadratic in Disguise, hints that the Quadratic Formula is alicable to a wider class of equations than those which are strictly quadratic. We give some general guidelines to recognizing these beasts in the wild on the next age. 9 Or, you ve solved the equation for you (u), now you have to solve it for your instructor (x). 10 We ll say more about this technique in Section 0.9.

20 9 Prerequisites Identifying Quadratics in Disguise An equation is a Quadratic in Disguise if it can be written in the form: AX m + BX m + C 0. In other words: There are exactly three terms, two with variables and one constant term. The exonent on the variable in one term is exactly twice the variable on the other term. To transform a Quadratic in Disguise to a quadratic equation, let u X m so u (X m ) X m. This transforms the equation into Au + Bu + C 0. For examle, x 6 x is a Quadratic in Disguise, since 6. If we let u x, we get u (x ) x 6, so the equation becomes u u However, x 6 x is not a Quadratic in Disguise, since 6 6. The substitution u x yields u (x ) x, not x 6 as required. We ll see more instances of Quadratics in Disguise in later sections. We close this section with a review of the discriminant of a quadratic equation as defined below. Definition The Discriminant: Given a quadratic equation AX + BX + C 0, the quantity B AC is called the discriminant of the equation. The discriminant is the radicand of the square root in the quadratic formula: X B ± B AC A It discriminates between the nature and number of solutions we get from a quadratic equation. The results are summarized below. Theorem Discriminant Theorem: Given a Quadratic Equation AX + BX + C 0, let D B AC be the discriminant. If D > 0, there are two distinct real number solutions to the equation. If D 0, there is one reeated real number solution. Note: Reeated here comes from the fact that both solutions B±0 A reduce to B A. If D < 0, there are no real solutions. For examle, x + x 1 0 has two real number solutions since the discriminant works out to be (1) (1)( 1) 5 > 0. This results in a ± 5 in the Quadratic Formula, generating two different answers. On the other hand, x + x has no real solutions since here, the discriminant is (1) (1)(1) < 0 which generates a ± in the Quadratic Formula. The equation x +x has discriminant () (1)(1) 0 so in the Quadratic Formula we get a ± 00 thereby generating just one solution. More can be said as well. For examle, the discriminant of 6x x 0 0 is 961. This is a erfect square, 961 1, which means our solutions are

21 0.7 Quadratic Equations 9 rational numbers. When our solutions are rational numbers, the quadratic actually factors nicely. In our examle 6x x 0 (x + 5)(x 8). Admittedly, if you ve already comuted the discriminant, you re most of the way done with the roblem and robably wouldn t take the time to exeriment with factoring the quadratic at this oint but we ll see another use for this analysis of the discriminant in the next section Secifically in Examle

22 9 Prerequisites Exercises In Exercises 1-1, find all real solutions. Check your answers, as directed by your instructor. 1. x (5t + ). (y ) 10. x + x w w 6. y(y + ) 1 7. z z v + 0.v x x t (t + 1) 11. (x ) x (y 1)(y + 1) 5y 1. w +w x + x 15. ( y) ( y) x +6x 15x v 7v v y 8y 18y 1 0. x x v +1 In Exercises - 7, find all real solutions and use a calculator to aroximate your answers, rounded to two decimal laces b 6. r r + r 5..9t + 100t x 1.65( x) 7. (0.5+A) 0.7(0.1 A) v In Exercises 8-0, use Theorem 0. along with the techniques in this section to find all real solutions to the following. 8. x x 9. x x x 1 0. x x + x 1. Prove that for every nonzero number, x + x + 0 has no real solutions.. Solve for t: 1 gt + vt + h 0. Assume g > 0, v 0 and h 0.

23 0.7 Quadratic Equations Answers 1. x ± 5 6. x 1 ± 5 7. z 1 ± t 5 ± r 1 1. w ±. t 5, 5 5. w 1,. y ±1, ± 5 6. y ± 5 8. v, 1 9. No real solution. 11. x 0 1. y ± x ±1 15. y ± x 0, 5 ± , ± 18. v 0, ±, ± y 5 ± 6 ± x 1. v, r b ± ±.0. r ± 50 ±.. r ± x 99 ± x 1,, ± 17, r 6.,.7 5. t 500 ± 10 91, t 5.68, 1.7 9, x 1.69, A 107 ± 7 70, A 0.50, x ±1, ± 0. x 1, 1, 7 1. The discriminant is: D < 0. Since D < 0, there are no real solutions.. t v ± v +gh g

24 16 Prerequisites 0.10 Comlex Numbers We conclude our Prerequisites chater with a review the set of Comlex Numbers. As you may recall, the comlex numbers fill an algebraic ga left by the real numbers. There is no real number x with x 1, since for any real number x 0. However, we could formally extract square roots and write x ± 1. We build the comlex numbers by relabeling the quantity 1 as i, the unfortunately misnamed imaginary unit. 1 The number i, while not a real number, is defined so that it lays along well with real numbers and acts very much like any other radical exression. For instance, (i) 6i, 7i i i, ( 7i) + ( + i) 5 i, and so forth. The key roerties which distinguish i from the real numbers are listed below. Definition The imaginary unit i satisfies the two following roerties: 1. i 1. If c is a real number with c 0 then c i c Proerty 1 in Definition 0.18 establishes that i does act as a square root of 1, and roerty establishes what we mean by the rincial square root of a negative real number. In roerty, it is imortant to remember the restriction on c. For examle, it is erfectly accetable to say i i() i. However, ( ) 6 i, otherwise, we d get ( ) i i(i) i ( 1), which is unaccetable. The moral of this story is that the general roerties of radicals do not aly for even roots of negative quantities. With Definition 0.18 in lace, we are now in osition to define the comlex numbers. Definition A comlex number is a number of the form a + bi, where a and b are real numbers and i is the imaginary unit. The set of comlex numbers is denoted C. Comlex numbers include things you d normally exect, like + i and 5 i. However, don t forget that a or b could be zero, which means numbers like i and 6 are also comlex numbers. In other words, don t forget that the comlex numbers include the real numbers, so 0 and 1 are both considered comlex numbers. The arithmetic of comlex numbers is as you would exect. The only things you need to remember are the two roerties in Definition The next examle should hel recall how these animals behave. 1 Some Technical Mathematics textbooks label it j. While it carries the adjective imaginary, these numbers have essential real-world imlications. For examle, every electronic device owes its existence to the study of imaginary numbers. Note the use of the indefinite article a. Whatever beast is chosen to be i, i is the other square root of 1. To use the language of Section 0.1., R C.

25 0.10 Comlex Numbers 17 Examle Perform the indicated oerations. 1. (1 i) ( + i). (1 i)( + i). 1 i i ( )( 1) 6. (x [1 + i])(x [1 i]) Solution. 1. As mentioned earlier, we treat exressions involving i as we would any other radical. We distribute and combine like terms: (1 i) ( + i) 1 i i Distribute 6i Gather like terms Technically, we d have to rewrite our answer 6i as ( ) + ( 6)i to be (in the strictest sense) in the form a + bi. That being said, even edants have their limits, and we ll consider 6i good enough.. Using the Distributive Proerty (a.k.a. F.O.I.L.), we get (1 i)( + i) (1)() + (1)(i) (i)() (i)(i) F.O.I.L. +i 6i 8i i 8( 1) i 1 i i. How in the world are we suosed to simlify 1 i i? Well, we deal with the denominator i as we would any other denominator containing two terms, one of which is a square root: we and multily both numerator and denominator by + i, the (comlex) conjugate of i. Doing so roduces 1 i i (1 i)( + i) ( i)( + i) Equivalent Fractions +i 6i 8i 9 16i F.O.I.L. i 8( 1) 9 16( 1) i 1 11 i i. We use roerty of Definition 0.18 first, then aly the rules of radicals alicable to real numbers to get 1 i i 1 i

26 18 Prerequisites 5. We adhere to the order of oerations here and erform the multilication before the radical to get ( )( 1) We can brute force multily using the distributive roerty and see that (x [1 + i])(x [1 i]) x x[1 i] x[1 + i] + [1 i][1 + i] F.O.I.L. x x +ix x ix +1 i +i i Distribute x x +1 ( 1) Gather like terms x x +5 i 1 This tye of factoring will be revisited in Section.. In the revious examle, we used the conjugate idea from Section 0.9 to divide two comlex numbers. More generally, the comlex conjugate of a comlex number a + bi is the number a bi. The notation commonly used for comlex conjugation is a bar : a + bi a bi. For examle, +i i and i +i. To find 6, we note that 66+0i 6 0i 6, so 6 6. Similarly, i i, since i 0+i 0 i i. Note that , not 5, since i + 5 0i + 5. Here, the conjugation secified by the bar notation involves reversing the sign before i 1, not before 5. The roerties of the conjugate are summarized in the following theorem. Theorem 0.1. Proerties of the Comlex Conjugate: Let z and w be comlex numbers. z z z + w z + w zw z w z n (z) n, for any natural number n z is a real number if and only if z z. Essentially, Theorem 0.1 says that comlex conjugation works well with addition, multilication and owers. The roofs of these roerties can best be achieved by writing out z a + bi and w c + di for real numbers a, b, c and d. Next, we comute the left and right sides of each equation and verify that they are the same. The roof of the first roerty is a very quick exercise. To rove the second roerty, we comare z + w with z + w. We have z + w a + bi + c + di a bi + c di. To find z + w, we first comute so z + w (a + bi)+(c + di) (a + c)+(b + d)i z + w (a + c)+(b + d)i (a + c) (b + d)i a + c bi di a bi + c di z + w Trust us on this.

27 0.10 Comlex Numbers 19 As such, we have established z + w z + w. The roof for multilication works similarly. The roof that the conjugate works well with owers can be viewed as a reeated alication of the roduct rule, and is best roved using a technique called Mathematical Induction. 5 The last roerty is a characterization of real numbers. If z is real, then z a +0i, so z a 0i a z. On the other hand, if z z, then a + bi a bi which means b b so b 0. Hence, z a +0i a and is real. We now return to the business of solving quadratic equations. Consider x x The discriminant b ac 16 is negative, so we know by Theorem 0.10 there are no real solutions, since the Quadratic Formula would involve the term 16. Comlex numbers, however, are built just for such situations, so we can go ahead and aly the Quadratic Formula to get: x ( ) ± ( ) (1)(5) (1) ± 16 ± i 1± i. Examle Find the comlex solutions to the following equations Solution. x x +1 x +. t 9t +5. z Clearing fractions yields a quadratic equation so we then roceed as in Section 0.7. x x +1 x + x (x + )(x + 1) Multily by (x + 1) to clear denominators x x + x +x + F.O.I.L. x x +x + Gather like terms 0 x +x + Subtract x From here, we aly the Quadratic Formula ± x (1)() (1) ± 8 ± i 8 ± i ( 1 ± i ) 1 ± i Quadratic Formula Simlify Definition of i Product Rule for Radicals Factor and reduce 5 See Section Remember, all real numbers are comlex numbers, so comlex solutions means both real and non-real answers.

28 10 Prerequisites We get two answers: x 1+i and its conjugate x 1 i. Checking both of these answers reviews all of the salient oints about comlex number arithmetic and is therefore strongly encouraged.. Since we have three terms, and the exonent on one term ( on t ) is exactly twice the exonent on the other ( on t ), we have a Quadratic in Disguise. We roceed accordingly. t 9t +5 t 9t 5 0 Subtract 9t and 5 (t + 1)(t 5) 0 Factor t or t 5 Zero Product Proerty From t we get t 1, or t 1. We extract square roots as follows: r r t ± ±i ±i 1 ±i ± i, where we have rationalized the denominator er convention. From t 5, we get t ± 5. In total, we have four comlex solutions - two real: t ± 5 and two non-real: t ± i.. To find the real solutions to z + 1 0, we can subtract the 1 from both sides and extract cube roots: z 1, so z 1 1. It turns out there are two more non-real comlex number solutions to this equation. To get at these, we factor: z +1 0 (z + 1)(z z + 1) 0 Factor (Sum of Two Cubes) z or z z +10 From z + 1 0, we get our real solution z 1. From z z + 1 0, we aly the Quadratic Formula to get: z ( 1) ± ( 1) (1)(1) (1) 1 ± 1 ± i Thus we get three solutions to z one real: z 1 and two non-real: z 1±i. As always, the reader is encouraged to test their algebraic mettle and check these solutions. It is no coincidence that the non-real solutions to the equations in Examle aear in comlex conjugate airs. Any time we use the Quadratic Formula to solve an equation with real coefficients, the answers will form a comlex conjugate air owing to the ± in the Quadratic Formula. This leads us to a generalization of Theorem 0.10 which we state on the next age.

29 0.10 Comlex Numbers 11 Theorem 0.1. Discriminant Theorem: Given a Quadratic Equation AX + BX + C 0, where A, B and C are real numbers, let D B AC be the discriminant. If D > 0, there are two distinct real number solutions to the equation. If D 0, there is one (reeated) real number solution. Note: Reeated here comes from the fact that both solutions B±0 A reduce to B A. If D < 0, there are two non-real solutions which form a comlex conjugate air. We will have much more to say about comlex solutions to equations in Section. and we will revisit Theorem 0.1 then.

30 1 Prerequisites Exercises In Exercises 1-10, use the given comlex numbers z and w to find and simlify the following. z + w zw z 1 z z w z zz (z) 1. z +i, w i. z 1+i, w i. z i, w 1+i. z i, w i 5. z 5i, w +7i 6. z 5+i, w +i 7. z i, w +i 8. z 1 i, w 1 i w z 9. z 1 + i, w 1 + i 10. z + i, w i In Exercises 11-18, simlify the quantity ( 5)( ) ( 9)( 16) 17. ( 9) 18. ( 9) We know that i 1 which means i i i ( 1) i i and i i i ( 1)( 1) 1. In Exercises 19-6, use this information to simlify the given ower of i. 19. i 5 0. i 6 1. i 7. i 8. i 15. i 6 5. i i 0 In Exercises 7-5, find all comlex solutions. 7. x +6x 8. 15t +t +5t(t + 1) 9. y +y 0. 1 w w 1. y y y. x x 1 x. x 5 x. 6. Multily and simlify: x [ i ] 5y +1 y 1 y 5. z 16 x [ + i ]

31 0.10 Comlex Numbers Answers 1. For z +i and w i z + w +7i zw 1 + 8i z 5 + 1i 1 z 1 1 i z w 1 i w z i z i zz 1 (z) 5 1i. For z 1+i and w i z + w 1 zw 1 i z i 1 z 1 1 i z w 1+i w z 1 1 i z 1 i zz (z) i. For z i and w 1+i z + w 1+i zw i z 1 1 z i z w i w z +i z i zz 1 (z) 1. For z i and w i z + w +i zw 8+8i z 16 1 z 1 i z w 1+i w z 1 1 i z i zz 16 (z) For z 5i and w +7i z + w 5+i zw i z 16 0i 1 z + 5 i z w i w z i z +5i zz (z) i

32 1 Prerequisites 6. For z 5+i and w +i z + w 1+i zw 6i z 10i 1 z i z w i w z i z 5 i zz 6 (z) + 10i 7. For z i and w +i z + w zw z i 1 z + i z w i w z i z +i zz (z) i 8. For z 1 i and w 1 i z + w i zw z i 1 z 1 + i z w 1 + i w z 1 i z 1+i zz (z) +i 9. For z 1 + i and w 1 + i z + w i zw 1 z 1 + i 1 z 1 i z w 1 i w z 1 + i z 1 i zz 1 (z) 1 i 10. For z + i and w i z + w zw 1 z i 1 z i z w i w z i z i zz 1 (z) i 11. 7i 1. i

33 0.10 Comlex Numbers i 19. i 5 i i 1 i i 0. i 6 i i 1 ( 1) 1 1. i 7 i i 1 ( i) i. i 8 i i i (1) 1. i 15 i i 1 ( i) i. i 6 i 6 i 1 ( 1) 1 5. i 117 i 9 i 1 i i 6. i 0 i x ± i 1 0. w 1 ± i 7 5 ± i. x 6. x 6x + 8. t 5, ± i 1. y ± i. y ±i, ± i 9. y ±, ±i. x 0, 1 ± i 5. z ±, ±i