Show polynomial division steps

This page demonstrates how to generate ECC blocks by performing polynomial divison on a message polynomial. Just enter the coefficients of the message polynomial and the desired number of ECC blocks, then click Perform Division.

If the steps are unclear, please read the error correction generation section of the tutorial, which includes a detailed explanation of these steps.


(example: 32,91,11,120,209,114,220,77,67,64,236,17,236 )

(example: 13 )

Polynomial Division Steps

The first step to the division is to prepare the message polynomial for the division. The full message polynomial is:

182x15 + 230x14 + 247x13 + 119x12 + 50x11 + 7x10 + 118x9 + 134x8 + 87x7 + 38x6 + 82x5 + 6x4 + 134x3 + 151x2 + 50x1 + 7

To make sure that the exponent of the lead term doesn't become too small during the division, multiply the message polynomial by xn where n is the number of error correction codewords that are needed. In this case n is 18, for 18 error correction codewords, so multiply the message polynomial by x18, which gives us:

182x33 + 230x32 + 247x31 + 119x30 + 50x29 + 7x28 + 118x27 + 134x26 + 87x25 + 38x24 + 82x23 + 6x22 + 134x21 + 151x20 + 50x19 + 7x18

The lead term of the generator polynomial should also have the same exponent, so multiply by x15 to get

α0x33 + α215x32 + α234x31 + α158x30 + α94x29 + α184x28 + α97x27 + α118x26 + α170x25 + α79x24 + α187x23 + α152x22 + α148x21 + α252x20 + α179x19 + α5x18 + α98x17 + α96x16 + α153x15

Now it is possible to perform the repeated division steps. The number of steps in the division must equal the number of terms in the message polynomial. In this case, the division will take 16 steps to complete. This will result in a remainder that has 18 terms. These terms will be the 18 error correction codewords that are required.

Step 1a: Multiply the Generator Polynomial by the Lead Term of the Message Polynomial

The first step is to multiply the generator polynomial by the lead term of the message polynomial. The lead term in this case is 182x33. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 182x33 to alpha notation. According to the log antilog table, for the integer value 182, the alpha exponent is 93. Therefore 182 = α93. Multiply the generator polynomial by α93:

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α93x33 + α(308 % 255)x32 + α(327 % 255)x31 + α251x30 + α187x29 + α(277 % 255)x28 + α190x27 + α211x26 + α(263 % 255)x25 + α172x24 + α(280 % 255)x23 + α245x22 + α241x21 + α(345 % 255)x20 + α(272 % 255)x19 + α98x18 + α191x17 + α189x16 + α246x15

The result is:

α93x33 + α53x32 + α72x31 + α251x30 + α187x29 + α22x28 + α190x27 + α211x26 + α8x25 + α172x24 + α25x23 + α245x22 + α241x21 + α90x20 + α17x19 + α98x18 + α191x17 + α189x16 + α246x15

Now, convert this to integer notation:

182x33 + 40x32 + 101x31 + 216x30 + 220x29 + 234x28 + 174x27 + 178x26 + 29x25 + 123x24 + 3x23 + 233x22 + 88x21 + 223x20 + 152x19 + 67x18 + 65x17 + 87x16 + 207x15

Step 1b: XOR the result with the message polynomial

Since this is the first division step, XOR the result from 1a with the message polynomial.

(182 ⊕ 182)x33 + (230 ⊕ 40)x32 + (247 ⊕ 101)x31 + (119 ⊕ 216)x30 + (50 ⊕ 220)x29 + (7 ⊕ 234)x28 + (118 ⊕ 174)x27 + (134 ⊕ 178)x26 + (87 ⊕ 29)x25 + (38 ⊕ 123)x24 + (82 ⊕ 3)x23 + (6 ⊕ 233)x22 + (134 ⊕ 88)x21 + (151 ⊕ 223)x20 + (50 ⊕ 152)x19 + (7 ⊕ 67)x18 + (0 ⊕ 65)x17 + (0 ⊕ 87)x16 + (0 ⊕ 207)x15

The result is:

0x33 + 206x32 + 146x31 + 175x30 + 238x29 + 237x28 + 216x27 + 52x26 + 74x25 + 93x24 + 81x23 + 239x22 + 222x21 + 72x20 + 170x19 + 68x18 + 65x17 + 87x16 + 207x15

Discard the lead 0 term to get:

206x32 + 146x31 + 175x30 + 238x29 + 237x28 + 216x27 + 52x26 + 74x25 + 93x24 + 81x23 + 239x22 + 222x21 + 72x20 + 170x19 + 68x18 + 65x17 + 87x16 + 207x15

Step 2a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 206x32. Convert 206x32 to alpha notation. According to the log antilog table, for the integer value 206, the alpha exponent is 111. Therefore 206 = α111. Multiply the generator polynomial by α111:

111 * α0)x32 + (α111 * α215)x31 + (α111 * α234)x30 + (α111 * α158)x29 + (α111 * α94)x28 + (α111 * α184)x27 + (α111 * α97)x26 + (α111 * α118)x25 + (α111 * α170)x24 + (α111 * α79)x23 + (α111 * α187)x22 + (α111 * α152)x21 + (α111 * α148)x20 + (α111 * α252)x19 + (α111 * α179)x18 + (α111 * α5)x17 + (α111 * α98)x16 + (α111 * α96)x15 + (α111 * α153)x14

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α111x32 + α(326 % 255)x31 + α(345 % 255)x30 + α(269 % 255)x29 + α205x28 + α(295 % 255)x27 + α208x26 + α229x25 + α(281 % 255)x24 + α190x23 + α(298 % 255)x22 + α(263 % 255)x21 + α(259 % 255)x20 + α(363 % 255)x19 + α(290 % 255)x18 + α116x17 + α209x16 + α207x15 + α(264 % 255)x14

The result is:

α111x32 + α71x31 + α90x30 + α14x29 + α205x28 + α40x27 + α208x26 + α229x25 + α26x24 + α190x23 + α43x22 + α8x21 + α4x20 + α108x19 + α35x18 + α116x17 + α209x16 + α207x15 + α9x14

Now, convert this to integer notation:

206x32 + 188x31 + 223x30 + 19x29 + 167x28 + 106x27 + 81x26 + 122x25 + 6x24 + 174x23 + 119x22 + 29x21 + 16x20 + 208x19 + 156x18 + 248x17 + 162x16 + 166x15 + 58x14

Step 2b: XOR the result with the result from step 1b

Use the result from step 1b to perform the next XOR.

(206 ⊕ 206)x32 + (146 ⊕ 188)x31 + (175 ⊕ 223)x30 + (238 ⊕ 19)x29 + (237 ⊕ 167)x28 + (216 ⊕ 106)x27 + (52 ⊕ 81)x26 + (74 ⊕ 122)x25 + (93 ⊕ 6)x24 + (81 ⊕ 174)x23 + (239 ⊕ 119)x22 + (222 ⊕ 29)x21 + (72 ⊕ 16)x20 + (170 ⊕ 208)x19 + (68 ⊕ 156)x18 + (65 ⊕ 248)x17 + (87 ⊕ 162)x16 + (207 ⊕ 166)x15 + (0 ⊕ 58)x14

The result is:

0x32 + 46x31 + 112x30 + 253x29 + 74x28 + 178x27 + 101x26 + 48x25 + 91x24 + 255x23 + 152x22 + 195x21 + 88x20 + 122x19 + 216x18 + 185x17 + 245x16 + 105x15 + 58x14

Discard the lead 0 term to get:

46x31 + 112x30 + 253x29 + 74x28 + 178x27 + 101x26 + 48x25 + 91x24 + 255x23 + 152x22 + 195x21 + 88x20 + 122x19 + 216x18 + 185x17 + 245x16 + 105x15 + 58x14

Step 3a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 46x31. Convert 46x31 to alpha notation. According to the log antilog table, for the integer value 46, the alpha exponent is 130. Therefore 46 = α130. Multiply the generator polynomial by α130:

130 * α0)x31 + (α130 * α215)x30 + (α130 * α234)x29 + (α130 * α158)x28 + (α130 * α94)x27 + (α130 * α184)x26 + (α130 * α97)x25 + (α130 * α118)x24 + (α130 * α170)x23 + (α130 * α79)x22 + (α130 * α187)x21 + (α130 * α152)x20 + (α130 * α148)x19 + (α130 * α252)x18 + (α130 * α179)x17 + (α130 * α5)x16 + (α130 * α98)x15 + (α130 * α96)x14 + (α130 * α153)x13

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α130x31 + α(345 % 255)x30 + α(364 % 255)x29 + α(288 % 255)x28 + α224x27 + α(314 % 255)x26 + α227x25 + α248x24 + α(300 % 255)x23 + α209x22 + α(317 % 255)x21 + α(282 % 255)x20 + α(278 % 255)x19 + α(382 % 255)x18 + α(309 % 255)x17 + α135x16 + α228x15 + α226x14 + α(283 % 255)x13

The result is:

α130x31 + α90x30 + α109x29 + α33x28 + α224x27 + α59x26 + α227x25 + α248x24 + α45x23 + α209x22 + α62x21 + α27x20 + α23x19 + α127x18 + α54x17 + α135x16 + α228x15 + α226x14 + α28x13

Now, convert this to integer notation:

46x31 + 223x30 + 189x29 + 39x28 + 18x27 + 210x26 + 144x25 + 27x24 + 193x23 + 162x22 + 222x21 + 12x20 + 201x19 + 204x18 + 80x17 + 169x16 + 61x15 + 72x14 + 24x13

Step 3b: XOR the result with the result from step 2b

Use the result from step 2b to perform the next XOR.

(46 ⊕ 46)x31 + (112 ⊕ 223)x30 + (253 ⊕ 189)x29 + (74 ⊕ 39)x28 + (178 ⊕ 18)x27 + (101 ⊕ 210)x26 + (48 ⊕ 144)x25 + (91 ⊕ 27)x24 + (255 ⊕ 193)x23 + (152 ⊕ 162)x22 + (195 ⊕ 222)x21 + (88 ⊕ 12)x20 + (122 ⊕ 201)x19 + (216 ⊕ 204)x18 + (185 ⊕ 80)x17 + (245 ⊕ 169)x16 + (105 ⊕ 61)x15 + (58 ⊕ 72)x14 + (0 ⊕ 24)x13

The result is:

0x31 + 175x30 + 64x29 + 109x28 + 160x27 + 183x26 + 160x25 + 64x24 + 62x23 + 58x22 + 29x21 + 84x20 + 179x19 + 20x18 + 233x17 + 92x16 + 84x15 + 114x14 + 24x13

Discard the lead 0 term to get:

175x30 + 64x29 + 109x28 + 160x27 + 183x26 + 160x25 + 64x24 + 62x23 + 58x22 + 29x21 + 84x20 + 179x19 + 20x18 + 233x17 + 92x16 + 84x15 + 114x14 + 24x13

Step 4a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 175x30. Convert 175x30 to alpha notation. According to the log antilog table, for the integer value 175, the alpha exponent is 97. Therefore 175 = α97. Multiply the generator polynomial by α97:

97 * α0)x30 + (α97 * α215)x29 + (α97 * α234)x28 + (α97 * α158)x27 + (α97 * α94)x26 + (α97 * α184)x25 + (α97 * α97)x24 + (α97 * α118)x23 + (α97 * α170)x22 + (α97 * α79)x21 + (α97 * α187)x20 + (α97 * α152)x19 + (α97 * α148)x18 + (α97 * α252)x17 + (α97 * α179)x16 + (α97 * α5)x15 + (α97 * α98)x14 + (α97 * α96)x13 + (α97 * α153)x12

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α97x30 + α(312 % 255)x29 + α(331 % 255)x28 + α255x27 + α191x26 + α(281 % 255)x25 + α194x24 + α215x23 + α(267 % 255)x22 + α176x21 + α(284 % 255)x20 + α249x19 + α245x18 + α(349 % 255)x17 + α(276 % 255)x16 + α102x15 + α195x14 + α193x13 + α250x12

The result is:

α97x30 + α57x29 + α76x28 + α0x27 + α191x26 + α26x25 + α194x24 + α215x23 + α12x22 + α176x21 + α29x20 + α249x19 + α245x18 + α94x17 + α21x16 + α102x15 + α195x14 + α193x13 + α250x12

Now, convert this to integer notation:

175x30 + 186x29 + 30x28 + 1x27 + 65x26 + 6x25 + 50x24 + 239x23 + 205x22 + 227x21 + 48x20 + 54x19 + 233x18 + 113x17 + 117x16 + 68x15 + 100x14 + 25x13 + 108x12

Step 4b: XOR the result with the result from step 3b

Use the result from step 3b to perform the next XOR.

(175 ⊕ 175)x30 + (64 ⊕ 186)x29 + (109 ⊕ 30)x28 + (160 ⊕ 1)x27 + (183 ⊕ 65)x26 + (160 ⊕ 6)x25 + (64 ⊕ 50)x24 + (62 ⊕ 239)x23 + (58 ⊕ 205)x22 + (29 ⊕ 227)x21 + (84 ⊕ 48)x20 + (179 ⊕ 54)x19 + (20 ⊕ 233)x18 + (233 ⊕ 113)x17 + (92 ⊕ 117)x16 + (84 ⊕ 68)x15 + (114 ⊕ 100)x14 + (24 ⊕ 25)x13 + (0 ⊕ 108)x12

The result is:

0x30 + 250x29 + 115x28 + 161x27 + 246x26 + 166x25 + 114x24 + 209x23 + 247x22 + 254x21 + 100x20 + 133x19 + 253x18 + 152x17 + 41x16 + 16x15 + 22x14 + 1x13 + 108x12

Discard the lead 0 term to get:

250x29 + 115x28 + 161x27 + 246x26 + 166x25 + 114x24 + 209x23 + 247x22 + 254x21 + 100x20 + 133x19 + 253x18 + 152x17 + 41x16 + 16x15 + 22x14 + 1x13 + 108x12

Step 5a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 250x29. Convert 250x29 to alpha notation. According to the log antilog table, for the integer value 250, the alpha exponent is 244. Therefore 250 = α244. Multiply the generator polynomial by α244:

244 * α0)x29 + (α244 * α215)x28 + (α244 * α234)x27 + (α244 * α158)x26 + (α244 * α94)x25 + (α244 * α184)x24 + (α244 * α97)x23 + (α244 * α118)x22 + (α244 * α170)x21 + (α244 * α79)x20 + (α244 * α187)x19 + (α244 * α152)x18 + (α244 * α148)x17 + (α244 * α252)x16 + (α244 * α179)x15 + (α244 * α5)x14 + (α244 * α98)x13 + (α244 * α96)x12 + (α244 * α153)x11

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α244x29 + α(459 % 255)x28 + α(478 % 255)x27 + α(402 % 255)x26 + α(338 % 255)x25 + α(428 % 255)x24 + α(341 % 255)x23 + α(362 % 255)x22 + α(414 % 255)x21 + α(323 % 255)x20 + α(431 % 255)x19 + α(396 % 255)x18 + α(392 % 255)x17 + α(496 % 255)x16 + α(423 % 255)x15 + α249x14 + α(342 % 255)x13 + α(340 % 255)x12 + α(397 % 255)x11

The result is:

α244x29 + α204x28 + α223x27 + α147x26 + α83x25 + α173x24 + α86x23 + α107x22 + α159x21 + α68x20 + α176x19 + α141x18 + α137x17 + α241x16 + α168x15 + α249x14 + α87x13 + α85x12 + α142x11

Now, convert this to integer notation:

250x29 + 221x28 + 9x27 + 41x26 + 187x25 + 246x24 + 177x23 + 104x22 + 115x21 + 153x20 + 227x19 + 21x18 + 158x17 + 88x16 + 252x15 + 54x14 + 127x13 + 214x12 + 42x11

Step 5b: XOR the result with the result from step 4b

Use the result from step 4b to perform the next XOR.

(250 ⊕ 250)x29 + (115 ⊕ 221)x28 + (161 ⊕ 9)x27 + (246 ⊕ 41)x26 + (166 ⊕ 187)x25 + (114 ⊕ 246)x24 + (209 ⊕ 177)x23 + (247 ⊕ 104)x22 + (254 ⊕ 115)x21 + (100 ⊕ 153)x20 + (133 ⊕ 227)x19 + (253 ⊕ 21)x18 + (152 ⊕ 158)x17 + (41 ⊕ 88)x16 + (16 ⊕ 252)x15 + (22 ⊕ 54)x14 + (1 ⊕ 127)x13 + (108 ⊕ 214)x12 + (0 ⊕ 42)x11

The result is:

0x29 + 174x28 + 168x27 + 223x26 + 29x25 + 132x24 + 96x23 + 159x22 + 141x21 + 253x20 + 102x19 + 232x18 + 6x17 + 113x16 + 236x15 + 32x14 + 126x13 + 186x12 + 42x11

Discard the lead 0 term to get:

174x28 + 168x27 + 223x26 + 29x25 + 132x24 + 96x23 + 159x22 + 141x21 + 253x20 + 102x19 + 232x18 + 6x17 + 113x16 + 236x15 + 32x14 + 126x13 + 186x12 + 42x11

Step 6a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 174x28. Convert 174x28 to alpha notation. According to the log antilog table, for the integer value 174, the alpha exponent is 190. Therefore 174 = α190. Multiply the generator polynomial by α190:

190 * α0)x28 + (α190 * α215)x27 + (α190 * α234)x26 + (α190 * α158)x25 + (α190 * α94)x24 + (α190 * α184)x23 + (α190 * α97)x22 + (α190 * α118)x21 + (α190 * α170)x20 + (α190 * α79)x19 + (α190 * α187)x18 + (α190 * α152)x17 + (α190 * α148)x16 + (α190 * α252)x15 + (α190 * α179)x14 + (α190 * α5)x13 + (α190 * α98)x12 + (α190 * α96)x11 + (α190 * α153)x10

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α190x28 + α(405 % 255)x27 + α(424 % 255)x26 + α(348 % 255)x25 + α(284 % 255)x24 + α(374 % 255)x23 + α(287 % 255)x22 + α(308 % 255)x21 + α(360 % 255)x20 + α(269 % 255)x19 + α(377 % 255)x18 + α(342 % 255)x17 + α(338 % 255)x16 + α(442 % 255)x15 + α(369 % 255)x14 + α195x13 + α(288 % 255)x12 + α(286 % 255)x11 + α(343 % 255)x10

The result is:

α190x28 + α150x27 + α169x26 + α93x25 + α29x24 + α119x23 + α32x22 + α53x21 + α105x20 + α14x19 + α122x18 + α87x17 + α83x16 + α187x15 + α114x14 + α195x13 + α33x12 + α31x11 + α88x10

Now, convert this to integer notation:

174x28 + 85x27 + 229x26 + 182x25 + 48x24 + 147x23 + 157x22 + 40x21 + 26x20 + 19x19 + 236x18 + 127x17 + 187x16 + 220x15 + 62x14 + 100x13 + 39x12 + 192x11 + 254x10

Step 6b: XOR the result with the result from step 5b

Use the result from step 5b to perform the next XOR.

(174 ⊕ 174)x28 + (168 ⊕ 85)x27 + (223 ⊕ 229)x26 + (29 ⊕ 182)x25 + (132 ⊕ 48)x24 + (96 ⊕ 147)x23 + (159 ⊕ 157)x22 + (141 ⊕ 40)x21 + (253 ⊕ 26)x20 + (102 ⊕ 19)x19 + (232 ⊕ 236)x18 + (6 ⊕ 127)x17 + (113 ⊕ 187)x16 + (236 ⊕ 220)x15 + (32 ⊕ 62)x14 + (126 ⊕ 100)x13 + (186 ⊕ 39)x12 + (42 ⊕ 192)x11 + (0 ⊕ 254)x10

The result is:

0x28 + 253x27 + 58x26 + 171x25 + 180x24 + 243x23 + 2x22 + 165x21 + 231x20 + 117x19 + 4x18 + 121x17 + 202x16 + 48x15 + 30x14 + 26x13 + 157x12 + 234x11 + 254x10

Discard the lead 0 term to get:

253x27 + 58x26 + 171x25 + 180x24 + 243x23 + 2x22 + 165x21 + 231x20 + 117x19 + 4x18 + 121x17 + 202x16 + 48x15 + 30x14 + 26x13 + 157x12 + 234x11 + 254x10

Step 7a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 253x27. Convert 253x27 to alpha notation. According to the log antilog table, for the integer value 253, the alpha exponent is 80. Therefore 253 = α80. Multiply the generator polynomial by α80:

80 * α0)x27 + (α80 * α215)x26 + (α80 * α234)x25 + (α80 * α158)x24 + (α80 * α94)x23 + (α80 * α184)x22 + (α80 * α97)x21 + (α80 * α118)x20 + (α80 * α170)x19 + (α80 * α79)x18 + (α80 * α187)x17 + (α80 * α152)x16 + (α80 * α148)x15 + (α80 * α252)x14 + (α80 * α179)x13 + (α80 * α5)x12 + (α80 * α98)x11 + (α80 * α96)x10 + (α80 * α153)x9

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α80x27 + α(295 % 255)x26 + α(314 % 255)x25 + α238x24 + α174x23 + α(264 % 255)x22 + α177x21 + α198x20 + α250x19 + α159x18 + α(267 % 255)x17 + α232x16 + α228x15 + α(332 % 255)x14 + α(259 % 255)x13 + α85x12 + α178x11 + α176x10 + α233x9

The result is:

α80x27 + α40x26 + α59x25 + α238x24 + α174x23 + α9x22 + α177x21 + α198x20 + α250x19 + α159x18 + α12x17 + α232x16 + α228x15 + α77x14 + α4x13 + α85x12 + α178x11 + α176x10 + α233x9

Now, convert this to integer notation:

253x27 + 106x26 + 210x25 + 11x24 + 241x23 + 58x22 + 219x21 + 7x20 + 108x19 + 115x18 + 205x17 + 247x16 + 61x15 + 60x14 + 16x13 + 214x12 + 171x11 + 227x10 + 243x9

Step 7b: XOR the result with the result from step 6b

Use the result from step 6b to perform the next XOR.

(253 ⊕ 253)x27 + (58 ⊕ 106)x26 + (171 ⊕ 210)x25 + (180 ⊕ 11)x24 + (243 ⊕ 241)x23 + (2 ⊕ 58)x22 + (165 ⊕ 219)x21 + (231 ⊕ 7)x20 + (117 ⊕ 108)x19 + (4 ⊕ 115)x18 + (121 ⊕ 205)x17 + (202 ⊕ 247)x16 + (48 ⊕ 61)x15 + (30 ⊕ 60)x14 + (26 ⊕ 16)x13 + (157 ⊕ 214)x12 + (234 ⊕ 171)x11 + (254 ⊕ 227)x10 + (0 ⊕ 243)x9

The result is:

0x27 + 80x26 + 121x25 + 191x24 + 2x23 + 56x22 + 126x21 + 224x20 + 25x19 + 119x18 + 180x17 + 61x16 + 13x15 + 34x14 + 10x13 + 75x12 + 65x11 + 29x10 + 243x9

Discard the lead 0 term to get:

80x26 + 121x25 + 191x24 + 2x23 + 56x22 + 126x21 + 224x20 + 25x19 + 119x18 + 180x17 + 61x16 + 13x15 + 34x14 + 10x13 + 75x12 + 65x11 + 29x10 + 243x9

Step 8a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 80x26. Convert 80x26 to alpha notation. According to the log antilog table, for the integer value 80, the alpha exponent is 54. Therefore 80 = α54. Multiply the generator polynomial by α54:

54 * α0)x26 + (α54 * α215)x25 + (α54 * α234)x24 + (α54 * α158)x23 + (α54 * α94)x22 + (α54 * α184)x21 + (α54 * α97)x20 + (α54 * α118)x19 + (α54 * α170)x18 + (α54 * α79)x17 + (α54 * α187)x16 + (α54 * α152)x15 + (α54 * α148)x14 + (α54 * α252)x13 + (α54 * α179)x12 + (α54 * α5)x11 + (α54 * α98)x10 + (α54 * α96)x9 + (α54 * α153)x8

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α54x26 + α(269 % 255)x25 + α(288 % 255)x24 + α212x23 + α148x22 + α238x21 + α151x20 + α172x19 + α224x18 + α133x17 + α241x16 + α206x15 + α202x14 + α(306 % 255)x13 + α233x12 + α59x11 + α152x10 + α150x9 + α207x8

The result is:

α54x26 + α14x25 + α33x24 + α212x23 + α148x22 + α238x21 + α151x20 + α172x19 + α224x18 + α133x17 + α241x16 + α206x15 + α202x14 + α51x13 + α233x12 + α59x11 + α152x10 + α150x9 + α207x8

Now, convert this to integer notation:

80x26 + 19x25 + 39x24 + 121x23 + 82x22 + 11x21 + 170x20 + 123x19 + 18x18 + 109x17 + 88x16 + 83x15 + 112x14 + 10x13 + 243x12 + 210x11 + 73x10 + 85x9 + 166x8

Step 8b: XOR the result with the result from step 7b

Use the result from step 7b to perform the next XOR.

(80 ⊕ 80)x26 + (121 ⊕ 19)x25 + (191 ⊕ 39)x24 + (2 ⊕ 121)x23 + (56 ⊕ 82)x22 + (126 ⊕ 11)x21 + (224 ⊕ 170)x20 + (25 ⊕ 123)x19 + (119 ⊕ 18)x18 + (180 ⊕ 109)x17 + (61 ⊕ 88)x16 + (13 ⊕ 83)x15 + (34 ⊕ 112)x14 + (10 ⊕ 10)x13 + (75 ⊕ 243)x12 + (65 ⊕ 210)x11 + (29 ⊕ 73)x10 + (243 ⊕ 85)x9 + (0 ⊕ 166)x8

The result is:

0x26 + 106x25 + 152x24 + 123x23 + 106x22 + 117x21 + 74x20 + 98x19 + 101x18 + 217x17 + 101x16 + 94x15 + 82x14 + 0x13 + 184x12 + 147x11 + 84x10 + 166x9 + 166x8

Discard the lead 0 term to get:

106x25 + 152x24 + 123x23 + 106x22 + 117x21 + 74x20 + 98x19 + 101x18 + 217x17 + 101x16 + 94x15 + 82x14 + 0x13 + 184x12 + 147x11 + 84x10 + 166x9 + 166x8

Step 9a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 106x25. Convert 106x25 to alpha notation. According to the log antilog table, for the integer value 106, the alpha exponent is 40. Therefore 106 = α40. Multiply the generator polynomial by α40:

40 * α0)x25 + (α40 * α215)x24 + (α40 * α234)x23 + (α40 * α158)x22 + (α40 * α94)x21 + (α40 * α184)x20 + (α40 * α97)x19 + (α40 * α118)x18 + (α40 * α170)x17 + (α40 * α79)x16 + (α40 * α187)x15 + (α40 * α152)x14 + (α40 * α148)x13 + (α40 * α252)x12 + (α40 * α179)x11 + (α40 * α5)x10 + (α40 * α98)x9 + (α40 * α96)x8 + (α40 * α153)x7

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α40x25 + α255x24 + α(274 % 255)x23 + α198x22 + α134x21 + α224x20 + α137x19 + α158x18 + α210x17 + α119x16 + α227x15 + α192x14 + α188x13 + α(292 % 255)x12 + α219x11 + α45x10 + α138x9 + α136x8 + α193x7

The result is:

α40x25 + α0x24 + α19x23 + α198x22 + α134x21 + α224x20 + α137x19 + α158x18 + α210x17 + α119x16 + α227x15 + α192x14 + α188x13 + α37x12 + α219x11 + α45x10 + α138x9 + α136x8 + α193x7

Now, convert this to integer notation:

106x25 + 1x24 + 90x23 + 7x22 + 218x21 + 18x20 + 158x19 + 183x18 + 89x17 + 147x16 + 144x15 + 130x14 + 165x13 + 74x12 + 86x11 + 193x10 + 33x9 + 79x8 + 25x7

Step 9b: XOR the result with the result from step 8b

Use the result from step 8b to perform the next XOR.

(106 ⊕ 106)x25 + (152 ⊕ 1)x24 + (123 ⊕ 90)x23 + (106 ⊕ 7)x22 + (117 ⊕ 218)x21 + (74 ⊕ 18)x20 + (98 ⊕ 158)x19 + (101 ⊕ 183)x18 + (217 ⊕ 89)x17 + (101 ⊕ 147)x16 + (94 ⊕ 144)x15 + (82 ⊕ 130)x14 + (0 ⊕ 165)x13 + (184 ⊕ 74)x12 + (147 ⊕ 86)x11 + (84 ⊕ 193)x10 + (166 ⊕ 33)x9 + (166 ⊕ 79)x8 + (0 ⊕ 25)x7

The result is:

0x25 + 153x24 + 33x23 + 109x22 + 175x21 + 88x20 + 252x19 + 210x18 + 128x17 + 246x16 + 206x15 + 208x14 + 165x13 + 242x12 + 197x11 + 149x10 + 135x9 + 233x8 + 25x7

Discard the lead 0 term to get:

153x24 + 33x23 + 109x22 + 175x21 + 88x20 + 252x19 + 210x18 + 128x17 + 246x16 + 206x15 + 208x14 + 165x13 + 242x12 + 197x11 + 149x10 + 135x9 + 233x8 + 25x7

Step 10a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 153x24. Convert 153x24 to alpha notation. According to the log antilog table, for the integer value 153, the alpha exponent is 68. Therefore 153 = α68. Multiply the generator polynomial by α68:

68 * α0)x24 + (α68 * α215)x23 + (α68 * α234)x22 + (α68 * α158)x21 + (α68 * α94)x20 + (α68 * α184)x19 + (α68 * α97)x18 + (α68 * α118)x17 + (α68 * α170)x16 + (α68 * α79)x15 + (α68 * α187)x14 + (α68 * α152)x13 + (α68 * α148)x12 + (α68 * α252)x11 + (α68 * α179)x10 + (α68 * α5)x9 + (α68 * α98)x8 + (α68 * α96)x7 + (α68 * α153)x6

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α68x24 + α(283 % 255)x23 + α(302 % 255)x22 + α226x21 + α162x20 + α252x19 + α165x18 + α186x17 + α238x16 + α147x15 + α255x14 + α220x13 + α216x12 + α(320 % 255)x11 + α247x10 + α73x9 + α166x8 + α164x7 + α221x6

The result is:

α68x24 + α28x23 + α47x22 + α226x21 + α162x20 + α252x19 + α165x18 + α186x17 + α238x16 + α147x15 + α0x14 + α220x13 + α216x12 + α65x11 + α247x10 + α73x9 + α166x8 + α164x7 + α221x6

Now, convert this to integer notation:

153x24 + 24x23 + 35x22 + 72x21 + 191x20 + 173x19 + 145x18 + 110x17 + 11x16 + 41x15 + 1x14 + 172x13 + 195x12 + 190x11 + 131x10 + 202x9 + 63x8 + 198x7 + 69x6

Step 10b: XOR the result with the result from step 9b

Use the result from step 9b to perform the next XOR.

(153 ⊕ 153)x24 + (33 ⊕ 24)x23 + (109 ⊕ 35)x22 + (175 ⊕ 72)x21 + (88 ⊕ 191)x20 + (252 ⊕ 173)x19 + (210 ⊕ 145)x18 + (128 ⊕ 110)x17 + (246 ⊕ 11)x16 + (206 ⊕ 41)x15 + (208 ⊕ 1)x14 + (165 ⊕ 172)x13 + (242 ⊕ 195)x12 + (197 ⊕ 190)x11 + (149 ⊕ 131)x10 + (135 ⊕ 202)x9 + (233 ⊕ 63)x8 + (25 ⊕ 198)x7 + (0 ⊕ 69)x6

The result is:

0x24 + 57x23 + 78x22 + 231x21 + 231x20 + 81x19 + 67x18 + 238x17 + 253x16 + 231x15 + 209x14 + 9x13 + 49x12 + 123x11 + 22x10 + 77x9 + 214x8 + 223x7 + 69x6

Discard the lead 0 term to get:

57x23 + 78x22 + 231x21 + 231x20 + 81x19 + 67x18 + 238x17 + 253x16 + 231x15 + 209x14 + 9x13 + 49x12 + 123x11 + 22x10 + 77x9 + 214x8 + 223x7 + 69x6

Step 11a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 57x23. Convert 57x23 to alpha notation. According to the log antilog table, for the integer value 57, the alpha exponent is 154. Therefore 57 = α154. Multiply the generator polynomial by α154:

154 * α0)x23 + (α154 * α215)x22 + (α154 * α234)x21 + (α154 * α158)x20 + (α154 * α94)x19 + (α154 * α184)x18 + (α154 * α97)x17 + (α154 * α118)x16 + (α154 * α170)x15 + (α154 * α79)x14 + (α154 * α187)x13 + (α154 * α152)x12 + (α154 * α148)x11 + (α154 * α252)x10 + (α154 * α179)x9 + (α154 * α5)x8 + (α154 * α98)x7 + (α154 * α96)x6 + (α154 * α153)x5

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α154x23 + α(369 % 255)x22 + α(388 % 255)x21 + α(312 % 255)x20 + α248x19 + α(338 % 255)x18 + α251x17 + α(272 % 255)x16 + α(324 % 255)x15 + α233x14 + α(341 % 255)x13 + α(306 % 255)x12 + α(302 % 255)x11 + α(406 % 255)x10 + α(333 % 255)x9 + α159x8 + α252x7 + α250x6 + α(307 % 255)x5

The result is:

α154x23 + α114x22 + α133x21 + α57x20 + α248x19 + α83x18 + α251x17 + α17x16 + α69x15 + α233x14 + α86x13 + α51x12 + α47x11 + α151x10 + α78x9 + α159x8 + α252x7 + α250x6 + α52x5

Now, convert this to integer notation:

57x23 + 62x22 + 109x21 + 186x20 + 27x19 + 187x18 + 216x17 + 152x16 + 47x15 + 243x14 + 177x13 + 10x12 + 35x11 + 170x10 + 120x9 + 115x8 + 173x7 + 108x6 + 20x5

Step 11b: XOR the result with the result from step 10b

Use the result from step 10b to perform the next XOR.

(57 ⊕ 57)x23 + (78 ⊕ 62)x22 + (231 ⊕ 109)x21 + (231 ⊕ 186)x20 + (81 ⊕ 27)x19 + (67 ⊕ 187)x18 + (238 ⊕ 216)x17 + (253 ⊕ 152)x16 + (231 ⊕ 47)x15 + (209 ⊕ 243)x14 + (9 ⊕ 177)x13 + (49 ⊕ 10)x12 + (123 ⊕ 35)x11 + (22 ⊕ 170)x10 + (77 ⊕ 120)x9 + (214 ⊕ 115)x8 + (223 ⊕ 173)x7 + (69 ⊕ 108)x6 + (0 ⊕ 20)x5

The result is:

0x23 + 112x22 + 138x21 + 93x20 + 74x19 + 248x18 + 54x17 + 101x16 + 200x15 + 34x14 + 184x13 + 59x12 + 88x11 + 188x10 + 53x9 + 165x8 + 114x7 + 41x6 + 20x5

Discard the lead 0 term to get:

112x22 + 138x21 + 93x20 + 74x19 + 248x18 + 54x17 + 101x16 + 200x15 + 34x14 + 184x13 + 59x12 + 88x11 + 188x10 + 53x9 + 165x8 + 114x7 + 41x6 + 20x5

Step 12a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 112x22. Convert 112x22 to alpha notation. According to the log antilog table, for the integer value 112, the alpha exponent is 202. Therefore 112 = α202. Multiply the generator polynomial by α202:

202 * α0)x22 + (α202 * α215)x21 + (α202 * α234)x20 + (α202 * α158)x19 + (α202 * α94)x18 + (α202 * α184)x17 + (α202 * α97)x16 + (α202 * α118)x15 + (α202 * α170)x14 + (α202 * α79)x13 + (α202 * α187)x12 + (α202 * α152)x11 + (α202 * α148)x10 + (α202 * α252)x9 + (α202 * α179)x8 + (α202 * α5)x7 + (α202 * α98)x6 + (α202 * α96)x5 + (α202 * α153)x4

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α202x22 + α(417 % 255)x21 + α(436 % 255)x20 + α(360 % 255)x19 + α(296 % 255)x18 + α(386 % 255)x17 + α(299 % 255)x16 + α(320 % 255)x15 + α(372 % 255)x14 + α(281 % 255)x13 + α(389 % 255)x12 + α(354 % 255)x11 + α(350 % 255)x10 + α(454 % 255)x9 + α(381 % 255)x8 + α207x7 + α(300 % 255)x6 + α(298 % 255)x5 + α(355 % 255)x4

The result is:

α202x22 + α162x21 + α181x20 + α105x19 + α41x18 + α131x17 + α44x16 + α65x15 + α117x14 + α26x13 + α134x12 + α99x11 + α95x10 + α199x9 + α126x8 + α207x7 + α45x6 + α43x5 + α100x4

Now, convert this to integer notation:

112x22 + 191x21 + 49x20 + 26x19 + 212x18 + 92x17 + 238x16 + 190x15 + 237x14 + 6x13 + 218x12 + 134x11 + 226x10 + 14x9 + 102x8 + 166x7 + 193x6 + 119x5 + 17x4

Step 12b: XOR the result with the result from step 11b

Use the result from step 11b to perform the next XOR.

(112 ⊕ 112)x22 + (138 ⊕ 191)x21 + (93 ⊕ 49)x20 + (74 ⊕ 26)x19 + (248 ⊕ 212)x18 + (54 ⊕ 92)x17 + (101 ⊕ 238)x16 + (200 ⊕ 190)x15 + (34 ⊕ 237)x14 + (184 ⊕ 6)x13 + (59 ⊕ 218)x12 + (88 ⊕ 134)x11 + (188 ⊕ 226)x10 + (53 ⊕ 14)x9 + (165 ⊕ 102)x8 + (114 ⊕ 166)x7 + (41 ⊕ 193)x6 + (20 ⊕ 119)x5 + (0 ⊕ 17)x4

The result is:

0x22 + 53x21 + 108x20 + 80x19 + 44x18 + 106x17 + 139x16 + 118x15 + 207x14 + 190x13 + 225x12 + 222x11 + 94x10 + 59x9 + 195x8 + 212x7 + 232x6 + 99x5 + 17x4

Discard the lead 0 term to get:

53x21 + 108x20 + 80x19 + 44x18 + 106x17 + 139x16 + 118x15 + 207x14 + 190x13 + 225x12 + 222x11 + 94x10 + 59x9 + 195x8 + 212x7 + 232x6 + 99x5 + 17x4

Step 13a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 53x21. Convert 53x21 to alpha notation. According to the log antilog table, for the integer value 53, the alpha exponent is 39. Therefore 53 = α39. Multiply the generator polynomial by α39:

39 * α0)x21 + (α39 * α215)x20 + (α39 * α234)x19 + (α39 * α158)x18 + (α39 * α94)x17 + (α39 * α184)x16 + (α39 * α97)x15 + (α39 * α118)x14 + (α39 * α170)x13 + (α39 * α79)x12 + (α39 * α187)x11 + (α39 * α152)x10 + (α39 * α148)x9 + (α39 * α252)x8 + (α39 * α179)x7 + (α39 * α5)x6 + (α39 * α98)x5 + (α39 * α96)x4 + (α39 * α153)x3

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α39x21 + α254x20 + α(273 % 255)x19 + α197x18 + α133x17 + α223x16 + α136x15 + α157x14 + α209x13 + α118x12 + α226x11 + α191x10 + α187x9 + α(291 % 255)x8 + α218x7 + α44x6 + α137x5 + α135x4 + α192x3

The result is:

α39x21 + α254x20 + α18x19 + α197x18 + α133x17 + α223x16 + α136x15 + α157x14 + α209x13 + α118x12 + α226x11 + α191x10 + α187x9 + α36x8 + α218x7 + α44x6 + α137x5 + α135x4 + α192x3

Now, convert this to integer notation:

53x21 + 142x20 + 45x19 + 141x18 + 109x17 + 9x16 + 79x15 + 213x14 + 162x13 + 199x12 + 72x11 + 65x10 + 220x9 + 37x8 + 43x7 + 238x6 + 158x5 + 169x4 + 130x3

Step 13b: XOR the result with the result from step 12b

Use the result from step 12b to perform the next XOR.

(53 ⊕ 53)x21 + (108 ⊕ 142)x20 + (80 ⊕ 45)x19 + (44 ⊕ 141)x18 + (106 ⊕ 109)x17 + (139 ⊕ 9)x16 + (118 ⊕ 79)x15 + (207 ⊕ 213)x14 + (190 ⊕ 162)x13 + (225 ⊕ 199)x12 + (222 ⊕ 72)x11 + (94 ⊕ 65)x10 + (59 ⊕ 220)x9 + (195 ⊕ 37)x8 + (212 ⊕ 43)x7 + (232 ⊕ 238)x6 + (99 ⊕ 158)x5 + (17 ⊕ 169)x4 + (0 ⊕ 130)x3

The result is:

0x21 + 226x20 + 125x19 + 161x18 + 7x17 + 130x16 + 57x15 + 26x14 + 28x13 + 38x12 + 150x11 + 31x10 + 231x9 + 230x8 + 255x7 + 6x6 + 253x5 + 184x4 + 130x3

Discard the lead 0 term to get:

226x20 + 125x19 + 161x18 + 7x17 + 130x16 + 57x15 + 26x14 + 28x13 + 38x12 + 150x11 + 31x10 + 231x9 + 230x8 + 255x7 + 6x6 + 253x5 + 184x4 + 130x3

Step 14a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 226x20. Convert 226x20 to alpha notation. According to the log antilog table, for the integer value 226, the alpha exponent is 95. Therefore 226 = α95. Multiply the generator polynomial by α95:

95 * α0)x20 + (α95 * α215)x19 + (α95 * α234)x18 + (α95 * α158)x17 + (α95 * α94)x16 + (α95 * α184)x15 + (α95 * α97)x14 + (α95 * α118)x13 + (α95 * α170)x12 + (α95 * α79)x11 + (α95 * α187)x10 + (α95 * α152)x9 + (α95 * α148)x8 + (α95 * α252)x7 + (α95 * α179)x6 + (α95 * α5)x5 + (α95 * α98)x4 + (α95 * α96)x3 + (α95 * α153)x2

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α95x20 + α(310 % 255)x19 + α(329 % 255)x18 + α253x17 + α189x16 + α(279 % 255)x15 + α192x14 + α213x13 + α(265 % 255)x12 + α174x11 + α(282 % 255)x10 + α247x9 + α243x8 + α(347 % 255)x7 + α(274 % 255)x6 + α100x5 + α193x4 + α191x3 + α248x2

The result is:

α95x20 + α55x19 + α74x18 + α253x17 + α189x16 + α24x15 + α192x14 + α213x13 + α10x12 + α174x11 + α27x10 + α247x9 + α243x8 + α92x7 + α19x6 + α100x5 + α193x4 + α191x3 + α248x2

Now, convert this to integer notation:

226x20 + 160x19 + 137x18 + 71x17 + 87x16 + 143x15 + 130x14 + 242x13 + 116x12 + 241x11 + 12x10 + 131x9 + 125x8 + 91x7 + 90x6 + 17x5 + 25x4 + 65x3 + 27x2

Step 14b: XOR the result with the result from step 13b

Use the result from step 13b to perform the next XOR.

(226 ⊕ 226)x20 + (125 ⊕ 160)x19 + (161 ⊕ 137)x18 + (7 ⊕ 71)x17 + (130 ⊕ 87)x16 + (57 ⊕ 143)x15 + (26 ⊕ 130)x14 + (28 ⊕ 242)x13 + (38 ⊕ 116)x12 + (150 ⊕ 241)x11 + (31 ⊕ 12)x10 + (231 ⊕ 131)x9 + (230 ⊕ 125)x8 + (255 ⊕ 91)x7 + (6 ⊕ 90)x6 + (253 ⊕ 17)x5 + (184 ⊕ 25)x4 + (130 ⊕ 65)x3 + (0 ⊕ 27)x2

The result is:

0x20 + 221x19 + 40x18 + 64x17 + 213x16 + 182x15 + 152x14 + 238x13 + 82x12 + 103x11 + 19x10 + 100x9 + 155x8 + 164x7 + 92x6 + 236x5 + 161x4 + 195x3 + 27x2

Discard the lead 0 term to get:

221x19 + 40x18 + 64x17 + 213x16 + 182x15 + 152x14 + 238x13 + 82x12 + 103x11 + 19x10 + 100x9 + 155x8 + 164x7 + 92x6 + 236x5 + 161x4 + 195x3 + 27x2

Step 15a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 221x19. Convert 221x19 to alpha notation. According to the log antilog table, for the integer value 221, the alpha exponent is 204. Therefore 221 = α204. Multiply the generator polynomial by α204:

204 * α0)x19 + (α204 * α215)x18 + (α204 * α234)x17 + (α204 * α158)x16 + (α204 * α94)x15 + (α204 * α184)x14 + (α204 * α97)x13 + (α204 * α118)x12 + (α204 * α170)x11 + (α204 * α79)x10 + (α204 * α187)x9 + (α204 * α152)x8 + (α204 * α148)x7 + (α204 * α252)x6 + (α204 * α179)x5 + (α204 * α5)x4 + (α204 * α98)x3 + (α204 * α96)x2 + (α204 * α153)x1

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α204x19 + α(419 % 255)x18 + α(438 % 255)x17 + α(362 % 255)x16 + α(298 % 255)x15 + α(388 % 255)x14 + α(301 % 255)x13 + α(322 % 255)x12 + α(374 % 255)x11 + α(283 % 255)x10 + α(391 % 255)x9 + α(356 % 255)x8 + α(352 % 255)x7 + α(456 % 255)x6 + α(383 % 255)x5 + α209x4 + α(302 % 255)x3 + α(300 % 255)x2 + α(357 % 255)x1

The result is:

α204x19 + α164x18 + α183x17 + α107x16 + α43x15 + α133x14 + α46x13 + α67x12 + α119x11 + α28x10 + α136x9 + α101x8 + α97x7 + α201x6 + α128x5 + α209x4 + α47x3 + α45x2 + α102x1

Now, convert this to integer notation:

221x19 + 198x18 + 196x17 + 104x16 + 119x15 + 109x14 + 159x13 + 194x12 + 147x11 + 24x10 + 79x9 + 34x8 + 175x7 + 56x6 + 133x5 + 162x4 + 35x3 + 193x2 + 68x1

Step 15b: XOR the result with the result from step 14b

Use the result from step 14b to perform the next XOR.

(221 ⊕ 221)x19 + (40 ⊕ 198)x18 + (64 ⊕ 196)x17 + (213 ⊕ 104)x16 + (182 ⊕ 119)x15 + (152 ⊕ 109)x14 + (238 ⊕ 159)x13 + (82 ⊕ 194)x12 + (103 ⊕ 147)x11 + (19 ⊕ 24)x10 + (100 ⊕ 79)x9 + (155 ⊕ 34)x8 + (164 ⊕ 175)x7 + (92 ⊕ 56)x6 + (236 ⊕ 133)x5 + (161 ⊕ 162)x4 + (195 ⊕ 35)x3 + (27 ⊕ 193)x2 + (0 ⊕ 68)x1

The result is:

0x19 + 238x18 + 132x17 + 189x16 + 193x15 + 245x14 + 113x13 + 144x12 + 244x11 + 11x10 + 43x9 + 185x8 + 11x7 + 100x6 + 105x5 + 3x4 + 224x3 + 218x2 + 68x1

Discard the lead 0 term to get:

238x18 + 132x17 + 189x16 + 193x15 + 245x14 + 113x13 + 144x12 + 244x11 + 11x10 + 43x9 + 185x8 + 11x7 + 100x6 + 105x5 + 3x4 + 224x3 + 218x2 + 68x1

Step 16a: Multiply the Generator Polynomial by the Lead Term of the XOR result from the previous step

Next, multiply the generator polynomial by the lead term of the XOR result from the previous step. The lead term in this case is 238x18. Convert 238x18 to alpha notation. According to the log antilog table, for the integer value 238, the alpha exponent is 44. Therefore 238 = α44. Multiply the generator polynomial by α44:

44 * α0)x18 + (α44 * α215)x17 + (α44 * α234)x16 + (α44 * α158)x15 + (α44 * α94)x14 + (α44 * α184)x13 + (α44 * α97)x12 + (α44 * α118)x11 + (α44 * α170)x10 + (α44 * α79)x9 + (α44 * α187)x8 + (α44 * α152)x7 + (α44 * α148)x6 + (α44 * α252)x5 + (α44 * α179)x4 + (α44 * α5)x3 + (α44 * α98)x2 + (α44 * α96)x1 + (α44 * α153)x0

The exponents of the alphas are added together. In this case, at least one of the exponents is larger than 255, so perform modulo 255 as follows:

α44x18 + α(259 % 255)x17 + α(278 % 255)x16 + α202x15 + α138x14 + α228x13 + α141x12 + α162x11 + α214x10 + α123x9 + α231x8 + α196x7 + α192x6 + α(296 % 255)x5 + α223x4 + α49x3 + α142x2 + α140x1 + α197x0

The result is:

α44x18 + α4x17 + α23x16 + α202x15 + α138x14 + α228x13 + α141x12 + α162x11 + α214x10 + α123x9 + α231x8 + α196x7 + α192x6 + α41x5 + α223x4 + α49x3 + α142x2 + α140x1 + α197x0

Now, convert this to integer notation:

238x18 + 16x17 + 201x16 + 112x15 + 33x14 + 61x13 + 21x12 + 191x11 + 249x10 + 197x9 + 245x8 + 200x7 + 130x6 + 212x5 + 9x4 + 140x3 + 42x2 + 132x1 + 141

Step 16b: XOR the result with the result from step 15b

Use the result from step 15b to perform the next XOR.

(238 ⊕ 238)x18 + (132 ⊕ 16)x17 + (189 ⊕ 201)x16 + (193 ⊕ 112)x15 + (245 ⊕ 33)x14 + (113 ⊕ 61)x13 + (144 ⊕ 21)x12 + (244 ⊕ 191)x11 + (11 ⊕ 249)x10 + (43 ⊕ 197)x9 + (185 ⊕ 245)x8 + (11 ⊕ 200)x7 + (100 ⊕ 130)x6 + (105 ⊕ 212)x5 + (3 ⊕ 9)x4 + (224 ⊕ 140)x3 + (218 ⊕ 42)x2 + (68 ⊕ 132)x1 + (0 ⊕ 141)x0

The result is:

0x18 + 148x17 + 116x16 + 177x15 + 212x14 + 76x13 + 133x12 + 75x11 + 242x10 + 238x9 + 76x8 + 195x7 + 230x6 + 189x5 + 10x4 + 108x3 + 240x2 + 192x1 + 141

Discard the lead 0 term to get:

148x17 + 116x16 + 177x15 + 212x14 + 76x13 + 133x12 + 75x11 + 242x10 + 238x9 + 76x8 + 195x7 + 230x6 + 189x5 + 10x4 + 108x3 + 240x2 + 192x1 + 141

Use the terms of the remainder as the error correction codewords

The division has been performed 16 times, which is the number of terms in the message polynomial. This means that the division is complete and the terms of the above polynomial are the error correction codewords to use for the original message polynomial:

148  116  177  212  76  133  75  242  238  76  195  230  189  10  108  240  192  141