1 00:00:00,000 --> 00:00:00,550 2 00:00:00,550 --> 00:00:02,950 I've been sent is pretty interesting algebra problem, 3 00:00:02,950 --> 00:00:06,759 and I think the question isn't properly formed, or it might be 4 00:00:06,759 --> 00:00:08,400 missing some piece of information. 5 00:00:08,400 --> 00:00:10,050 But it's an interesting question nonetheless so I 6 00:00:10,050 --> 00:00:12,339 thought I'd work it out, and we can talk about why 7 00:00:12,339 --> 00:00:14,150 it's not properly formed. 8 00:00:14,150 --> 00:00:16,519 So they give us a function, f of x. 9 00:00:16,519 --> 00:00:18,910 They say it's a third degree polynomial of the form 10 00:00:18,910 --> 00:00:28,019 ax to the third plus bx squared plus cx plus d. 11 00:00:28,019 --> 00:00:32,280 And they tell us a couple of the 0's of this polynomial. 12 00:00:32,280 --> 00:00:40,329 They say that we have 0's at the point minus 1, 0 and 13 00:00:40,329 --> 00:00:42,149 the point 2 comma 0. 14 00:00:42,149 --> 00:00:46,359 And they tell us we have a y-intercept, y-intercept, 15 00:00:46,359 --> 00:00:50,759 at the point 0, minus 2. 16 00:00:50,759 --> 00:00:58,320 And what they ask us is, what is a plus b plus c plus d? 17 00:00:58,320 --> 00:01:01,329 So let's see if we can make some headway on this. 18 00:01:01,329 --> 00:01:03,829 So the first thing is to think about, is what would a third 19 00:01:03,829 --> 00:01:06,010 degree polynomial look like and what are we even talking 20 00:01:06,010 --> 00:01:07,070 about when we say 0's. 21 00:01:07,069 --> 00:01:09,429 So let me just draw a little bit of a graph. 22 00:01:09,430 --> 00:01:12,280 And I don't know exactly what this third degree polynomial 23 00:01:12,280 --> 00:01:14,150 looks like just yet. 24 00:01:14,150 --> 00:01:16,840 Let me draw my axis. 25 00:01:16,840 --> 00:01:20,880 Now a third degree polynomial can have as many as three 0's. 26 00:01:20,879 --> 00:01:23,349 And 0's are just the points where the 27 00:01:23,349 --> 00:01:24,750 function is equal to 0. 28 00:01:24,750 --> 00:01:27,730 So, we have a 0 at minus 1, 0. 29 00:01:27,730 --> 00:01:30,320 So that's minus 1, 0, that's right there. 30 00:01:30,319 --> 00:01:32,639 We have a zero 2, 0. 31 00:01:32,640 --> 00:01:34,109 1, 2, 0. 32 00:01:34,109 --> 00:01:36,530 So those are the two 0's that they've given us. 33 00:01:36,530 --> 00:01:40,290 They also told us the y-intercept at 0, minus 2. 34 00:01:40,290 --> 00:01:42,720 So it intersects the y-axis right there. 35 00:01:42,719 --> 00:01:46,200 But this guy can have as many as three 0's. 36 00:01:46,200 --> 00:01:49,269 Now some of them might be complex. 37 00:01:49,269 --> 00:01:52,750 But complex 0's come in pairs of two, the two 38 00:01:52,750 --> 00:01:53,980 complex conjugates. 39 00:01:53,980 --> 00:01:55,710 I won't go into too much detail here. 40 00:01:55,709 --> 00:01:59,629 So this one must have a third real 0, because if the third 41 00:01:59,629 --> 00:02:02,060 one was complex, you'd need another complex 0, and you 42 00:02:02,060 --> 00:02:06,920 can't have four 0's for a third degree polynomial. 43 00:02:06,920 --> 00:02:09,759 So that third root, it could be sitting some place out 44 00:02:09,759 --> 00:02:11,459 here, maybe over here. 45 00:02:11,460 --> 00:02:13,909 Or it could even be a repeat of one of these 46 00:02:13,909 --> 00:02:14,849 two roots over here. 47 00:02:14,849 --> 00:02:16,889 But let's just assume that there's some third root, we 48 00:02:16,889 --> 00:02:18,259 don't know what it looks like. 49 00:02:18,259 --> 00:02:20,609 Let's say it looks something like-- let's say that third 50 00:02:20,610 --> 00:02:22,410 root is sitting out here some place. 51 00:02:22,409 --> 00:02:25,549 And then a potential graph, and I could be completely wrong, 52 00:02:25,550 --> 00:02:27,750 I'm just guessing, but just to get a sense of what third 53 00:02:27,750 --> 00:02:30,979 degree polynomials look like, and how can a graph intersect 54 00:02:30,979 --> 00:02:34,159 the x-axis three times, a potential graph might look 55 00:02:34,159 --> 00:02:35,879 something like this. 56 00:02:35,879 --> 00:02:39,305 Curves and bam, hits the other 0, hits the y-intercept, and 57 00:02:39,305 --> 00:02:40,530 then goes back up like that. 58 00:02:40,530 --> 00:02:43,039 It might go like that, it might go the other way, go like that. 59 00:02:43,039 --> 00:02:45,120 Up, and then down like that. 60 00:02:45,120 --> 00:02:47,920 Well, there's many ways you could draw something that 61 00:02:47,919 --> 00:02:51,579 essentially curves twice to intersect these three points 62 00:02:51,580 --> 00:02:54,870 and also that y-intercept, but I'm not going to go into 63 00:02:54,870 --> 00:02:55,719 all of those right now. 64 00:02:55,719 --> 00:02:59,189 But let's see if we can figure out the coefficients here. 65 00:02:59,189 --> 00:03:01,590 So the key point is me telling you that there 66 00:03:01,590 --> 00:03:03,250 must be a third 0 here. 67 00:03:03,250 --> 00:03:06,599 Let's call that third 0, it's at the point, it's at the 68 00:03:06,599 --> 00:03:10,469 point-- let's call it --r3 comma 0. 69 00:03:10,469 --> 00:03:13,389 And I'm using the letter r for roots. 70 00:03:13,389 --> 00:03:16,889 Roots are the x values of the 0's. 71 00:03:16,889 --> 00:03:23,949 So if we say f of minus 1 is equal to 0, we say that x is 72 00:03:23,949 --> 00:03:26,719 equal to minus 1 is a root. 73 00:03:26,719 --> 00:03:32,359 Similarly, we know that f of 2, f of 2, is equal to 0. 74 00:03:32,360 --> 00:03:36,780 Or we could say x equals 2 is a root, these are roots. 75 00:03:36,780 --> 00:03:40,159 So when someone tells you a 0 they're also essentially 76 00:03:40,159 --> 00:03:43,359 telling you the roots, and then we know there must be a third 77 00:03:43,360 --> 00:03:46,300 root at x is equal to r3. 78 00:03:46,300 --> 00:03:49,340 So x is equal to r3. 79 00:03:49,340 --> 00:03:53,979 Now, if we have a third degree polynomial where these three x 80 00:03:53,979 --> 00:03:57,569 values make it 0, we can rewrite this third degree 81 00:03:57,569 --> 00:04:01,269 polynomial as-- we can rewrite it as-- I'll do it in a 82 00:04:01,270 --> 00:04:07,820 slightly different color --f of x is equal to x plus 1-- you'll 83 00:04:07,819 --> 00:04:10,299 see why I'm doing x plus 1 instead of x minus 1 in a 84 00:04:10,300 --> 00:04:20,230 second --plus 1 times x minus 2 times x minus r3. 85 00:04:20,230 --> 00:04:24,015 And we want to put an a out front, and I'll tell you in 86 00:04:24,014 --> 00:04:26,550 a second why I'm putting this a out front. 87 00:04:26,550 --> 00:04:30,400 Because if you were to multiply just the x terms here, when 88 00:04:30,399 --> 00:04:33,370 actually do your distributed property and multiply out these 89 00:04:33,370 --> 00:04:36,259 binomials, you'll just get x to the third, but we had 90 00:04:36,259 --> 00:04:37,389 an ax to the third. 91 00:04:37,389 --> 00:04:40,360 So you just needed a constant a there to make this 92 00:04:40,360 --> 00:04:42,100 an ax to the third. 93 00:04:42,100 --> 00:04:44,490 Now, why did I write out like this? 94 00:04:44,490 --> 00:04:47,519 Well, if you put a minus 1 in here what's going to happen? 95 00:04:47,519 --> 00:04:49,659 This term is going to be minus 1 plus 1. 96 00:04:49,660 --> 00:04:50,850 It's going to be 0. 97 00:04:50,850 --> 00:04:53,210 Who cares what these are or what a is, the whole 98 00:04:53,209 --> 00:04:54,379 thing's going to be 0. 99 00:04:54,379 --> 00:04:57,170 If we put x is equal to 2, this term right 100 00:04:57,170 --> 00:04:58,610 here is going to be 0. 101 00:04:58,610 --> 00:05:01,189 Who cares what the other terms or the a is, f 102 00:05:01,189 --> 00:05:03,069 of 2 is going to be 0. 103 00:05:03,069 --> 00:05:06,360 Similarly, when x is equal to r3, this last term is 0, who 104 00:05:06,360 --> 00:05:08,170 cares about anything over here. 105 00:05:08,170 --> 00:05:11,550 So we know that this guy up here can be rewritten 106 00:05:11,550 --> 00:05:13,520 in this form over here. 107 00:05:13,519 --> 00:05:17,490 And so if we can solve for r3 we can just multiply this thing 108 00:05:17,490 --> 00:05:20,370 out and try to do a little bit of pattern matching to figure 109 00:05:20,370 --> 00:05:23,759 out what these coefficients are going to be. 110 00:05:23,759 --> 00:05:28,099 Now, the other big clue they gave us is a y-intercept. 111 00:05:28,100 --> 00:05:31,640 That point right there, the point 0, minus 2. 112 00:05:31,639 --> 00:05:36,159 They're telling us that f of 0 is equal to minus 2. 113 00:05:36,160 --> 00:05:37,939 Well what's f of 0? 114 00:05:37,939 --> 00:05:42,730 f of 0-- I'll write it here --f of 0-- If you put 0 in here, 115 00:05:42,730 --> 00:05:46,420 this term becomes 0, this term becomes 0, that term becomes 0, 116 00:05:46,420 --> 00:05:47,840 you are just left with the d. 117 00:05:47,839 --> 00:05:49,939 So f of 0 is equal to d. 118 00:05:49,939 --> 00:05:53,759 Or we could say that d must be equal to minus 2. 119 00:05:53,759 --> 00:05:57,529 d is equal to minus 2, so we solved at least one 120 00:05:57,529 --> 00:05:59,089 of the call efficient. 121 00:05:59,089 --> 00:06:02,449 The constant term. d is equal to minus 2. 122 00:06:02,449 --> 00:06:05,969 Now, if d is equal to minus 2, that means all of these 123 00:06:05,970 --> 00:06:08,740 constant terms, when you were to multiply out this 124 00:06:08,740 --> 00:06:12,340 expression, and I would include the a, because the a is scaling 125 00:06:12,339 --> 00:06:15,619 everything, that must be equal to minus 2. 126 00:06:15,620 --> 00:06:16,730 Let me be clear here. 127 00:06:16,730 --> 00:06:19,910 So let me rewrite this expression. 128 00:06:19,910 --> 00:06:22,970 This is equal to, I'll just multiply this a times 129 00:06:22,970 --> 00:06:23,740 this last term. 130 00:06:23,740 --> 00:06:27,400 We do it in any order you want, so this is the same thing as x 131 00:06:27,399 --> 00:06:36,449 plus 1 times x minus 2 to times a a minus r, let me 132 00:06:36,449 --> 00:06:40,870 write, minus ar3. 133 00:06:40,870 --> 00:06:43,399 This is the exact same thing as this thing up here, I'll just 134 00:06:43,399 --> 00:06:46,859 write this is with f of x is equal to, and this is the exact 135 00:06:46,860 --> 00:06:48,460 same thing is that up there. 136 00:06:48,459 --> 00:06:53,560 Now, to solve for an r3, or attempt to solve for an r3, you 137 00:06:53,560 --> 00:06:56,540 just have to realize that when you multiply this thing out, 138 00:06:56,540 --> 00:06:59,689 the way to get the constant term over here, or the d over 139 00:06:59,689 --> 00:07:02,839 here, that's going to be a product of the constant 140 00:07:02,839 --> 00:07:04,310 terms in our expressions. 141 00:07:04,310 --> 00:07:06,920 Because if I introduced any products with any x terms, 142 00:07:06,920 --> 00:07:09,870 you're going to get one of these other terms over here. 143 00:07:09,870 --> 00:07:13,300 The ax cubed term is generated from that term, that term, 144 00:07:13,300 --> 00:07:15,319 and that term, being multiplied together. 145 00:07:15,319 --> 00:07:17,680 and then the constant term is multiplied by the 146 00:07:17,680 --> 00:07:19,790 constant terms to be multiplied together. 147 00:07:19,790 --> 00:07:22,080 And then the two in the middle are multiplied by different 148 00:07:22,079 --> 00:07:24,259 combinations of constant terms. 149 00:07:24,259 --> 00:07:26,459 And you'll see that if you actually want to 150 00:07:26,459 --> 00:07:28,259 multiply this out. 151 00:07:28,259 --> 00:07:35,000 But if you take my word for it, that times that times that 152 00:07:35,000 --> 00:07:37,339 times, actually let me write it this way. 153 00:07:37,339 --> 00:07:43,819 That times that times that, have got to be equal to date we 154 00:07:43,819 --> 00:07:47,550 can then attempt to find some relationship between a and d, 155 00:07:47,550 --> 00:07:49,500 which is mine is minus 2. 156 00:07:49,500 --> 00:07:57,889 So we could say 1 times minus 2 times minus ar3, these two 157 00:07:57,889 --> 00:08:00,050 minuses will cancel out. 158 00:08:00,050 --> 00:08:02,170 It is going to be equal to our d terms. 159 00:08:02,170 --> 00:08:07,110 It's going to be equal to minus 2, and so if we simplify that a 160 00:08:07,110 --> 00:08:14,720 little bit we get 2ar3 is equal to minus 2, divide both sides 161 00:08:14,720 --> 00:08:18,755 by 2a and you get r3 is equal to minus 1/a. 162 00:08:18,754 --> 00:08:21,319 163 00:08:21,319 --> 00:08:24,009 So we haven't solved for the third root, we've just gotten 164 00:08:24,009 --> 00:08:27,990 it in terms of, in terms of our coefficient a. 165 00:08:27,990 --> 00:08:33,060 But letls see if we can still use that in some useful way. 166 00:08:33,059 --> 00:08:37,709 So if r3 is minus 1/a, this term right here becomes what? 167 00:08:37,710 --> 00:08:39,210 Let me rewrite it. 168 00:08:39,210 --> 00:08:41,259 Let me rewrite it. 169 00:08:41,259 --> 00:08:43,029 All right, I'll do it in blue. 170 00:08:43,029 --> 00:08:50,699 f of x is going to be equal to x plus 1 times x minus 2 times 171 00:08:50,700 --> 00:08:54,610 ax-- now minus a times r3. 172 00:08:54,610 --> 00:08:55,250 Let me do it over here. 173 00:08:55,250 --> 00:08:59,779 Minus a times r3 three is minus 1/a. 174 00:08:59,779 --> 00:09:01,339 So that's a minus times a minus. 175 00:09:01,340 --> 00:09:02,420 That's a plus. 176 00:09:02,419 --> 00:09:03,419 The a's cancel out. 177 00:09:03,419 --> 00:09:04,209 You just get a 1. 178 00:09:04,210 --> 00:09:07,300 So this term now if we substitute r3 with this will 179 00:09:07,299 --> 00:09:11,139 just turn into a this will just turn into a plus 1. 180 00:09:11,139 --> 00:09:15,639 Minus a times minus 1/a that's plus 1. 181 00:09:15,639 --> 00:09:16,449 Plus 1. 182 00:09:16,450 --> 00:09:20,160 We just substituted that for that to get this right here. 183 00:09:20,159 --> 00:09:21,789 Now let's multiply this thing out. 184 00:09:21,789 --> 00:09:24,120 Let's actually do the algebraic multiplication 185 00:09:24,120 --> 00:09:26,129 and see what we get. 186 00:09:26,129 --> 00:09:27,980 So first I'm going to multiply these two terms. 187 00:09:27,980 --> 00:09:30,580 I'm going to do it in a different color. 188 00:09:30,580 --> 00:09:34,050 Let me multiply those two terms right there. 189 00:09:34,049 --> 00:09:39,740 So this is going to be equal to x squared, x squared plus 1x 190 00:09:39,740 --> 00:09:43,399 minus 2x So that's minus x, right? 191 00:09:43,399 --> 00:09:48,235 1 times x is x, minus 2 times x is x minus 2x. 192 00:09:48,235 --> 00:09:53,110 So it's minus x, and then minus 2, just multiplied these two 193 00:09:53,110 --> 00:09:54,940 binomials out right here. 194 00:09:54,940 --> 00:09:59,000 And then I'm going to have to multiply that times ax plus 1, 195 00:09:59,000 --> 00:10:00,879 so I'm going to write the ax plus 1 down here 196 00:10:00,879 --> 00:10:02,720 just to save space. 197 00:10:02,720 --> 00:10:05,810 ax plus 1. 198 00:10:05,809 --> 00:10:09,599 And now we just do a little bit of algebraic multiplication. 199 00:10:09,600 --> 00:10:14,139 1 times minus 2 is-- I'll do it in magenta --minus 2. 200 00:10:14,139 --> 00:10:17,149 1 times minus x is minus x. 201 00:10:17,149 --> 00:10:20,689 1 times x squared is x squared. 202 00:10:20,690 --> 00:10:26,820 Now ax times minus 2 is minus 2ax. 203 00:10:26,820 --> 00:10:29,910 Minus 2ax. 204 00:10:29,909 --> 00:10:37,679 ax times minus x is minus ax squared, right? 205 00:10:37,679 --> 00:10:41,159 Minus ax squared and then ax times x squared 206 00:10:41,159 --> 00:10:43,019 is ax to the third. 207 00:10:43,019 --> 00:10:44,750 ax to the third. 208 00:10:44,750 --> 00:10:48,220 Then we add everything together, and we get our f of x 209 00:10:48,220 --> 00:10:52,379 is going to be equal to ax to the third-- let me scroll to 210 00:10:52,379 --> 00:10:54,710 the left a little bit --ax to the third. 211 00:10:54,710 --> 00:10:57,920 x squared minus ax squared. 212 00:10:57,919 --> 00:11:04,620 So that's plus 1 minus a, 1 minus a x squared. 213 00:11:04,620 --> 00:11:08,360 And then we have these two added together so this is minus 214 00:11:08,360 --> 00:11:17,659 1 minus 2a or we could call this equal to-- so minus 1 215 00:11:17,659 --> 00:11:25,889 minus 2a --or we could say minus 1 plus 2ax and then we 216 00:11:25,889 --> 00:11:30,090 have our minus 2, which makes sense because that is our d. 217 00:11:30,090 --> 00:11:31,300 That's our d. 218 00:11:31,299 --> 00:11:33,029 So this is really the solution. 219 00:11:33,029 --> 00:11:36,409 This is the form that are equation is going to take and 220 00:11:36,409 --> 00:11:38,589 the reason why I said in the beginning of the problem that 221 00:11:38,590 --> 00:11:42,269 the equation or the problem wasn't well defined is that I 222 00:11:42,269 --> 00:11:48,429 can pick any real number a and this equation will 223 00:11:48,429 --> 00:11:51,279 satisfy these conditions that we were given. 224 00:11:51,279 --> 00:11:54,269 So I suspect that I wasn't given all of the constraints 225 00:11:54,269 --> 00:11:54,939 on this problem. 226 00:11:54,940 --> 00:11:56,930 Maybe there was a third constraint that a is equal to 227 00:11:56,929 --> 00:11:59,989 1, or b is equal to something else, or a plus b is 228 00:11:59,990 --> 00:12:00,860 equal to something else. 229 00:12:00,860 --> 00:12:04,060 But just using these constraints, just using these 230 00:12:04,059 --> 00:12:07,049 constraints, the equation will take this form. 231 00:12:07,049 --> 00:12:11,689 So, if it was in this form, of course this is a, this is b, 232 00:12:11,690 --> 00:12:15,800 which would be 1 minus a, that's b, that's a, and then 233 00:12:15,799 --> 00:12:18,539 this right here is equal to c. 234 00:12:18,539 --> 00:12:21,730 So we could pick, you know, we could pick one particular 235 00:12:21,730 --> 00:12:24,320 choice if we just want to get an answer to their problem, but 236 00:12:24,320 --> 00:12:26,210 it might not be the answer they were looking for because they 237 00:12:26,210 --> 00:12:27,670 might have picked a different a. 238 00:12:27,669 --> 00:12:30,289 But if you just pick a simple a is equal to 1, than 239 00:12:30,289 --> 00:12:31,689 what's b equal to? 240 00:12:31,690 --> 00:12:34,750 Then b is equal to 1 minus a, which is equal to 0. 241 00:12:34,750 --> 00:12:38,730 Then b is equal to 0, and then c is equal-- actually c would 242 00:12:38,730 --> 00:12:42,529 include this minus sign right there --c would be one plus 2a. 243 00:12:42,529 --> 00:12:46,470 So 2a is 2, 1 plus 2 is 3. 244 00:12:46,470 --> 00:12:47,790 Put a minus sign. 245 00:12:47,789 --> 00:12:49,919 So c would be equal to minus 3. 246 00:12:49,919 --> 00:12:52,279 And of course no matter what we do, d is going 247 00:12:52,279 --> 00:12:55,480 to be equal to minus 2. 248 00:12:55,480 --> 00:13:00,230 So if we pick a is equal to 1, the answer here to what is a 249 00:13:00,230 --> 00:13:07,970 plus b plus c plus d, would be equal to 1 plus minus 3. 250 00:13:07,970 --> 00:13:11,410 1 plus minus 3 is minus 2. 251 00:13:11,409 --> 00:13:14,529 Minus another minus 2 is minus 4. 252 00:13:14,529 --> 00:13:18,110 That's another possibility, but we just as easily could have 253 00:13:18,110 --> 00:13:21,310 picked, you know a could have been something, it could be a 254 00:13:21,309 --> 00:13:22,629 fraction, it could be anything. 255 00:13:22,629 --> 00:13:26,309 It could be, you know a could be minus 1. 256 00:13:26,309 --> 00:13:29,989 a could be equal to minus 1, if a is equal to minus 1, then b 257 00:13:29,990 --> 00:13:33,930 is equal to 2, then b is equal to 2, right? 258 00:13:33,929 --> 00:13:35,399 1 minus minus 1. 259 00:13:35,399 --> 00:13:38,789 And then c is equal to 1 minus 2, which is minus 1, but then 260 00:13:38,789 --> 00:13:40,399 you have a minus out front. 261 00:13:40,399 --> 00:13:44,039 So then c is equal to 1, and then d is always going 262 00:13:44,039 --> 00:13:46,230 to be equal to minus 2. 263 00:13:46,230 --> 00:13:47,930 d is equal to minus 2. 264 00:13:47,929 --> 00:13:53,019 And then you have a plus b plus c plus d would be equal to., 265 00:13:53,019 --> 00:13:57,350 well these would cancel out, you'd get them equal to 0. 266 00:13:57,350 --> 00:14:01,110 So, we don't know exactly what answer the problem writer 267 00:14:01,110 --> 00:14:03,500 was looking for, but it is a pretty interesting problem. 268 00:14:03,500 --> 00:14:06,529 And then you know, just out of curiosity, it-- let's take this 269 00:14:06,529 --> 00:14:09,750 first example that we took --where a is equal to 1, we 270 00:14:09,750 --> 00:14:12,279 could then look at what our root must be. 271 00:14:12,279 --> 00:14:17,470 If a is equal to 1, then our third root is that r3 would be 272 00:14:17,470 --> 00:14:20,790 equal-- that's just for this situation right here --than r3 273 00:14:20,789 --> 00:14:23,329 would be equal to negative 1 divided by 1 would be equal to 274 00:14:23,330 --> 00:14:27,410 minus 1, which means that we would have a repeated root, we 275 00:14:27,409 --> 00:14:30,939 would have a repeated root, right there. 276 00:14:30,940 --> 00:14:34,830 We wouldn't necessarily have a distinct root out there. 277 00:14:34,830 --> 00:14:39,730 And if we're curious about what that would look like, let's 278 00:14:39,730 --> 00:14:43,129 look at the situation where a is equal to 1. 279 00:14:43,129 --> 00:14:44,279 Actually maybe we wil look at both of them. 280 00:14:44,279 --> 00:14:48,709 When a is equal to 1, our equation is x to the third-- 281 00:14:48,710 --> 00:14:54,139 where is the x button --x to the third, there's no x 282 00:14:54,139 --> 00:15:03,039 squared term, minus 3x minus 3x, minus 3x, minus 2. 283 00:15:03,039 --> 00:15:05,879 So let's grab both of them simultaneously. 284 00:15:05,879 --> 00:15:08,159 So that's this one, where we picked this choice 285 00:15:08,159 --> 00:15:10,209 of coefficients. 286 00:15:10,210 --> 00:15:13,759 And then let's do this choice of coefficients, if we have, if 287 00:15:13,759 --> 00:15:23,269 we have minus x to the third, right? a is minus 1 plus 2 x 288 00:15:23,269 --> 00:15:31,350 squared, and then we have, plus x, c is 1, plus x minus 2. 289 00:15:31,350 --> 00:15:32,519 So those are 2 graps. 290 00:15:32,519 --> 00:15:34,370 Let's see what they look like. 291 00:15:34,370 --> 00:15:36,639 Let's graph them. 292 00:15:36,639 --> 00:15:39,500 So our first one, there we go, we hit that point, 293 00:15:39,500 --> 00:15:41,500 that's our double root. 294 00:15:41,500 --> 00:15:44,200 Notice, both of these two graphs meet our 295 00:15:44,200 --> 00:15:45,480 two constraints. 296 00:15:45,480 --> 00:15:47,789 They both have a root there at minus 1. 297 00:15:47,789 --> 00:15:51,309 They both have a y-intercept at y is equal to minus 2. 298 00:15:51,309 --> 00:15:54,779 And they both have a root at x is equal to 299 00:15:54,779 --> 00:15:56,419 positive 2 right there. 300 00:15:56,419 --> 00:15:59,479 Now the second one right here had a separate distinct third 301 00:15:59,480 --> 00:16:02,960 root, while the first one, here, has a double root 302 00:16:02,960 --> 00:16:04,000 right over there. 303 00:16:04,000 --> 00:16:06,230 Anyway, I actually think this problem was somewhat more 304 00:16:06,230 --> 00:16:08,850 interesting by the fact we had to actually look at the 305 00:16:08,850 --> 00:16:10,240 different solutions to it. 306 00:16:10,240 --> 00:16:12,029 And there's actually an infinite number of solutions 307 00:16:12,029 --> 00:16:15,179 to this based on what you choose for your a.