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.

Enter the coefficients of the message polynomial, separated by commas (no spaces):
(example: 32,91,11,120,209,114,220,77,67,64,236,17,236 )

Enter the desired number of ECC blocks:
(example: 13 )

Perform Division

Polynomial Division Steps

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

246x¹⁴ + 246x¹³ + 66x¹² + 7x¹¹ + 118x¹⁰ + 134x⁹ + 242x⁸ + 7x⁷ + 38x⁶ + 86x⁵ + 22x⁴ + 198x³ + 199x² + 146x¹ + 6

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:

246x³² + 246x³¹ + 66x³⁰ + 7x²⁹ + 118x²⁸ + 134x²⁷ + 242x²⁶ + 7x²⁵ + 38x²⁴ + 86x²³ + 22x²² + 198x²¹ + 199x²⁰ + 146x¹⁹ + 6x¹⁸

The lead term of the generator polynomial should also have the same exponent, so multiply by x¹⁴ to get

α⁰x³² + α²¹⁵x³¹ + α²³⁴x³⁰ + α¹⁵⁸x²⁹ + α⁹⁴x²⁸ + α¹⁸⁴x²⁷ + α⁹⁷x²⁶ + α¹¹⁸x²⁵ + α¹⁷⁰x²⁴ + α⁷⁹x²³ + α¹⁸⁷x²² + α¹⁵²x²¹ + α¹⁴⁸x²⁰ + α²⁵²x¹⁹ + α¹⁷⁹x¹⁸ + α⁵x¹⁷ + α⁹⁸x¹⁶ + α⁹⁶x¹⁵ + α¹⁵³x¹⁴

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 15 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 246x³². Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 246x³² to alpha notation. According to the log antilog table, for the integer value 246, the alpha exponent is 173. Therefore 246 = α¹⁷³. Multiply the generator polynomial by α¹⁷³:

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:

α¹⁷³x³² + α^{(388 % 255)}x³¹ + α^{(407 % 255)}x³⁰ + α^{(331 % 255)}x²⁹ + α^{(267 % 255)}x²⁸ + α^{(357 % 255)}x²⁷ + α^{(270 % 255)}x²⁶ + α^{(291 % 255)}x²⁵ + α^{(343 % 255)}x²⁴ + α²⁵²x²³ + α^{(360 % 255)}x²² + α^{(325 % 255)}x²¹ + α^{(321 % 255)}x²⁰ + α^{(425 % 255)}x¹⁹ + α^{(352 % 255)}x¹⁸ + α¹⁷⁸x¹⁷ + α^{(271 % 255)}x¹⁶ + α^{(269 % 255)}x¹⁵ + α^{(326 % 255)}x¹⁴

The result is:

α¹⁷³x³² + α¹³³x³¹ + α¹⁵²x³⁰ + α⁷⁶x²⁹ + α¹²x²⁸ + α¹⁰²x²⁷ + α¹⁵x²⁶ + α³⁶x²⁵ + α⁸⁸x²⁴ + α²⁵²x²³ + α¹⁰⁵x²² + α⁷⁰x²¹ + α⁶⁶x²⁰ + α¹⁷⁰x¹⁹ + α⁹⁷x¹⁸ + α¹⁷⁸x¹⁷ + α¹⁶x¹⁶ + α¹⁴x¹⁵ + α⁷¹x¹⁴

Now, convert this to integer notation:

246x³² + 109x³¹ + 73x³⁰ + 30x²⁹ + 205x²⁸ + 68x²⁷ + 38x²⁶ + 37x²⁵ + 254x²⁴ + 173x²³ + 26x²² + 94x²¹ + 97x²⁰ + 215x¹⁹ + 175x¹⁸ + 171x¹⁷ + 76x¹⁶ + 19x¹⁵ + 188x¹⁴

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.

(246 ⊕ 246)x³² + (246 ⊕ 109)x³¹ + (66 ⊕ 73)x³⁰ + (7 ⊕ 30)x²⁹ + (118 ⊕ 205)x²⁸ + (134 ⊕ 68)x²⁷ + (242 ⊕ 38)x²⁶ + (7 ⊕ 37)x²⁵ + (38 ⊕ 254)x²⁴ + (86 ⊕ 173)x²³ + (22 ⊕ 26)x²² + (198 ⊕ 94)x²¹ + (199 ⊕ 97)x²⁰ + (146 ⊕ 215)x¹⁹ + (6 ⊕ 175)x¹⁸ + (0 ⊕ 171)x¹⁷ + (0 ⊕ 76)x¹⁶ + (0 ⊕ 19)x¹⁵ + (0 ⊕ 188)x¹⁴

The result is:

0x³² + 155x³¹ + 11x³⁰ + 25x²⁹ + 187x²⁸ + 194x²⁷ + 212x²⁶ + 34x²⁵ + 216x²⁴ + 251x²³ + 12x²² + 152x²¹ + 166x²⁰ + 69x¹⁹ + 169x¹⁸ + 171x¹⁷ + 76x¹⁶ + 19x¹⁵ + 188x¹⁴

Discard the lead 0 term to get:

155x³¹ + 11x³⁰ + 25x²⁹ + 187x²⁸ + 194x²⁷ + 212x²⁶ + 34x²⁵ + 216x²⁴ + 251x²³ + 12x²² + 152x²¹ + 166x²⁰ + 69x¹⁹ + 169x¹⁸ + 171x¹⁷ + 76x¹⁶ + 19x¹⁵ + 188x¹⁴

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 155x³¹. Convert 155x³¹ to alpha notation. According to the log antilog table, for the integer value 155, the alpha exponent is 217. Therefore 155 = α²¹⁷. Multiply the generator polynomial by α²¹⁷:

(α²¹⁷ * α⁰)x³¹ + (α²¹⁷ * α²¹⁵)x³⁰ + (α²¹⁷ * α²³⁴)x²⁹ + (α²¹⁷ * α¹⁵⁸)x²⁸ + (α²¹⁷ * α⁹⁴)x²⁷ + (α²¹⁷ * α¹⁸⁴)x²⁶ + (α²¹⁷ * α⁹⁷)x²⁵ + (α²¹⁷ * α¹¹⁸)x²⁴ + (α²¹⁷ * α¹⁷⁰)x²³ + (α²¹⁷ * α⁷⁹)x²² + (α²¹⁷ * α¹⁸⁷)x²¹ + (α²¹⁷ * α¹⁵²)x²⁰ + (α²¹⁷ * α¹⁴⁸)x¹⁹ + (α²¹⁷ * α²⁵²)x¹⁸ + (α²¹⁷ * α¹⁷⁹)x¹⁷ + (α²¹⁷ * α⁵)x¹⁶ + (α²¹⁷ * α⁹⁸)x¹⁵ + (α²¹⁷ * α⁹⁶)x¹⁴ + (α²¹⁷ * 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:

α²¹⁷x³¹ + α^{(432 % 255)}x³⁰ + α^{(451 % 255)}x²⁹ + α^{(375 % 255)}x²⁸ + α^{(311 % 255)}x²⁷ + α^{(401 % 255)}x²⁶ + α^{(314 % 255)}x²⁵ + α^{(335 % 255)}x²⁴ + α^{(387 % 255)}x²³ + α^{(296 % 255)}x²² + α^{(404 % 255)}x²¹ + α^{(369 % 255)}x²⁰ + α^{(365 % 255)}x¹⁹ + α^{(469 % 255)}x¹⁸ + α^{(396 % 255)}x¹⁷ + α²²²x¹⁶ + α^{(315 % 255)}x¹⁵ + α^{(313 % 255)}x¹⁴ + α^{(370 % 255)}x¹³

The result is:

α²¹⁷x³¹ + α¹⁷⁷x³⁰ + α¹⁹⁶x²⁹ + α¹²⁰x²⁸ + α⁵⁶x²⁷ + α¹⁴⁶x²⁶ + α⁵⁹x²⁵ + α⁸⁰x²⁴ + α¹³²x²³ + α⁴¹x²² + α¹⁴⁹x²¹ + α¹¹⁴x²⁰ + α¹¹⁰x¹⁹ + α²¹⁴x¹⁸ + α¹⁴¹x¹⁷ + α²²²x¹⁶ + α⁶⁰x¹⁵ + α⁵⁸x¹⁴ + α¹¹⁵x¹³

Now, convert this to integer notation:

155x³¹ + 219x³⁰ + 200x²⁹ + 59x²⁸ + 93x²⁷ + 154x²⁶ + 210x²⁵ + 253x²⁴ + 184x²³ + 212x²² + 164x²¹ + 62x²⁰ + 103x¹⁹ + 249x¹⁸ + 21x¹⁷ + 138x¹⁶ + 185x¹⁵ + 105x¹⁴ + 124x¹³

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

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

(155 ⊕ 155)x³¹ + (11 ⊕ 219)x³⁰ + (25 ⊕ 200)x²⁹ + (187 ⊕ 59)x²⁸ + (194 ⊕ 93)x²⁷ + (212 ⊕ 154)x²⁶ + (34 ⊕ 210)x²⁵ + (216 ⊕ 253)x²⁴ + (251 ⊕ 184)x²³ + (12 ⊕ 212)x²² + (152 ⊕ 164)x²¹ + (166 ⊕ 62)x²⁰ + (69 ⊕ 103)x¹⁹ + (169 ⊕ 249)x¹⁸ + (171 ⊕ 21)x¹⁷ + (76 ⊕ 138)x¹⁶ + (19 ⊕ 185)x¹⁵ + (188 ⊕ 105)x¹⁴ + (0 ⊕ 124)x¹³

The result is:

0x³¹ + 208x³⁰ + 209x²⁹ + 128x²⁸ + 159x²⁷ + 78x²⁶ + 240x²⁵ + 37x²⁴ + 67x²³ + 216x²² + 60x²¹ + 152x²⁰ + 34x¹⁹ + 80x¹⁸ + 190x¹⁷ + 198x¹⁶ + 170x¹⁵ + 213x¹⁴ + 124x¹³

Discard the lead 0 term to get:

208x³⁰ + 209x²⁹ + 128x²⁸ + 159x²⁷ + 78x²⁶ + 240x²⁵ + 37x²⁴ + 67x²³ + 216x²² + 60x²¹ + 152x²⁰ + 34x¹⁹ + 80x¹⁸ + 190x¹⁷ + 198x¹⁶ + 170x¹⁵ + 213x¹⁴ + 124x¹³

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 208x³⁰. Convert 208x³⁰ to alpha notation. According to the log antilog table, for the integer value 208, the alpha exponent is 108. Therefore 208 = α¹⁰⁸. Multiply the generator polynomial by α¹⁰⁸:

(α¹⁰⁸ * α⁰)x³⁰ + (α¹⁰⁸ * α²¹⁵)x²⁹ + (α¹⁰⁸ * α²³⁴)x²⁸ + (α¹⁰⁸ * α¹⁵⁸)x²⁷ + (α¹⁰⁸ * α⁹⁴)x²⁶ + (α¹⁰⁸ * α¹⁸⁴)x²⁵ + (α¹⁰⁸ * α⁹⁷)x²⁴ + (α¹⁰⁸ * α¹¹⁸)x²³ + (α¹⁰⁸ * α¹⁷⁰)x²² + (α¹⁰⁸ * α⁷⁹)x²¹ + (α¹⁰⁸ * α¹⁸⁷)x²⁰ + (α¹⁰⁸ * α¹⁵²)x¹⁹ + (α¹⁰⁸ * α¹⁴⁸)x¹⁸ + (α¹⁰⁸ * α²⁵²)x¹⁷ + (α¹⁰⁸ * α¹⁷⁹)x¹⁶ + (α¹⁰⁸ * α⁵)x¹⁵ + (α¹⁰⁸ * α⁹⁸)x¹⁴ + (α¹⁰⁸ * α⁹⁶)x¹³ + (α¹⁰⁸ * 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:

α¹⁰⁸x³⁰ + α^{(323 % 255)}x²⁹ + α^{(342 % 255)}x²⁸ + α^{(266 % 255)}x²⁷ + α²⁰²x²⁶ + α^{(292 % 255)}x²⁵ + α²⁰⁵x²⁴ + α²²⁶x²³ + α^{(278 % 255)}x²² + α¹⁸⁷x²¹ + α^{(295 % 255)}x²⁰ + α^{(260 % 255)}x¹⁹ + α^{(256 % 255)}x¹⁸ + α^{(360 % 255)}x¹⁷ + α^{(287 % 255)}x¹⁶ + α¹¹³x¹⁵ + α²⁰⁶x¹⁴ + α²⁰⁴x¹³ + α^{(261 % 255)}x¹²

The result is:

α¹⁰⁸x³⁰ + α⁶⁸x²⁹ + α⁸⁷x²⁸ + α¹¹x²⁷ + α²⁰²x²⁶ + α³⁷x²⁵ + α²⁰⁵x²⁴ + α²²⁶x²³ + α²³x²² + α¹⁸⁷x²¹ + α⁴⁰x²⁰ + α⁵x¹⁹ + α¹x¹⁸ + α¹⁰⁵x¹⁷ + α³²x¹⁶ + α¹¹³x¹⁵ + α²⁰⁶x¹⁴ + α²⁰⁴x¹³ + α⁶x¹²

Now, convert this to integer notation:

208x³⁰ + 153x²⁹ + 127x²⁸ + 232x²⁷ + 112x²⁶ + 74x²⁵ + 167x²⁴ + 72x²³ + 201x²² + 220x²¹ + 106x²⁰ + 32x¹⁹ + 2x¹⁸ + 26x¹⁷ + 157x¹⁶ + 31x¹⁵ + 83x¹⁴ + 221x¹³ + 64x¹²

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

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

(208 ⊕ 208)x³⁰ + (209 ⊕ 153)x²⁹ + (128 ⊕ 127)x²⁸ + (159 ⊕ 232)x²⁷ + (78 ⊕ 112)x²⁶ + (240 ⊕ 74)x²⁵ + (37 ⊕ 167)x²⁴ + (67 ⊕ 72)x²³ + (216 ⊕ 201)x²² + (60 ⊕ 220)x²¹ + (152 ⊕ 106)x²⁰ + (34 ⊕ 32)x¹⁹ + (80 ⊕ 2)x¹⁸ + (190 ⊕ 26)x¹⁷ + (198 ⊕ 157)x¹⁶ + (170 ⊕ 31)x¹⁵ + (213 ⊕ 83)x¹⁴ + (124 ⊕ 221)x¹³ + (0 ⊕ 64)x¹²

The result is:

0x³⁰ + 72x²⁹ + 255x²⁸ + 119x²⁷ + 62x²⁶ + 186x²⁵ + 130x²⁴ + 11x²³ + 17x²² + 224x²¹ + 242x²⁰ + 2x¹⁹ + 82x¹⁸ + 164x¹⁷ + 91x¹⁶ + 181x¹⁵ + 134x¹⁴ + 161x¹³ + 64x¹²

Discard the lead 0 term to get:

72x²⁹ + 255x²⁸ + 119x²⁷ + 62x²⁶ + 186x²⁵ + 130x²⁴ + 11x²³ + 17x²² + 224x²¹ + 242x²⁰ + 2x¹⁹ + 82x¹⁸ + 164x¹⁷ + 91x¹⁶ + 181x¹⁵ + 134x¹⁴ + 161x¹³ + 64x¹²

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 72x²⁹. Convert 72x²⁹ to alpha notation. According to the log antilog table, for the integer value 72, the alpha exponent is 226. Therefore 72 = α²²⁶. Multiply the generator polynomial by α²²⁶:

(α²²⁶ * α⁰)x²⁹ + (α²²⁶ * α²¹⁵)x²⁸ + (α²²⁶ * α²³⁴)x²⁷ + (α²²⁶ * α¹⁵⁸)x²⁶ + (α²²⁶ * α⁹⁴)x²⁵ + (α²²⁶ * α¹⁸⁴)x²⁴ + (α²²⁶ * α⁹⁷)x²³ + (α²²⁶ * α¹¹⁸)x²² + (α²²⁶ * α¹⁷⁰)x²¹ + (α²²⁶ * α⁷⁹)x²⁰ + (α²²⁶ * α¹⁸⁷)x¹⁹ + (α²²⁶ * α¹⁵²)x¹⁸ + (α²²⁶ * α¹⁴⁸)x¹⁷ + (α²²⁶ * α²⁵²)x¹⁶ + (α²²⁶ * α¹⁷⁹)x¹⁵ + (α²²⁶ * α⁵)x¹⁴ + (α²²⁶ * α⁹⁸)x¹³ + (α²²⁶ * α⁹⁶)x¹² + (α²²⁶ * 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:

α²²⁶x²⁹ + α^{(441 % 255)}x²⁸ + α^{(460 % 255)}x²⁷ + α^{(384 % 255)}x²⁶ + α^{(320 % 255)}x²⁵ + α^{(410 % 255)}x²⁴ + α^{(323 % 255)}x²³ + α^{(344 % 255)}x²² + α^{(396 % 255)}x²¹ + α^{(305 % 255)}x²⁰ + α^{(413 % 255)}x¹⁹ + α^{(378 % 255)}x¹⁸ + α^{(374 % 255)}x¹⁷ + α^{(478 % 255)}x¹⁶ + α^{(405 % 255)}x¹⁵ + α²³¹x¹⁴ + α^{(324 % 255)}x¹³ + α^{(322 % 255)}x¹² + α^{(379 % 255)}x¹¹

The result is:

α²²⁶x²⁹ + α¹⁸⁶x²⁸ + α²⁰⁵x²⁷ + α¹²⁹x²⁶ + α⁶⁵x²⁵ + α¹⁵⁵x²⁴ + α⁶⁸x²³ + α⁸⁹x²² + α¹⁴¹x²¹ + α⁵⁰x²⁰ + α¹⁵⁸x¹⁹ + α¹²³x¹⁸ + α¹¹⁹x¹⁷ + α²²³x¹⁶ + α¹⁵⁰x¹⁵ + α²³¹x¹⁴ + α⁶⁹x¹³ + α⁶⁷x¹² + α¹²⁴x¹¹

Now, convert this to integer notation:

72x²⁹ + 110x²⁸ + 167x²⁷ + 23x²⁶ + 190x²⁵ + 114x²⁴ + 153x²³ + 225x²² + 21x²¹ + 5x²⁰ + 183x¹⁹ + 197x¹⁸ + 147x¹⁷ + 9x¹⁶ + 85x¹⁵ + 245x¹⁴ + 47x¹³ + 194x¹² + 151x¹¹

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

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

(72 ⊕ 72)x²⁹ + (255 ⊕ 110)x²⁸ + (119 ⊕ 167)x²⁷ + (62 ⊕ 23)x²⁶ + (186 ⊕ 190)x²⁵ + (130 ⊕ 114)x²⁴ + (11 ⊕ 153)x²³ + (17 ⊕ 225)x²² + (224 ⊕ 21)x²¹ + (242 ⊕ 5)x²⁰ + (2 ⊕ 183)x¹⁹ + (82 ⊕ 197)x¹⁸ + (164 ⊕ 147)x¹⁷ + (91 ⊕ 9)x¹⁶ + (181 ⊕ 85)x¹⁵ + (134 ⊕ 245)x¹⁴ + (161 ⊕ 47)x¹³ + (64 ⊕ 194)x¹² + (0 ⊕ 151)x¹¹

The result is:

0x²⁹ + 145x²⁸ + 208x²⁷ + 41x²⁶ + 4x²⁵ + 240x²⁴ + 146x²³ + 240x²² + 245x²¹ + 247x²⁰ + 181x¹⁹ + 151x¹⁸ + 55x¹⁷ + 82x¹⁶ + 224x¹⁵ + 115x¹⁴ + 142x¹³ + 130x¹² + 151x¹¹

Discard the lead 0 term to get:

145x²⁸ + 208x²⁷ + 41x²⁶ + 4x²⁵ + 240x²⁴ + 146x²³ + 240x²² + 245x²¹ + 247x²⁰ + 181x¹⁹ + 151x¹⁸ + 55x¹⁷ + 82x¹⁶ + 224x¹⁵ + 115x¹⁴ + 142x¹³ + 130x¹² + 151x¹¹

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 145x²⁸. Convert 145x²⁸ to alpha notation. According to the log antilog table, for the integer value 145, the alpha exponent is 165. Therefore 145 = α¹⁶⁵. Multiply the generator polynomial by α¹⁶⁵:

(α¹⁶⁵ * α⁰)x²⁸ + (α¹⁶⁵ * α²¹⁵)x²⁷ + (α¹⁶⁵ * α²³⁴)x²⁶ + (α¹⁶⁵ * α¹⁵⁸)x²⁵ + (α¹⁶⁵ * α⁹⁴)x²⁴ + (α¹⁶⁵ * α¹⁸⁴)x²³ + (α¹⁶⁵ * α⁹⁷)x²² + (α¹⁶⁵ * α¹¹⁸)x²¹ + (α¹⁶⁵ * α¹⁷⁰)x²⁰ + (α¹⁶⁵ * α⁷⁹)x¹⁹ + (α¹⁶⁵ * α¹⁸⁷)x¹⁸ + (α¹⁶⁵ * α¹⁵²)x¹⁷ + (α¹⁶⁵ * α¹⁴⁸)x¹⁶ + (α¹⁶⁵ * α²⁵²)x¹⁵ + (α¹⁶⁵ * α¹⁷⁹)x¹⁴ + (α¹⁶⁵ * α⁵)x¹³ + (α¹⁶⁵ * α⁹⁸)x¹² + (α¹⁶⁵ * α⁹⁶)x¹¹ + (α¹⁶⁵ * 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:

α¹⁶⁵x²⁸ + α^{(380 % 255)}x²⁷ + α^{(399 % 255)}x²⁶ + α^{(323 % 255)}x²⁵ + α^{(259 % 255)}x²⁴ + α^{(349 % 255)}x²³ + α^{(262 % 255)}x²² + α^{(283 % 255)}x²¹ + α^{(335 % 255)}x²⁰ + α²⁴⁴x¹⁹ + α^{(352 % 255)}x¹⁸ + α^{(317 % 255)}x¹⁷ + α^{(313 % 255)}x¹⁶ + α^{(417 % 255)}x¹⁵ + α^{(344 % 255)}x¹⁴ + α¹⁷⁰x¹³ + α^{(263 % 255)}x¹² + α^{(261 % 255)}x¹¹ + α^{(318 % 255)}x¹⁰

The result is:

α¹⁶⁵x²⁸ + α¹²⁵x²⁷ + α¹⁴⁴x²⁶ + α⁶⁸x²⁵ + α⁴x²⁴ + α⁹⁴x²³ + α⁷x²² + α²⁸x²¹ + α⁸⁰x²⁰ + α²⁴⁴x¹⁹ + α⁹⁷x¹⁸ + α⁶²x¹⁷ + α⁵⁸x¹⁶ + α¹⁶²x¹⁵ + α⁸⁹x¹⁴ + α¹⁷⁰x¹³ + α⁸x¹² + α⁶x¹¹ + α⁶³x¹⁰

Now, convert this to integer notation:

145x²⁸ + 51x²⁷ + 168x²⁶ + 153x²⁵ + 16x²⁴ + 113x²³ + 128x²² + 24x²¹ + 253x²⁰ + 250x¹⁹ + 175x¹⁸ + 222x¹⁷ + 105x¹⁶ + 191x¹⁵ + 225x¹⁴ + 215x¹³ + 29x¹² + 64x¹¹ + 161x¹⁰

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

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

(145 ⊕ 145)x²⁸ + (208 ⊕ 51)x²⁷ + (41 ⊕ 168)x²⁶ + (4 ⊕ 153)x²⁵ + (240 ⊕ 16)x²⁴ + (146 ⊕ 113)x²³ + (240 ⊕ 128)x²² + (245 ⊕ 24)x²¹ + (247 ⊕ 253)x²⁰ + (181 ⊕ 250)x¹⁹ + (151 ⊕ 175)x¹⁸ + (55 ⊕ 222)x¹⁷ + (82 ⊕ 105)x¹⁶ + (224 ⊕ 191)x¹⁵ + (115 ⊕ 225)x¹⁴ + (142 ⊕ 215)x¹³ + (130 ⊕ 29)x¹² + (151 ⊕ 64)x¹¹ + (0 ⊕ 161)x¹⁰

The result is:

0x²⁸ + 227x²⁷ + 129x²⁶ + 157x²⁵ + 224x²⁴ + 227x²³ + 112x²² + 237x²¹ + 10x²⁰ + 79x¹⁹ + 56x¹⁸ + 233x¹⁷ + 59x¹⁶ + 95x¹⁵ + 146x¹⁴ + 89x¹³ + 159x¹² + 215x¹¹ + 161x¹⁰

Discard the lead 0 term to get:

227x²⁷ + 129x²⁶ + 157x²⁵ + 224x²⁴ + 227x²³ + 112x²² + 237x²¹ + 10x²⁰ + 79x¹⁹ + 56x¹⁸ + 233x¹⁷ + 59x¹⁶ + 95x¹⁵ + 146x¹⁴ + 89x¹³ + 159x¹² + 215x¹¹ + 161x¹⁰

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 227x²⁷. Convert 227x²⁷ to alpha notation. According to the log antilog table, for the integer value 227, the alpha exponent is 176. Therefore 227 = α¹⁷⁶. Multiply the generator polynomial by α¹⁷⁶:

(α¹⁷⁶ * α⁰)x²⁷ + (α¹⁷⁶ * α²¹⁵)x²⁶ + (α¹⁷⁶ * α²³⁴)x²⁵ + (α¹⁷⁶ * α¹⁵⁸)x²⁴ + (α¹⁷⁶ * α⁹⁴)x²³ + (α¹⁷⁶ * α¹⁸⁴)x²² + (α¹⁷⁶ * α⁹⁷)x²¹ + (α¹⁷⁶ * α¹¹⁸)x²⁰ + (α¹⁷⁶ * α¹⁷⁰)x¹⁹ + (α¹⁷⁶ * α⁷⁹)x¹⁸ + (α¹⁷⁶ * α¹⁸⁷)x¹⁷ + (α¹⁷⁶ * α¹⁵²)x¹⁶ + (α¹⁷⁶ * α¹⁴⁸)x¹⁵ + (α¹⁷⁶ * α²⁵²)x¹⁴ + (α¹⁷⁶ * α¹⁷⁹)x¹³ + (α¹⁷⁶ * α⁵)x¹² + (α¹⁷⁶ * α⁹⁸)x¹¹ + (α¹⁷⁶ * α⁹⁶)x¹⁰ + (α¹⁷⁶ * 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:

α¹⁷⁶x²⁷ + α^{(391 % 255)}x²⁶ + α^{(410 % 255)}x²⁵ + α^{(334 % 255)}x²⁴ + α^{(270 % 255)}x²³ + α^{(360 % 255)}x²² + α^{(273 % 255)}x²¹ + α^{(294 % 255)}x²⁰ + α^{(346 % 255)}x¹⁹ + α²⁵⁵x¹⁸ + α^{(363 % 255)}x¹⁷ + α^{(328 % 255)}x¹⁶ + α^{(324 % 255)}x¹⁵ + α^{(428 % 255)}x¹⁴ + α^{(355 % 255)}x¹³ + α¹⁸¹x¹² + α^{(274 % 255)}x¹¹ + α^{(272 % 255)}x¹⁰ + α^{(329 % 255)}x⁹

The result is:

α¹⁷⁶x²⁷ + α¹³⁶x²⁶ + α¹⁵⁵x²⁵ + α⁷⁹x²⁴ + α¹⁵x²³ + α¹⁰⁵x²² + α¹⁸x²¹ + α³⁹x²⁰ + α⁹¹x¹⁹ + α⁰x¹⁸ + α¹⁰⁸x¹⁷ + α⁷³x¹⁶ + α⁶⁹x¹⁵ + α¹⁷³x¹⁴ + α¹⁰⁰x¹³ + α¹⁸¹x¹² + α¹⁹x¹¹ + α¹⁷x¹⁰ + α⁷⁴x⁹

Now, convert this to integer notation:

227x²⁷ + 79x²⁶ + 114x²⁵ + 240x²⁴ + 38x²³ + 26x²² + 45x²¹ + 53x²⁰ + 163x¹⁹ + 1x¹⁸ + 208x¹⁷ + 202x¹⁶ + 47x¹⁵ + 246x¹⁴ + 17x¹³ + 49x¹² + 90x¹¹ + 152x¹⁰ + 137x⁹

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

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

(227 ⊕ 227)x²⁷ + (129 ⊕ 79)x²⁶ + (157 ⊕ 114)x²⁵ + (224 ⊕ 240)x²⁴ + (227 ⊕ 38)x²³ + (112 ⊕ 26)x²² + (237 ⊕ 45)x²¹ + (10 ⊕ 53)x²⁰ + (79 ⊕ 163)x¹⁹ + (56 ⊕ 1)x¹⁸ + (233 ⊕ 208)x¹⁷ + (59 ⊕ 202)x¹⁶ + (95 ⊕ 47)x¹⁵ + (146 ⊕ 246)x¹⁴ + (89 ⊕ 17)x¹³ + (159 ⊕ 49)x¹² + (215 ⊕ 90)x¹¹ + (161 ⊕ 152)x¹⁰ + (0 ⊕ 137)x⁹

The result is:

0x²⁷ + 206x²⁶ + 239x²⁵ + 16x²⁴ + 197x²³ + 106x²² + 192x²¹ + 63x²⁰ + 236x¹⁹ + 57x¹⁸ + 57x¹⁷ + 241x¹⁶ + 112x¹⁵ + 100x¹⁴ + 72x¹³ + 174x¹² + 141x¹¹ + 57x¹⁰ + 137x⁹

Discard the lead 0 term to get:

206x²⁶ + 239x²⁵ + 16x²⁴ + 197x²³ + 106x²² + 192x²¹ + 63x²⁰ + 236x¹⁹ + 57x¹⁸ + 57x¹⁷ + 241x¹⁶ + 112x¹⁵ + 100x¹⁴ + 72x¹³ + 174x¹² + 141x¹¹ + 57x¹⁰ + 137x⁹

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 206x²⁶. Convert 206x²⁶ 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:

(α¹¹¹ * α⁰)x²⁶ + (α¹¹¹ * α²¹⁵)x²⁵ + (α¹¹¹ * α²³⁴)x²⁴ + (α¹¹¹ * α¹⁵⁸)x²³ + (α¹¹¹ * α⁹⁴)x²² + (α¹¹¹ * α¹⁸⁴)x²¹ + (α¹¹¹ * α⁹⁷)x²⁰ + (α¹¹¹ * α¹¹⁸)x¹⁹ + (α¹¹¹ * α¹⁷⁰)x¹⁸ + (α¹¹¹ * α⁷⁹)x¹⁷ + (α¹¹¹ * α¹⁸⁷)x¹⁶ + (α¹¹¹ * α¹⁵²)x¹⁵ + (α¹¹¹ * α¹⁴⁸)x¹⁴ + (α¹¹¹ * α²⁵²)x¹³ + (α¹¹¹ * α¹⁷⁹)x¹² + (α¹¹¹ * α⁵)x¹¹ + (α¹¹¹ * α⁹⁸)x¹⁰ + (α¹¹¹ * α⁹⁶)x⁹ + (α¹¹¹ * 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:

α¹¹¹x²⁶ + α^{(326 % 255)}x²⁵ + α^{(345 % 255)}x²⁴ + α^{(269 % 255)}x²³ + α²⁰⁵x²² + α^{(295 % 255)}x²¹ + α²⁰⁸x²⁰ + α²²⁹x¹⁹ + α^{(281 % 255)}x¹⁸ + α¹⁹⁰x¹⁷ + α^{(298 % 255)}x¹⁶ + α^{(263 % 255)}x¹⁵ + α^{(259 % 255)}x¹⁴ + α^{(363 % 255)}x¹³ + α^{(290 % 255)}x¹² + α¹¹⁶x¹¹ + α²⁰⁹x¹⁰ + α²⁰⁷x⁹ + α^{(264 % 255)}x⁸

The result is:

α¹¹¹x²⁶ + α⁷¹x²⁵ + α⁹⁰x²⁴ + α¹⁴x²³ + α²⁰⁵x²² + α⁴⁰x²¹ + α²⁰⁸x²⁰ + α²²⁹x¹⁹ + α²⁶x¹⁸ + α¹⁹⁰x¹⁷ + α⁴³x¹⁶ + α⁸x¹⁵ + α⁴x¹⁴ + α¹⁰⁸x¹³ + α³⁵x¹² + α¹¹⁶x¹¹ + α²⁰⁹x¹⁰ + α²⁰⁷x⁹ + α⁹x⁸

Now, convert this to integer notation:

206x²⁶ + 188x²⁵ + 223x²⁴ + 19x²³ + 167x²² + 106x²¹ + 81x²⁰ + 122x¹⁹ + 6x¹⁸ + 174x¹⁷ + 119x¹⁶ + 29x¹⁵ + 16x¹⁴ + 208x¹³ + 156x¹² + 248x¹¹ + 162x¹⁰ + 166x⁹ + 58x⁸

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

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

(206 ⊕ 206)x²⁶ + (239 ⊕ 188)x²⁵ + (16 ⊕ 223)x²⁴ + (197 ⊕ 19)x²³ + (106 ⊕ 167)x²² + (192 ⊕ 106)x²¹ + (63 ⊕ 81)x²⁰ + (236 ⊕ 122)x¹⁹ + (57 ⊕ 6)x¹⁸ + (57 ⊕ 174)x¹⁷ + (241 ⊕ 119)x¹⁶ + (112 ⊕ 29)x¹⁵ + (100 ⊕ 16)x¹⁴ + (72 ⊕ 208)x¹³ + (174 ⊕ 156)x¹² + (141 ⊕ 248)x¹¹ + (57 ⊕ 162)x¹⁰ + (137 ⊕ 166)x⁹ + (0 ⊕ 58)x⁸

The result is:

0x²⁶ + 83x²⁵ + 207x²⁴ + 214x²³ + 205x²² + 170x²¹ + 110x²⁰ + 150x¹⁹ + 63x¹⁸ + 151x¹⁷ + 134x¹⁶ + 109x¹⁵ + 116x¹⁴ + 152x¹³ + 50x¹² + 117x¹¹ + 155x¹⁰ + 47x⁹ + 58x⁸

Discard the lead 0 term to get:

83x²⁵ + 207x²⁴ + 214x²³ + 205x²² + 170x²¹ + 110x²⁰ + 150x¹⁹ + 63x¹⁸ + 151x¹⁷ + 134x¹⁶ + 109x¹⁵ + 116x¹⁴ + 152x¹³ + 50x¹² + 117x¹¹ + 155x¹⁰ + 47x⁹ + 58x⁸

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 83x²⁵. Convert 83x²⁵ to alpha notation. According to the log antilog table, for the integer value 83, the alpha exponent is 206. Therefore 83 = α²⁰⁶. Multiply the generator polynomial by α²⁰⁶:

(α²⁰⁶ * α⁰)x²⁵ + (α²⁰⁶ * α²¹⁵)x²⁴ + (α²⁰⁶ * α²³⁴)x²³ + (α²⁰⁶ * α¹⁵⁸)x²² + (α²⁰⁶ * α⁹⁴)x²¹ + (α²⁰⁶ * α¹⁸⁴)x²⁰ + (α²⁰⁶ * α⁹⁷)x¹⁹ + (α²⁰⁶ * α¹¹⁸)x¹⁸ + (α²⁰⁶ * α¹⁷⁰)x¹⁷ + (α²⁰⁶ * α⁷⁹)x¹⁶ + (α²⁰⁶ * α¹⁸⁷)x¹⁵ + (α²⁰⁶ * α¹⁵²)x¹⁴ + (α²⁰⁶ * α¹⁴⁸)x¹³ + (α²⁰⁶ * α²⁵²)x¹² + (α²⁰⁶ * α¹⁷⁹)x¹¹ + (α²⁰⁶ * α⁵)x¹⁰ + (α²⁰⁶ * α⁹⁸)x⁹ + (α²⁰⁶ * α⁹⁶)x⁸ + (α²⁰⁶ * 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:

α²⁰⁶x²⁵ + α^{(421 % 255)}x²⁴ + α^{(440 % 255)}x²³ + α^{(364 % 255)}x²² + α^{(300 % 255)}x²¹ + α^{(390 % 255)}x²⁰ + α^{(303 % 255)}x¹⁹ + α^{(324 % 255)}x¹⁸ + α^{(376 % 255)}x¹⁷ + α^{(285 % 255)}x¹⁶ + α^{(393 % 255)}x¹⁵ + α^{(358 % 255)}x¹⁴ + α^{(354 % 255)}x¹³ + α^{(458 % 255)}x¹² + α^{(385 % 255)}x¹¹ + α²¹¹x¹⁰ + α^{(304 % 255)}x⁹ + α^{(302 % 255)}x⁸ + α^{(359 % 255)}x⁷

The result is:

α²⁰⁶x²⁵ + α¹⁶⁶x²⁴ + α¹⁸⁵x²³ + α¹⁰⁹x²² + α⁴⁵x²¹ + α¹³⁵x²⁰ + α⁴⁸x¹⁹ + α⁶⁹x¹⁸ + α¹²¹x¹⁷ + α³⁰x¹⁶ + α¹³⁸x¹⁵ + α¹⁰³x¹⁴ + α⁹⁹x¹³ + α²⁰³x¹² + α¹³⁰x¹¹ + α²¹¹x¹⁰ + α⁴⁹x⁹ + α⁴⁷x⁸ + α¹⁰⁴x⁷

Now, convert this to integer notation:

83x²⁵ + 63x²⁴ + 55x²³ + 189x²² + 193x²¹ + 169x²⁰ + 70x¹⁹ + 47x¹⁸ + 118x¹⁷ + 96x¹⁶ + 33x¹⁵ + 136x¹⁴ + 134x¹³ + 224x¹² + 46x¹¹ + 178x¹⁰ + 140x⁹ + 35x⁸ + 13x⁷

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

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

(83 ⊕ 83)x²⁵ + (207 ⊕ 63)x²⁴ + (214 ⊕ 55)x²³ + (205 ⊕ 189)x²² + (170 ⊕ 193)x²¹ + (110 ⊕ 169)x²⁰ + (150 ⊕ 70)x¹⁹ + (63 ⊕ 47)x¹⁸ + (151 ⊕ 118)x¹⁷ + (134 ⊕ 96)x¹⁶ + (109 ⊕ 33)x¹⁵ + (116 ⊕ 136)x¹⁴ + (152 ⊕ 134)x¹³ + (50 ⊕ 224)x¹² + (117 ⊕ 46)x¹¹ + (155 ⊕ 178)x¹⁰ + (47 ⊕ 140)x⁹ + (58 ⊕ 35)x⁸ + (0 ⊕ 13)x⁷

The result is:

0x²⁵ + 240x²⁴ + 225x²³ + 112x²² + 107x²¹ + 199x²⁰ + 208x¹⁹ + 16x¹⁸ + 225x¹⁷ + 230x¹⁶ + 76x¹⁵ + 252x¹⁴ + 30x¹³ + 210x¹² + 91x¹¹ + 41x¹⁰ + 163x⁹ + 25x⁸ + 13x⁷

Discard the lead 0 term to get:

240x²⁴ + 225x²³ + 112x²² + 107x²¹ + 199x²⁰ + 208x¹⁹ + 16x¹⁸ + 225x¹⁷ + 230x¹⁶ + 76x¹⁵ + 252x¹⁴ + 30x¹³ + 210x¹² + 91x¹¹ + 41x¹⁰ + 163x⁹ + 25x⁸ + 13x⁷

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 240x²⁴. Convert 240x²⁴ to alpha notation. According to the log antilog table, for the integer value 240, the alpha exponent is 79. Therefore 240 = α⁷⁹. Multiply the generator polynomial by α⁷⁹:

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

α⁷⁹x²⁴ + α^{(294 % 255)}x²³ + α^{(313 % 255)}x²² + α²³⁷x²¹ + α¹⁷³x²⁰ + α^{(263 % 255)}x¹⁹ + α¹⁷⁶x¹⁸ + α¹⁹⁷x¹⁷ + α²⁴⁹x¹⁶ + α¹⁵⁸x¹⁵ + α^{(266 % 255)}x¹⁴ + α²³¹x¹³ + α²²⁷x¹² + α^{(331 % 255)}x¹¹ + α^{(258 % 255)}x¹⁰ + α⁸⁴x⁹ + α¹⁷⁷x⁸ + α¹⁷⁵x⁷ + α²³²x⁶

The result is:

α⁷⁹x²⁴ + α³⁹x²³ + α⁵⁸x²² + α²³⁷x²¹ + α¹⁷³x²⁰ + α⁸x¹⁹ + α¹⁷⁶x¹⁸ + α¹⁹⁷x¹⁷ + α²⁴⁹x¹⁶ + α¹⁵⁸x¹⁵ + α¹¹x¹⁴ + α²³¹x¹³ + α²²⁷x¹² + α⁷⁶x¹¹ + α³x¹⁰ + α⁸⁴x⁹ + α¹⁷⁷x⁸ + α¹⁷⁵x⁷ + α²³²x⁶

Now, convert this to integer notation:

240x²⁴ + 53x²³ + 105x²² + 139x²¹ + 246x²⁰ + 29x¹⁹ + 227x¹⁸ + 141x¹⁷ + 54x¹⁶ + 183x¹⁵ + 232x¹⁴ + 245x¹³ + 144x¹² + 30x¹¹ + 8x¹⁰ + 107x⁹ + 219x⁸ + 255x⁷ + 247x⁶

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

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

(240 ⊕ 240)x²⁴ + (225 ⊕ 53)x²³ + (112 ⊕ 105)x²² + (107 ⊕ 139)x²¹ + (199 ⊕ 246)x²⁰ + (208 ⊕ 29)x¹⁹ + (16 ⊕ 227)x¹⁸ + (225 ⊕ 141)x¹⁷ + (230 ⊕ 54)x¹⁶ + (76 ⊕ 183)x¹⁵ + (252 ⊕ 232)x¹⁴ + (30 ⊕ 245)x¹³ + (210 ⊕ 144)x¹² + (91 ⊕ 30)x¹¹ + (41 ⊕ 8)x¹⁰ + (163 ⊕ 107)x⁹ + (25 ⊕ 219)x⁸ + (13 ⊕ 255)x⁷ + (0 ⊕ 247)x⁶

The result is:

0x²⁴ + 212x²³ + 25x²² + 224x²¹ + 49x²⁰ + 205x¹⁹ + 243x¹⁸ + 108x¹⁷ + 208x¹⁶ + 251x¹⁵ + 20x¹⁴ + 235x¹³ + 66x¹² + 69x¹¹ + 33x¹⁰ + 200x⁹ + 194x⁸ + 242x⁷ + 247x⁶

Discard the lead 0 term to get:

212x²³ + 25x²² + 224x²¹ + 49x²⁰ + 205x¹⁹ + 243x¹⁸ + 108x¹⁷ + 208x¹⁶ + 251x¹⁵ + 20x¹⁴ + 235x¹³ + 66x¹² + 69x¹¹ + 33x¹⁰ + 200x⁹ + 194x⁸ + 242x⁷ + 247x⁶

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 212x²³. Convert 212x²³ to alpha notation. According to the log antilog table, for the integer value 212, the alpha exponent is 41. Therefore 212 = α⁴¹. Multiply the generator polynomial by α⁴¹:

(α⁴¹ * α⁰)x²³ + (α⁴¹ * α²¹⁵)x²² + (α⁴¹ * α²³⁴)x²¹ + (α⁴¹ * α¹⁵⁸)x²⁰ + (α⁴¹ * α⁹⁴)x¹⁹ + (α⁴¹ * α¹⁸⁴)x¹⁸ + (α⁴¹ * α⁹⁷)x¹⁷ + (α⁴¹ * α¹¹⁸)x¹⁶ + (α⁴¹ * α¹⁷⁰)x¹⁵ + (α⁴¹ * α⁷⁹)x¹⁴ + (α⁴¹ * α¹⁸⁷)x¹³ + (α⁴¹ * α¹⁵²)x¹² + (α⁴¹ * α¹⁴⁸)x¹¹ + (α⁴¹ * α²⁵²)x¹⁰ + (α⁴¹ * α¹⁷⁹)x⁹ + (α⁴¹ * α⁵)x⁸ + (α⁴¹ * α⁹⁸)x⁷ + (α⁴¹ * α⁹⁶)x⁶ + (α⁴¹ * 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:

α⁴¹x²³ + α^{(256 % 255)}x²² + α^{(275 % 255)}x²¹ + α¹⁹⁹x²⁰ + α¹³⁵x¹⁹ + α²²⁵x¹⁸ + α¹³⁸x¹⁷ + α¹⁵⁹x¹⁶ + α²¹¹x¹⁵ + α¹²⁰x¹⁴ + α²²⁸x¹³ + α¹⁹³x¹² + α¹⁸⁹x¹¹ + α^{(293 % 255)}x¹⁰ + α²²⁰x⁹ + α⁴⁶x⁸ + α¹³⁹x⁷ + α¹³⁷x⁶ + α¹⁹⁴x⁵

The result is:

α⁴¹x²³ + α¹x²² + α²⁰x²¹ + α¹⁹⁹x²⁰ + α¹³⁵x¹⁹ + α²²⁵x¹⁸ + α¹³⁸x¹⁷ + α¹⁵⁹x¹⁶ + α²¹¹x¹⁵ + α¹²⁰x¹⁴ + α²²⁸x¹³ + α¹⁹³x¹² + α¹⁸⁹x¹¹ + α³⁸x¹⁰ + α²²⁰x⁹ + α⁴⁶x⁸ + α¹³⁹x⁷ + α¹³⁷x⁶ + α¹⁹⁴x⁵

Now, convert this to integer notation:

212x²³ + 2x²² + 180x²¹ + 14x²⁰ + 169x¹⁹ + 36x¹⁸ + 33x¹⁷ + 115x¹⁶ + 178x¹⁵ + 59x¹⁴ + 61x¹³ + 25x¹² + 87x¹¹ + 148x¹⁰ + 172x⁹ + 159x⁸ + 66x⁷ + 158x⁶ + 50x⁵

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

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

(212 ⊕ 212)x²³ + (25 ⊕ 2)x²² + (224 ⊕ 180)x²¹ + (49 ⊕ 14)x²⁰ + (205 ⊕ 169)x¹⁹ + (243 ⊕ 36)x¹⁸ + (108 ⊕ 33)x¹⁷ + (208 ⊕ 115)x¹⁶ + (251 ⊕ 178)x¹⁵ + (20 ⊕ 59)x¹⁴ + (235 ⊕ 61)x¹³ + (66 ⊕ 25)x¹² + (69 ⊕ 87)x¹¹ + (33 ⊕ 148)x¹⁰ + (200 ⊕ 172)x⁹ + (194 ⊕ 159)x⁸ + (242 ⊕ 66)x⁷ + (247 ⊕ 158)x⁶ + (0 ⊕ 50)x⁵

The result is:

0x²³ + 27x²² + 84x²¹ + 63x²⁰ + 100x¹⁹ + 215x¹⁸ + 77x¹⁷ + 163x¹⁶ + 73x¹⁵ + 47x¹⁴ + 214x¹³ + 91x¹² + 18x¹¹ + 181x¹⁰ + 100x⁹ + 93x⁸ + 176x⁷ + 105x⁶ + 50x⁵

Discard the lead 0 term to get:

27x²² + 84x²¹ + 63x²⁰ + 100x¹⁹ + 215x¹⁸ + 77x¹⁷ + 163x¹⁶ + 73x¹⁵ + 47x¹⁴ + 214x¹³ + 91x¹² + 18x¹¹ + 181x¹⁰ + 100x⁹ + 93x⁸ + 176x⁷ + 105x⁶ + 50x⁵

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 27x²². Convert 27x²² to alpha notation. According to the log antilog table, for the integer value 27, the alpha exponent is 248. Therefore 27 = α²⁴⁸. Multiply the generator polynomial by α²⁴⁸:

(α²⁴⁸ * α⁰)x²² + (α²⁴⁸ * α²¹⁵)x²¹ + (α²⁴⁸ * α²³⁴)x²⁰ + (α²⁴⁸ * α¹⁵⁸)x¹⁹ + (α²⁴⁸ * α⁹⁴)x¹⁸ + (α²⁴⁸ * α¹⁸⁴)x¹⁷ + (α²⁴⁸ * α⁹⁷)x¹⁶ + (α²⁴⁸ * α¹¹⁸)x¹⁵ + (α²⁴⁸ * α¹⁷⁰)x¹⁴ + (α²⁴⁸ * α⁷⁹)x¹³ + (α²⁴⁸ * α¹⁸⁷)x¹² + (α²⁴⁸ * α¹⁵²)x¹¹ + (α²⁴⁸ * α¹⁴⁸)x¹⁰ + (α²⁴⁸ * α²⁵²)x⁹ + (α²⁴⁸ * α¹⁷⁹)x⁸ + (α²⁴⁸ * α⁵)x⁷ + (α²⁴⁸ * α⁹⁸)x⁶ + (α²⁴⁸ * α⁹⁶)x⁵ + (α²⁴⁸ * 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:

α²⁴⁸x²² + α^{(463 % 255)}x²¹ + α^{(482 % 255)}x²⁰ + α^{(406 % 255)}x¹⁹ + α^{(342 % 255)}x¹⁸ + α^{(432 % 255)}x¹⁷ + α^{(345 % 255)}x¹⁶ + α^{(366 % 255)}x¹⁵ + α^{(418 % 255)}x¹⁴ + α^{(327 % 255)}x¹³ + α^{(435 % 255)}x¹² + α^{(400 % 255)}x¹¹ + α^{(396 % 255)}x¹⁰ + α^{(500 % 255)}x⁹ + α^{(427 % 255)}x⁸ + α²⁵³x⁷ + α^{(346 % 255)}x⁶ + α^{(344 % 255)}x⁵ + α^{(401 % 255)}x⁴

The result is:

α²⁴⁸x²² + α²⁰⁸x²¹ + α²²⁷x²⁰ + α¹⁵¹x¹⁹ + α⁸⁷x¹⁸ + α¹⁷⁷x¹⁷ + α⁹⁰x¹⁶ + α¹¹¹x¹⁵ + α¹⁶³x¹⁴ + α⁷²x¹³ + α¹⁸⁰x¹² + α¹⁴⁵x¹¹ + α¹⁴¹x¹⁰ + α²⁴⁵x⁹ + α¹⁷²x⁸ + α²⁵³x⁷ + α⁹¹x⁶ + α⁸⁹x⁵ + α¹⁴⁶x⁴

Now, convert this to integer notation:

27x²² + 81x²¹ + 144x²⁰ + 170x¹⁹ + 127x¹⁸ + 219x¹⁷ + 223x¹⁶ + 206x¹⁵ + 99x¹⁴ + 101x¹³ + 150x¹² + 77x¹¹ + 21x¹⁰ + 233x⁹ + 123x⁸ + 71x⁷ + 163x⁶ + 225x⁵ + 154x⁴

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

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

(27 ⊕ 27)x²² + (84 ⊕ 81)x²¹ + (63 ⊕ 144)x²⁰ + (100 ⊕ 170)x¹⁹ + (215 ⊕ 127)x¹⁸ + (77 ⊕ 219)x¹⁷ + (163 ⊕ 223)x¹⁶ + (73 ⊕ 206)x¹⁵ + (47 ⊕ 99)x¹⁴ + (214 ⊕ 101)x¹³ + (91 ⊕ 150)x¹² + (18 ⊕ 77)x¹¹ + (181 ⊕ 21)x¹⁰ + (100 ⊕ 233)x⁹ + (93 ⊕ 123)x⁸ + (176 ⊕ 71)x⁷ + (105 ⊕ 163)x⁶ + (50 ⊕ 225)x⁵ + (0 ⊕ 154)x⁴

The result is:

0x²² + 5x²¹ + 175x²⁰ + 206x¹⁹ + 168x¹⁸ + 150x¹⁷ + 124x¹⁶ + 135x¹⁵ + 76x¹⁴ + 179x¹³ + 205x¹² + 95x¹¹ + 160x¹⁰ + 141x⁹ + 38x⁸ + 247x⁷ + 202x⁶ + 211x⁵ + 154x⁴

Discard the lead 0 term to get:

5x²¹ + 175x²⁰ + 206x¹⁹ + 168x¹⁸ + 150x¹⁷ + 124x¹⁶ + 135x¹⁵ + 76x¹⁴ + 179x¹³ + 205x¹² + 95x¹¹ + 160x¹⁰ + 141x⁹ + 38x⁸ + 247x⁷ + 202x⁶ + 211x⁵ + 154x⁴

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 5x²¹. Convert 5x²¹ to alpha notation. According to the log antilog table, for the integer value 5, the alpha exponent is 50. Therefore 5 = α⁵⁰. Multiply the generator polynomial by α⁵⁰:

(α⁵⁰ * α⁰)x²¹ + (α⁵⁰ * α²¹⁵)x²⁰ + (α⁵⁰ * α²³⁴)x¹⁹ + (α⁵⁰ * α¹⁵⁸)x¹⁸ + (α⁵⁰ * α⁹⁴)x¹⁷ + (α⁵⁰ * α¹⁸⁴)x¹⁶ + (α⁵⁰ * α⁹⁷)x¹⁵ + (α⁵⁰ * α¹¹⁸)x¹⁴ + (α⁵⁰ * α¹⁷⁰)x¹³ + (α⁵⁰ * α⁷⁹)x¹² + (α⁵⁰ * α¹⁸⁷)x¹¹ + (α⁵⁰ * α¹⁵²)x¹⁰ + (α⁵⁰ * α¹⁴⁸)x⁹ + (α⁵⁰ * α²⁵²)x⁸ + (α⁵⁰ * α¹⁷⁹)x⁷ + (α⁵⁰ * α⁵)x⁶ + (α⁵⁰ * α⁹⁸)x⁵ + (α⁵⁰ * α⁹⁶)x⁴ + (α⁵⁰ * 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:

α⁵⁰x²¹ + α^{(265 % 255)}x²⁰ + α^{(284 % 255)}x¹⁹ + α²⁰⁸x¹⁸ + α¹⁴⁴x¹⁷ + α²³⁴x¹⁶ + α¹⁴⁷x¹⁵ + α¹⁶⁸x¹⁴ + α²²⁰x¹³ + α¹²⁹x¹² + α²³⁷x¹¹ + α²⁰²x¹⁰ + α¹⁹⁸x⁹ + α^{(302 % 255)}x⁸ + α²²⁹x⁷ + α⁵⁵x⁶ + α¹⁴⁸x⁵ + α¹⁴⁶x⁴ + α²⁰³x³

The result is:

α⁵⁰x²¹ + α¹⁰x²⁰ + α²⁹x¹⁹ + α²⁰⁸x¹⁸ + α¹⁴⁴x¹⁷ + α²³⁴x¹⁶ + α¹⁴⁷x¹⁵ + α¹⁶⁸x¹⁴ + α²²⁰x¹³ + α¹²⁹x¹² + α²³⁷x¹¹ + α²⁰²x¹⁰ + α¹⁹⁸x⁹ + α⁴⁷x⁸ + α²²⁹x⁷ + α⁵⁵x⁶ + α¹⁴⁸x⁵ + α¹⁴⁶x⁴ + α²⁰³x³

Now, convert this to integer notation:

5x²¹ + 116x²⁰ + 48x¹⁹ + 81x¹⁸ + 168x¹⁷ + 251x¹⁶ + 41x¹⁵ + 252x¹⁴ + 172x¹³ + 23x¹² + 139x¹¹ + 112x¹⁰ + 7x⁹ + 35x⁸ + 122x⁷ + 160x⁶ + 82x⁵ + 154x⁴ + 224x³

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

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

(5 ⊕ 5)x²¹ + (175 ⊕ 116)x²⁰ + (206 ⊕ 48)x¹⁹ + (168 ⊕ 81)x¹⁸ + (150 ⊕ 168)x¹⁷ + (124 ⊕ 251)x¹⁶ + (135 ⊕ 41)x¹⁵ + (76 ⊕ 252)x¹⁴ + (179 ⊕ 172)x¹³ + (205 ⊕ 23)x¹² + (95 ⊕ 139)x¹¹ + (160 ⊕ 112)x¹⁰ + (141 ⊕ 7)x⁹ + (38 ⊕ 35)x⁸ + (247 ⊕ 122)x⁷ + (202 ⊕ 160)x⁶ + (211 ⊕ 82)x⁵ + (154 ⊕ 154)x⁴ + (0 ⊕ 224)x³

The result is:

0x²¹ + 219x²⁰ + 254x¹⁹ + 249x¹⁸ + 62x¹⁷ + 135x¹⁶ + 174x¹⁵ + 176x¹⁴ + 31x¹³ + 218x¹² + 212x¹¹ + 208x¹⁰ + 138x⁹ + 5x⁸ + 141x⁷ + 106x⁶ + 129x⁵ + 0x⁴ + 224x³

Discard the lead 0 term to get:

219x²⁰ + 254x¹⁹ + 249x¹⁸ + 62x¹⁷ + 135x¹⁶ + 174x¹⁵ + 176x¹⁴ + 31x¹³ + 218x¹² + 212x¹¹ + 208x¹⁰ + 138x⁹ + 5x⁸ + 141x⁷ + 106x⁶ + 129x⁵ + 0x⁴ + 224x³

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 219x²⁰. Convert 219x²⁰ to alpha notation. According to the log antilog table, for the integer value 219, the alpha exponent is 177. Therefore 219 = α¹⁷⁷. Multiply the generator polynomial by α¹⁷⁷:

(α¹⁷⁷ * α⁰)x²⁰ + (α¹⁷⁷ * α²¹⁵)x¹⁹ + (α¹⁷⁷ * α²³⁴)x¹⁸ + (α¹⁷⁷ * α¹⁵⁸)x¹⁷ + (α¹⁷⁷ * α⁹⁴)x¹⁶ + (α¹⁷⁷ * α¹⁸⁴)x¹⁵ + (α¹⁷⁷ * α⁹⁷)x¹⁴ + (α¹⁷⁷ * α¹¹⁸)x¹³ + (α¹⁷⁷ * α¹⁷⁰)x¹² + (α¹⁷⁷ * α⁷⁹)x¹¹ + (α¹⁷⁷ * α¹⁸⁷)x¹⁰ + (α¹⁷⁷ * α¹⁵²)x⁹ + (α¹⁷⁷ * α¹⁴⁸)x⁸ + (α¹⁷⁷ * α²⁵²)x⁷ + (α¹⁷⁷ * α¹⁷⁹)x⁶ + (α¹⁷⁷ * α⁵)x⁵ + (α¹⁷⁷ * α⁹⁸)x⁴ + (α¹⁷⁷ * α⁹⁶)x³ + (α¹⁷⁷ * 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:

α¹⁷⁷x²⁰ + α^{(392 % 255)}x¹⁹ + α^{(411 % 255)}x¹⁸ + α^{(335 % 255)}x¹⁷ + α^{(271 % 255)}x¹⁶ + α^{(361 % 255)}x¹⁵ + α^{(274 % 255)}x¹⁴ + α^{(295 % 255)}x¹³ + α^{(347 % 255)}x¹² + α^{(256 % 255)}x¹¹ + α^{(364 % 255)}x¹⁰ + α^{(329 % 255)}x⁹ + α^{(325 % 255)}x⁸ + α^{(429 % 255)}x⁷ + α^{(356 % 255)}x⁶ + α¹⁸²x⁵ + α^{(275 % 255)}x⁴ + α^{(273 % 255)}x³ + α^{(330 % 255)}x²

The result is:

α¹⁷⁷x²⁰ + α¹³⁷x¹⁹ + α¹⁵⁶x¹⁸ + α⁸⁰x¹⁷ + α¹⁶x¹⁶ + α¹⁰⁶x¹⁵ + α¹⁹x¹⁴ + α⁴⁰x¹³ + α⁹²x¹² + α¹x¹¹ + α¹⁰⁹x¹⁰ + α⁷⁴x⁹ + α⁷⁰x⁸ + α¹⁷⁴x⁷ + α¹⁰¹x⁶ + α¹⁸²x⁵ + α²⁰x⁴ + α¹⁸x³ + α⁷⁵x²

Now, convert this to integer notation:

219x²⁰ + 158x¹⁹ + 228x¹⁸ + 253x¹⁷ + 76x¹⁶ + 52x¹⁵ + 90x¹⁴ + 106x¹³ + 91x¹² + 2x¹¹ + 189x¹⁰ + 137x⁹ + 94x⁸ + 241x⁷ + 34x⁶ + 98x⁵ + 180x⁴ + 45x³ + 15x²

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

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

(219 ⊕ 219)x²⁰ + (254 ⊕ 158)x¹⁹ + (249 ⊕ 228)x¹⁸ + (62 ⊕ 253)x¹⁷ + (135 ⊕ 76)x¹⁶ + (174 ⊕ 52)x¹⁵ + (176 ⊕ 90)x¹⁴ + (31 ⊕ 106)x¹³ + (218 ⊕ 91)x¹² + (212 ⊕ 2)x¹¹ + (208 ⊕ 189)x¹⁰ + (138 ⊕ 137)x⁹ + (5 ⊕ 94)x⁸ + (141 ⊕ 241)x⁷ + (106 ⊕ 34)x⁶ + (129 ⊕ 98)x⁵ + (0 ⊕ 180)x⁴ + (224 ⊕ 45)x³ + (0 ⊕ 15)x²

The result is:

0x²⁰ + 96x¹⁹ + 29x¹⁸ + 195x¹⁷ + 203x¹⁶ + 154x¹⁵ + 234x¹⁴ + 117x¹³ + 129x¹² + 214x¹¹ + 109x¹⁰ + 3x⁹ + 91x⁸ + 124x⁷ + 72x⁶ + 227x⁵ + 180x⁴ + 205x³ + 15x²

Discard the lead 0 term to get:

96x¹⁹ + 29x¹⁸ + 195x¹⁷ + 203x¹⁶ + 154x¹⁵ + 234x¹⁴ + 117x¹³ + 129x¹² + 214x¹¹ + 109x¹⁰ + 3x⁹ + 91x⁸ + 124x⁷ + 72x⁶ + 227x⁵ + 180x⁴ + 205x³ + 15x²

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 96x¹⁹. Convert 96x¹⁹ to alpha notation. According to the log antilog table, for the integer value 96, the alpha exponent is 30. Therefore 96 = α³⁰. Multiply the generator polynomial by α³⁰:

(α³⁰ * α⁰)x¹⁹ + (α³⁰ * α²¹⁵)x¹⁸ + (α³⁰ * α²³⁴)x¹⁷ + (α³⁰ * α¹⁵⁸)x¹⁶ + (α³⁰ * α⁹⁴)x¹⁵ + (α³⁰ * α¹⁸⁴)x¹⁴ + (α³⁰ * α⁹⁷)x¹³ + (α³⁰ * α¹¹⁸)x¹² + (α³⁰ * α¹⁷⁰)x¹¹ + (α³⁰ * α⁷⁹)x¹⁰ + (α³⁰ * α¹⁸⁷)x⁹ + (α³⁰ * α¹⁵²)x⁸ + (α³⁰ * α¹⁴⁸)x⁷ + (α³⁰ * α²⁵²)x⁶ + (α³⁰ * α¹⁷⁹)x⁵ + (α³⁰ * α⁵)x⁴ + (α³⁰ * α⁹⁸)x³ + (α³⁰ * α⁹⁶)x² + (α³⁰ * 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:

α³⁰x¹⁹ + α²⁴⁵x¹⁸ + α^{(264 % 255)}x¹⁷ + α¹⁸⁸x¹⁶ + α¹²⁴x¹⁵ + α²¹⁴x¹⁴ + α¹²⁷x¹³ + α¹⁴⁸x¹² + α²⁰⁰x¹¹ + α¹⁰⁹x¹⁰ + α²¹⁷x⁹ + α¹⁸²x⁸ + α¹⁷⁸x⁷ + α^{(282 % 255)}x⁶ + α²⁰⁹x⁵ + α³⁵x⁴ + α¹²⁸x³ + α¹²⁶x² + α¹⁸³x¹

The result is:

α³⁰x¹⁹ + α²⁴⁵x¹⁸ + α⁹x¹⁷ + α¹⁸⁸x¹⁶ + α¹²⁴x¹⁵ + α²¹⁴x¹⁴ + α¹²⁷x¹³ + α¹⁴⁸x¹² + α²⁰⁰x¹¹ + α¹⁰⁹x¹⁰ + α²¹⁷x⁹ + α¹⁸²x⁸ + α¹⁷⁸x⁷ + α²⁷x⁶ + α²⁰⁹x⁵ + α³⁵x⁴ + α¹²⁸x³ + α¹²⁶x² + α¹⁸³x¹

Now, convert this to integer notation:

96x¹⁹ + 233x¹⁸ + 58x¹⁷ + 165x¹⁶ + 151x¹⁵ + 249x¹⁴ + 204x¹³ + 82x¹² + 28x¹¹ + 189x¹⁰ + 155x⁹ + 98x⁸ + 171x⁷ + 12x⁶ + 162x⁵ + 156x⁴ + 133x³ + 102x² + 196x¹

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

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

(96 ⊕ 96)x¹⁹ + (29 ⊕ 233)x¹⁸ + (195 ⊕ 58)x¹⁷ + (203 ⊕ 165)x¹⁶ + (154 ⊕ 151)x¹⁵ + (234 ⊕ 249)x¹⁴ + (117 ⊕ 204)x¹³ + (129 ⊕ 82)x¹² + (214 ⊕ 28)x¹¹ + (109 ⊕ 189)x¹⁰ + (3 ⊕ 155)x⁹ + (91 ⊕ 98)x⁸ + (124 ⊕ 171)x⁷ + (72 ⊕ 12)x⁶ + (227 ⊕ 162)x⁵ + (180 ⊕ 156)x⁴ + (205 ⊕ 133)x³ + (15 ⊕ 102)x² + (0 ⊕ 196)x¹

The result is:

0x¹⁹ + 244x¹⁸ + 249x¹⁷ + 110x¹⁶ + 13x¹⁵ + 19x¹⁴ + 185x¹³ + 211x¹² + 202x¹¹ + 208x¹⁰ + 152x⁹ + 57x⁸ + 215x⁷ + 68x⁶ + 65x⁵ + 40x⁴ + 72x³ + 105x² + 196x¹

Discard the lead 0 term to get:

244x¹⁸ + 249x¹⁷ + 110x¹⁶ + 13x¹⁵ + 19x¹⁴ + 185x¹³ + 211x¹² + 202x¹¹ + 208x¹⁰ + 152x⁹ + 57x⁸ + 215x⁷ + 68x⁶ + 65x⁵ + 40x⁴ + 72x³ + 105x² + 196x¹

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 244x¹⁸. Convert 244x¹⁸ to alpha notation. According to the log antilog table, for the integer value 244, the alpha exponent is 230. Therefore 244 = α²³⁰. Multiply the generator polynomial by α²³⁰:

(α²³⁰ * α⁰)x¹⁸ + (α²³⁰ * α²¹⁵)x¹⁷ + (α²³⁰ * α²³⁴)x¹⁶ + (α²³⁰ * α¹⁵⁸)x¹⁵ + (α²³⁰ * α⁹⁴)x¹⁴ + (α²³⁰ * α¹⁸⁴)x¹³ + (α²³⁰ * α⁹⁷)x¹² + (α²³⁰ * α¹¹⁸)x¹¹ + (α²³⁰ * α¹⁷⁰)x¹⁰ + (α²³⁰ * α⁷⁹)x⁹ + (α²³⁰ * α¹⁸⁷)x⁸ + (α²³⁰ * α¹⁵²)x⁷ + (α²³⁰ * α¹⁴⁸)x⁶ + (α²³⁰ * α²⁵²)x⁵ + (α²³⁰ * α¹⁷⁹)x⁴ + (α²³⁰ * α⁵)x³ + (α²³⁰ * α⁹⁸)x² + (α²³⁰ * α⁹⁶)x¹ + (α²³⁰ * 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:

α²³⁰x¹⁸ + α^{(445 % 255)}x¹⁷ + α^{(464 % 255)}x¹⁶ + α^{(388 % 255)}x¹⁵ + α^{(324 % 255)}x¹⁴ + α^{(414 % 255)}x¹³ + α^{(327 % 255)}x¹² + α^{(348 % 255)}x¹¹ + α^{(400 % 255)}x¹⁰ + α^{(309 % 255)}x⁹ + α^{(417 % 255)}x⁸ + α^{(382 % 255)}x⁷ + α^{(378 % 255)}x⁶ + α^{(482 % 255)}x⁵ + α^{(409 % 255)}x⁴ + α²³⁵x³ + α^{(328 % 255)}x² + α^{(326 % 255)}x¹ + α^{(383 % 255)}x⁰

The result is:

α²³⁰x¹⁸ + α¹⁹⁰x¹⁷ + α²⁰⁹x¹⁶ + α¹³³x¹⁵ + α⁶⁹x¹⁴ + α¹⁵⁹x¹³ + α⁷²x¹² + α⁹³x¹¹ + α¹⁴⁵x¹⁰ + α⁵⁴x⁹ + α¹⁶²x⁸ + α¹²⁷x⁷ + α¹²³x⁶ + α²²⁷x⁵ + α¹⁵⁴x⁴ + α²³⁵x³ + α⁷³x² + α⁷¹x¹ + α¹²⁸x⁰

Now, convert this to integer notation:

244x¹⁸ + 174x¹⁷ + 162x¹⁶ + 109x¹⁵ + 47x¹⁴ + 115x¹³ + 101x¹² + 182x¹¹ + 77x¹⁰ + 80x⁹ + 191x⁸ + 204x⁷ + 197x⁶ + 144x⁵ + 57x⁴ + 235x³ + 202x² + 188x¹ + 133

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

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

(244 ⊕ 244)x¹⁸ + (249 ⊕ 174)x¹⁷ + (110 ⊕ 162)x¹⁶ + (13 ⊕ 109)x¹⁵ + (19 ⊕ 47)x¹⁴ + (185 ⊕ 115)x¹³ + (211 ⊕ 101)x¹² + (202 ⊕ 182)x¹¹ + (208 ⊕ 77)x¹⁰ + (152 ⊕ 80)x⁹ + (57 ⊕ 191)x⁸ + (215 ⊕ 204)x⁷ + (68 ⊕ 197)x⁶ + (65 ⊕ 144)x⁵ + (40 ⊕ 57)x⁴ + (72 ⊕ 235)x³ + (105 ⊕ 202)x² + (196 ⊕ 188)x¹ + (0 ⊕ 133)x⁰

The result is:

0x¹⁸ + 87x¹⁷ + 204x¹⁶ + 96x¹⁵ + 60x¹⁴ + 202x¹³ + 182x¹² + 124x¹¹ + 157x¹⁰ + 200x⁹ + 134x⁸ + 27x⁷ + 129x⁶ + 209x⁵ + 17x⁴ + 163x³ + 163x² + 120x¹ + 133

Discard the lead 0 term to get:

87x¹⁷ + 204x¹⁶ + 96x¹⁵ + 60x¹⁴ + 202x¹³ + 182x¹² + 124x¹¹ + 157x¹⁰ + 200x⁹ + 134x⁸ + 27x⁷ + 129x⁶ + 209x⁵ + 17x⁴ + 163x³ + 163x² + 120x¹ + 133

Use the terms of the remainder as the error correction codewords

The division has been performed 15 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:

87  204  96  60  202  182  124  157  200  134  27  129  209  17  163  163  120  133