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:

32x¹² + 91x¹¹ + 11x¹⁰ + 120x⁹ + 209x⁸ + 114x⁷ + 220x⁶ + 77x⁵ + 67x⁴ + 64x³ + 236x² + 17x¹ + 236

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 13, for 13 error correction codewords, so multiply the message polynomial by x¹³, which gives us:

32x²⁵ + 91x²⁴ + 11x²³ + 120x²² + 209x²¹ + 114x²⁰ + 220x¹⁹ + 77x¹⁸ + 67x¹⁷ + 64x¹⁶ + 236x¹⁵ + 17x¹⁴ + 236x¹³

The lead term of the generator polynomial should also have the same exponent, so multiply by x¹² to get

α⁰x²⁵ + α⁷⁴x²⁴ + α¹⁵²x²³ + α¹⁷⁶x²² + α¹⁰⁰x²¹ + α⁸⁶x²⁰ + α¹⁰⁰x¹⁹ + α¹⁰⁶x¹⁸ + α¹⁰⁴x¹⁷ + α¹³⁰x¹⁶ + α²¹⁸x¹⁵ + α²⁰⁶x¹⁴ + α¹⁴⁰x¹³ + α⁷⁸x¹²

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 13 steps to complete. This will result in a remainder that has 13 terms. These terms will be the 13 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 32x²⁵. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 32x²⁵ to alpha notation. According to the log antilog table, for the integer value 32, the alpha exponent is 5. Therefore 32 = α⁵. Multiply the generator polynomial by α⁵:

The exponents of the alphas are added together. The result is:

α⁵x²⁵ + α⁷⁹x²⁴ + α¹⁵⁷x²³ + α¹⁸¹x²² + α¹⁰⁵x²¹ + α⁹¹x²⁰ + α¹⁰⁵x¹⁹ + α¹¹¹x¹⁸ + α¹⁰⁹x¹⁷ + α¹³⁵x¹⁶ + α²²³x¹⁵ + α²¹¹x¹⁴ + α¹⁴⁵x¹³ + α⁸³x¹²

Now, convert this to integer notation:

32x²⁵ + 240x²⁴ + 213x²³ + 49x²² + 26x²¹ + 163x²⁰ + 26x¹⁹ + 206x¹⁸ + 189x¹⁷ + 169x¹⁶ + 9x¹⁵ + 178x¹⁴ + 77x¹³ + 187x¹²

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.

(32 ⊕ 32)x²⁵ + (91 ⊕ 240)x²⁴ + (11 ⊕ 213)x²³ + (120 ⊕ 49)x²² + (209 ⊕ 26)x²¹ + (114 ⊕ 163)x²⁰ + (220 ⊕ 26)x¹⁹ + (77 ⊕ 206)x¹⁸ + (67 ⊕ 189)x¹⁷ + (64 ⊕ 169)x¹⁶ + (236 ⊕ 9)x¹⁵ + (17 ⊕ 178)x¹⁴ + (236 ⊕ 77)x¹³ + (0 ⊕ 187)x¹²

The result is:

0x²⁵ + 171x²⁴ + 222x²³ + 73x²² + 203x²¹ + 209x²⁰ + 198x¹⁹ + 131x¹⁸ + 254x¹⁷ + 233x¹⁶ + 229x¹⁵ + 163x¹⁴ + 161x¹³ + 187x¹²

Discard the lead 0 term to get:

171x²⁴ + 222x²³ + 73x²² + 203x²¹ + 209x²⁰ + 198x¹⁹ + 131x¹⁸ + 254x¹⁷ + 233x¹⁶ + 229x¹⁵ + 163x¹⁴ + 161x¹³ + 187x¹²

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 171x²⁴. Convert 171x²⁴ to alpha notation. According to the log antilog table, for the integer value 171, the alpha exponent is 178. Therefore 171 = α¹⁷⁸. Multiply the generator polynomial by α¹⁷⁸:

(α¹⁷⁸ * α⁰)x²⁴ + (α¹⁷⁸ * α⁷⁴)x²³ + (α¹⁷⁸ * α¹⁵²)x²² + (α¹⁷⁸ * α¹⁷⁶)x²¹ + (α¹⁷⁸ * α¹⁰⁰)x²⁰ + (α¹⁷⁸ * α⁸⁶)x¹⁹ + (α¹⁷⁸ * α¹⁰⁰)x¹⁸ + (α¹⁷⁸ * α¹⁰⁶)x¹⁷ + (α¹⁷⁸ * α¹⁰⁴)x¹⁶ + (α¹⁷⁸ * α¹³⁰)x¹⁵ + (α¹⁷⁸ * α²¹⁸)x¹⁴ + (α¹⁷⁸ * α²⁰⁶)x¹³ + (α¹⁷⁸ * α¹⁴⁰)x¹² + (α¹⁷⁸ * α⁷⁸)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:

α¹⁷⁸x²⁴ + α²⁵²x²³ + α^{(330 % 255)}x²² + α^{(354 % 255)}x²¹ + α^{(278 % 255)}x²⁰ + α^{(264 % 255)}x¹⁹ + α^{(278 % 255)}x¹⁸ + α^{(284 % 255)}x¹⁷ + α^{(282 % 255)}x¹⁶ + α^{(308 % 255)}x¹⁵ + α^{(396 % 255)}x¹⁴ + α^{(384 % 255)}x¹³ + α^{(318 % 255)}x¹² + α^{(256 % 255)}x¹¹

The result is:

α¹⁷⁸x²⁴ + α²⁵²x²³ + α⁷⁵x²² + α⁹⁹x²¹ + α²³x²⁰ + α⁹x¹⁹ + α²³x¹⁸ + α²⁹x¹⁷ + α²⁷x¹⁶ + α⁵³x¹⁵ + α¹⁴¹x¹⁴ + α¹²⁹x¹³ + α⁶³x¹² + α¹x¹¹

Now, convert this to integer notation:

171x²⁴ + 173x²³ + 15x²² + 134x²¹ + 201x²⁰ + 58x¹⁹ + 201x¹⁸ + 48x¹⁷ + 12x¹⁶ + 40x¹⁵ + 21x¹⁴ + 23x¹³ + 161x¹² + 2x¹¹

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

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

(171 ⊕ 171)x²⁴ + (222 ⊕ 173)x²³ + (73 ⊕ 15)x²² + (203 ⊕ 134)x²¹ + (209 ⊕ 201)x²⁰ + (198 ⊕ 58)x¹⁹ + (131 ⊕ 201)x¹⁸ + (254 ⊕ 48)x¹⁷ + (233 ⊕ 12)x¹⁶ + (229 ⊕ 40)x¹⁵ + (163 ⊕ 21)x¹⁴ + (161 ⊕ 23)x¹³ + (187 ⊕ 161)x¹² + (0 ⊕ 2)x¹¹

The result is:

0x²⁴ + 115x²³ + 70x²² + 77x²¹ + 24x²⁰ + 252x¹⁹ + 74x¹⁸ + 206x¹⁷ + 229x¹⁶ + 205x¹⁵ + 182x¹⁴ + 182x¹³ + 26x¹² + 2x¹¹

Discard the lead 0 term to get:

115x²³ + 70x²² + 77x²¹ + 24x²⁰ + 252x¹⁹ + 74x¹⁸ + 206x¹⁷ + 229x¹⁶ + 205x¹⁵ + 182x¹⁴ + 182x¹³ + 26x¹² + 2x¹¹

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 115x²³. Convert 115x²³ to alpha notation. According to the log antilog table, for the integer value 115, the alpha exponent is 159. Therefore 115 = α¹⁵⁹. Multiply the generator polynomial by α¹⁵⁹:

(α¹⁵⁹ * α⁰)x²³ + (α¹⁵⁹ * α⁷⁴)x²² + (α¹⁵⁹ * α¹⁵²)x²¹ + (α¹⁵⁹ * α¹⁷⁶)x²⁰ + (α¹⁵⁹ * α¹⁰⁰)x¹⁹ + (α¹⁵⁹ * α⁸⁶)x¹⁸ + (α¹⁵⁹ * α¹⁰⁰)x¹⁷ + (α¹⁵⁹ * α¹⁰⁶)x¹⁶ + (α¹⁵⁹ * α¹⁰⁴)x¹⁵ + (α¹⁵⁹ * α¹³⁰)x¹⁴ + (α¹⁵⁹ * α²¹⁸)x¹³ + (α¹⁵⁹ * α²⁰⁶)x¹² + (α¹⁵⁹ * α¹⁴⁰)x¹¹ + (α¹⁵⁹ * α⁷⁸)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:

α¹⁵⁹x²³ + α²³³x²² + α^{(311 % 255)}x²¹ + α^{(335 % 255)}x²⁰ + α^{(259 % 255)}x¹⁹ + α²⁴⁵x¹⁸ + α^{(259 % 255)}x¹⁷ + α^{(265 % 255)}x¹⁶ + α^{(263 % 255)}x¹⁵ + α^{(289 % 255)}x¹⁴ + α^{(377 % 255)}x¹³ + α^{(365 % 255)}x¹² + α^{(299 % 255)}x¹¹ + α²³⁷x¹⁰

The result is:

α¹⁵⁹x²³ + α²³³x²² + α⁵⁶x²¹ + α⁸⁰x²⁰ + α⁴x¹⁹ + α²⁴⁵x¹⁸ + α⁴x¹⁷ + α¹⁰x¹⁶ + α⁸x¹⁵ + α³⁴x¹⁴ + α¹²²x¹³ + α¹¹⁰x¹² + α⁴⁴x¹¹ + α²³⁷x¹⁰

Now, convert this to integer notation:

115x²³ + 243x²² + 93x²¹ + 253x²⁰ + 16x¹⁹ + 233x¹⁸ + 16x¹⁷ + 116x¹⁶ + 29x¹⁵ + 78x¹⁴ + 236x¹³ + 103x¹² + 238x¹¹ + 139x¹⁰

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

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

(115 ⊕ 115)x²³ + (70 ⊕ 243)x²² + (77 ⊕ 93)x²¹ + (24 ⊕ 253)x²⁰ + (252 ⊕ 16)x¹⁹ + (74 ⊕ 233)x¹⁸ + (206 ⊕ 16)x¹⁷ + (229 ⊕ 116)x¹⁶ + (205 ⊕ 29)x¹⁵ + (182 ⊕ 78)x¹⁴ + (182 ⊕ 236)x¹³ + (26 ⊕ 103)x¹² + (2 ⊕ 238)x¹¹ + (0 ⊕ 139)x¹⁰

The result is:

0x²³ + 181x²² + 16x²¹ + 229x²⁰ + 236x¹⁹ + 163x¹⁸ + 222x¹⁷ + 145x¹⁶ + 208x¹⁵ + 248x¹⁴ + 90x¹³ + 125x¹² + 236x¹¹ + 139x¹⁰

Discard the lead 0 term to get:

181x²² + 16x²¹ + 229x²⁰ + 236x¹⁹ + 163x¹⁸ + 222x¹⁷ + 145x¹⁶ + 208x¹⁵ + 248x¹⁴ + 90x¹³ + 125x¹² + 236x¹¹ + 139x¹⁰

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 181x²². Convert 181x²² to alpha notation. According to the log antilog table, for the integer value 181, the alpha exponent is 42. Therefore 181 = α⁴². Multiply the generator polynomial by α⁴²:

(α⁴² * α⁰)x²² + (α⁴² * α⁷⁴)x²¹ + (α⁴² * α¹⁵²)x²⁰ + (α⁴² * α¹⁷⁶)x¹⁹ + (α⁴² * α¹⁰⁰)x¹⁸ + (α⁴² * α⁸⁶)x¹⁷ + (α⁴² * α¹⁰⁰)x¹⁶ + (α⁴² * α¹⁰⁶)x¹⁵ + (α⁴² * α¹⁰⁴)x¹⁴ + (α⁴² * α¹³⁰)x¹³ + (α⁴² * α²¹⁸)x¹² + (α⁴² * α²⁰⁶)x¹¹ + (α⁴² * α¹⁴⁰)x¹⁰ + (α⁴² * α⁷⁸)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:

α⁴²x²² + α¹¹⁶x²¹ + α¹⁹⁴x²⁰ + α²¹⁸x¹⁹ + α¹⁴²x¹⁸ + α¹²⁸x¹⁷ + α¹⁴²x¹⁶ + α¹⁴⁸x¹⁵ + α¹⁴⁶x¹⁴ + α¹⁷²x¹³ + α^{(260 % 255)}x¹² + α²⁴⁸x¹¹ + α¹⁸²x¹⁰ + α¹²⁰x⁹

The result is:

α⁴²x²² + α¹¹⁶x²¹ + α¹⁹⁴x²⁰ + α²¹⁸x¹⁹ + α¹⁴²x¹⁸ + α¹²⁸x¹⁷ + α¹⁴²x¹⁶ + α¹⁴⁸x¹⁵ + α¹⁴⁶x¹⁴ + α¹⁷²x¹³ + α⁵x¹² + α²⁴⁸x¹¹ + α¹⁸²x¹⁰ + α¹²⁰x⁹

Now, convert this to integer notation:

181x²² + 248x²¹ + 50x²⁰ + 43x¹⁹ + 42x¹⁸ + 133x¹⁷ + 42x¹⁶ + 82x¹⁵ + 154x¹⁴ + 123x¹³ + 32x¹² + 27x¹¹ + 98x¹⁰ + 59x⁹

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

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

(181 ⊕ 181)x²² + (16 ⊕ 248)x²¹ + (229 ⊕ 50)x²⁰ + (236 ⊕ 43)x¹⁹ + (163 ⊕ 42)x¹⁸ + (222 ⊕ 133)x¹⁷ + (145 ⊕ 42)x¹⁶ + (208 ⊕ 82)x¹⁵ + (248 ⊕ 154)x¹⁴ + (90 ⊕ 123)x¹³ + (125 ⊕ 32)x¹² + (236 ⊕ 27)x¹¹ + (139 ⊕ 98)x¹⁰ + (0 ⊕ 59)x⁹

The result is:

0x²² + 232x²¹ + 215x²⁰ + 199x¹⁹ + 137x¹⁸ + 91x¹⁷ + 187x¹⁶ + 130x¹⁵ + 98x¹⁴ + 33x¹³ + 93x¹² + 247x¹¹ + 233x¹⁰ + 59x⁹

Discard the lead 0 term to get:

232x²¹ + 215x²⁰ + 199x¹⁹ + 137x¹⁸ + 91x¹⁷ + 187x¹⁶ + 130x¹⁵ + 98x¹⁴ + 33x¹³ + 93x¹² + 247x¹¹ + 233x¹⁰ + 59x⁹

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 232x²¹. Convert 232x²¹ to alpha notation. According to the log antilog table, for the integer value 232, the alpha exponent is 11. Therefore 232 = α¹¹. Multiply the generator polynomial by α¹¹:

(α¹¹ * α⁰)x²¹ + (α¹¹ * α⁷⁴)x²⁰ + (α¹¹ * α¹⁵²)x¹⁹ + (α¹¹ * α¹⁷⁶)x¹⁸ + (α¹¹ * α¹⁰⁰)x¹⁷ + (α¹¹ * α⁸⁶)x¹⁶ + (α¹¹ * α¹⁰⁰)x¹⁵ + (α¹¹ * α¹⁰⁶)x¹⁴ + (α¹¹ * α¹⁰⁴)x¹³ + (α¹¹ * α¹³⁰)x¹² + (α¹¹ * α²¹⁸)x¹¹ + (α¹¹ * α²⁰⁶)x¹⁰ + (α¹¹ * α¹⁴⁰)x⁹ + (α¹¹ * α⁷⁸)x⁸

The exponents of the alphas are added together. The result is:

α¹¹x²¹ + α⁸⁵x²⁰ + α¹⁶³x¹⁹ + α¹⁸⁷x¹⁸ + α¹¹¹x¹⁷ + α⁹⁷x¹⁶ + α¹¹¹x¹⁵ + α¹¹⁷x¹⁴ + α¹¹⁵x¹³ + α¹⁴¹x¹² + α²²⁹x¹¹ + α²¹⁷x¹⁰ + α¹⁵¹x⁹ + α⁸⁹x⁸

Now, convert this to integer notation:

232x²¹ + 214x²⁰ + 99x¹⁹ + 220x¹⁸ + 206x¹⁷ + 175x¹⁶ + 206x¹⁵ + 237x¹⁴ + 124x¹³ + 21x¹² + 122x¹¹ + 155x¹⁰ + 170x⁹ + 225x⁸

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

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

(232 ⊕ 232)x²¹ + (215 ⊕ 214)x²⁰ + (199 ⊕ 99)x¹⁹ + (137 ⊕ 220)x¹⁸ + (91 ⊕ 206)x¹⁷ + (187 ⊕ 175)x¹⁶ + (130 ⊕ 206)x¹⁵ + (98 ⊕ 237)x¹⁴ + (33 ⊕ 124)x¹³ + (93 ⊕ 21)x¹² + (247 ⊕ 122)x¹¹ + (233 ⊕ 155)x¹⁰ + (59 ⊕ 170)x⁹ + (0 ⊕ 225)x⁸

The result is:

0x²¹ + 1x²⁰ + 164x¹⁹ + 85x¹⁸ + 149x¹⁷ + 20x¹⁶ + 76x¹⁵ + 143x¹⁴ + 93x¹³ + 72x¹² + 141x¹¹ + 114x¹⁰ + 145x⁹ + 225x⁸

Discard the lead 0 term to get:

1x²⁰ + 164x¹⁹ + 85x¹⁸ + 149x¹⁷ + 20x¹⁶ + 76x¹⁵ + 143x¹⁴ + 93x¹³ + 72x¹² + 141x¹¹ + 114x¹⁰ + 145x⁹ + 225x⁸

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 1x²⁰. Convert 1x²⁰ to alpha notation. According to the log antilog table, for the integer value 1, the alpha exponent is 0. Therefore 1 = α⁰. Multiply the generator polynomial by α⁰:

(α⁰ * α⁰)x²⁰ + (α⁰ * α⁷⁴)x¹⁹ + (α⁰ * α¹⁵²)x¹⁸ + (α⁰ * α¹⁷⁶)x¹⁷ + (α⁰ * α¹⁰⁰)x¹⁶ + (α⁰ * α⁸⁶)x¹⁵ + (α⁰ * α¹⁰⁰)x¹⁴ + (α⁰ * α¹⁰⁶)x¹³ + (α⁰ * α¹⁰⁴)x¹² + (α⁰ * α¹³⁰)x¹¹ + (α⁰ * α²¹⁸)x¹⁰ + (α⁰ * α²⁰⁶)x⁹ + (α⁰ * α¹⁴⁰)x⁸ + (α⁰ * α⁷⁸)x⁷

The exponents of the alphas are added together. The result is:

α⁰x²⁰ + α⁷⁴x¹⁹ + α¹⁵²x¹⁸ + α¹⁷⁶x¹⁷ + α¹⁰⁰x¹⁶ + α⁸⁶x¹⁵ + α¹⁰⁰x¹⁴ + α¹⁰⁶x¹³ + α¹⁰⁴x¹² + α¹³⁰x¹¹ + α²¹⁸x¹⁰ + α²⁰⁶x⁹ + α¹⁴⁰x⁸ + α⁷⁸x⁷

Now, convert this to integer notation:

1x²⁰ + 137x¹⁹ + 73x¹⁸ + 227x¹⁷ + 17x¹⁶ + 177x¹⁵ + 17x¹⁴ + 52x¹³ + 13x¹² + 46x¹¹ + 43x¹⁰ + 83x⁹ + 132x⁸ + 120x⁷

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

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

(1 ⊕ 1)x²⁰ + (164 ⊕ 137)x¹⁹ + (85 ⊕ 73)x¹⁸ + (149 ⊕ 227)x¹⁷ + (20 ⊕ 17)x¹⁶ + (76 ⊕ 177)x¹⁵ + (143 ⊕ 17)x¹⁴ + (93 ⊕ 52)x¹³ + (72 ⊕ 13)x¹² + (141 ⊕ 46)x¹¹ + (114 ⊕ 43)x¹⁰ + (145 ⊕ 83)x⁹ + (225 ⊕ 132)x⁸ + (0 ⊕ 120)x⁷

The result is:

0x²⁰ + 45x¹⁹ + 28x¹⁸ + 118x¹⁷ + 5x¹⁶ + 253x¹⁵ + 158x¹⁴ + 105x¹³ + 69x¹² + 163x¹¹ + 89x¹⁰ + 194x⁹ + 101x⁸ + 120x⁷

Discard the lead 0 term to get:

45x¹⁹ + 28x¹⁸ + 118x¹⁷ + 5x¹⁶ + 253x¹⁵ + 158x¹⁴ + 105x¹³ + 69x¹² + 163x¹¹ + 89x¹⁰ + 194x⁹ + 101x⁸ + 120x⁷

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 45x¹⁹. Convert 45x¹⁹ to alpha notation. According to the log antilog table, for the integer value 45, the alpha exponent is 18. Therefore 45 = α¹⁸. Multiply the generator polynomial by α¹⁸:

(α¹⁸ * α⁰)x¹⁹ + (α¹⁸ * α⁷⁴)x¹⁸ + (α¹⁸ * α¹⁵²)x¹⁷ + (α¹⁸ * α¹⁷⁶)x¹⁶ + (α¹⁸ * α¹⁰⁰)x¹⁵ + (α¹⁸ * α⁸⁶)x¹⁴ + (α¹⁸ * α¹⁰⁰)x¹³ + (α¹⁸ * α¹⁰⁶)x¹² + (α¹⁸ * α¹⁰⁴)x¹¹ + (α¹⁸ * α¹³⁰)x¹⁰ + (α¹⁸ * α²¹⁸)x⁹ + (α¹⁸ * α²⁰⁶)x⁸ + (α¹⁸ * α¹⁴⁰)x⁷ + (α¹⁸ * α⁷⁸)x⁶

The exponents of the alphas are added together. The result is:

α¹⁸x¹⁹ + α⁹²x¹⁸ + α¹⁷⁰x¹⁷ + α¹⁹⁴x¹⁶ + α¹¹⁸x¹⁵ + α¹⁰⁴x¹⁴ + α¹¹⁸x¹³ + α¹²⁴x¹² + α¹²²x¹¹ + α¹⁴⁸x¹⁰ + α²³⁶x⁹ + α²²⁴x⁸ + α¹⁵⁸x⁷ + α⁹⁶x⁶

Now, convert this to integer notation:

45x¹⁹ + 91x¹⁸ + 215x¹⁷ + 50x¹⁶ + 199x¹⁵ + 13x¹⁴ + 199x¹³ + 151x¹² + 236x¹¹ + 82x¹⁰ + 203x⁹ + 18x⁸ + 183x⁷ + 217x⁶

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

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

(45 ⊕ 45)x¹⁹ + (28 ⊕ 91)x¹⁸ + (118 ⊕ 215)x¹⁷ + (5 ⊕ 50)x¹⁶ + (253 ⊕ 199)x¹⁵ + (158 ⊕ 13)x¹⁴ + (105 ⊕ 199)x¹³ + (69 ⊕ 151)x¹² + (163 ⊕ 236)x¹¹ + (89 ⊕ 82)x¹⁰ + (194 ⊕ 203)x⁹ + (101 ⊕ 18)x⁸ + (120 ⊕ 183)x⁷ + (0 ⊕ 217)x⁶

The result is:

0x¹⁹ + 71x¹⁸ + 161x¹⁷ + 55x¹⁶ + 58x¹⁵ + 147x¹⁴ + 174x¹³ + 210x¹² + 79x¹¹ + 11x¹⁰ + 9x⁹ + 119x⁸ + 207x⁷ + 217x⁶

Discard the lead 0 term to get:

71x¹⁸ + 161x¹⁷ + 55x¹⁶ + 58x¹⁵ + 147x¹⁴ + 174x¹³ + 210x¹² + 79x¹¹ + 11x¹⁰ + 9x⁹ + 119x⁸ + 207x⁷ + 217x⁶

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 71x¹⁸. Convert 71x¹⁸ to alpha notation. According to the log antilog table, for the integer value 71, the alpha exponent is 253. Therefore 71 = α²⁵³. Multiply the generator polynomial by α²⁵³:

(α²⁵³ * α⁰)x¹⁸ + (α²⁵³ * α⁷⁴)x¹⁷ + (α²⁵³ * α¹⁵²)x¹⁶ + (α²⁵³ * α¹⁷⁶)x¹⁵ + (α²⁵³ * α¹⁰⁰)x¹⁴ + (α²⁵³ * α⁸⁶)x¹³ + (α²⁵³ * α¹⁰⁰)x¹² + (α²⁵³ * α¹⁰⁶)x¹¹ + (α²⁵³ * α¹⁰⁴)x¹⁰ + (α²⁵³ * α¹³⁰)x⁹ + (α²⁵³ * α²¹⁸)x⁸ + (α²⁵³ * α²⁰⁶)x⁷ + (α²⁵³ * α¹⁴⁰)x⁶ + (α²⁵³ * α⁷⁸)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:

α²⁵³x¹⁸ + α^{(327 % 255)}x¹⁷ + α^{(405 % 255)}x¹⁶ + α^{(429 % 255)}x¹⁵ + α^{(353 % 255)}x¹⁴ + α^{(339 % 255)}x¹³ + α^{(353 % 255)}x¹² + α^{(359 % 255)}x¹¹ + α^{(357 % 255)}x¹⁰ + α^{(383 % 255)}x⁹ + α^{(471 % 255)}x⁸ + α^{(459 % 255)}x⁷ + α^{(393 % 255)}x⁶ + α^{(331 % 255)}x⁵

The result is:

α²⁵³x¹⁸ + α⁷²x¹⁷ + α¹⁵⁰x¹⁶ + α¹⁷⁴x¹⁵ + α⁹⁸x¹⁴ + α⁸⁴x¹³ + α⁹⁸x¹² + α¹⁰⁴x¹¹ + α¹⁰²x¹⁰ + α¹²⁸x⁹ + α²¹⁶x⁸ + α²⁰⁴x⁷ + α¹³⁸x⁶ + α⁷⁶x⁵

Now, convert this to integer notation:

71x¹⁸ + 101x¹⁷ + 85x¹⁶ + 241x¹⁵ + 67x¹⁴ + 107x¹³ + 67x¹² + 13x¹¹ + 68x¹⁰ + 133x⁹ + 195x⁸ + 221x⁷ + 33x⁶ + 30x⁵

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

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

(71 ⊕ 71)x¹⁸ + (161 ⊕ 101)x¹⁷ + (55 ⊕ 85)x¹⁶ + (58 ⊕ 241)x¹⁵ + (147 ⊕ 67)x¹⁴ + (174 ⊕ 107)x¹³ + (210 ⊕ 67)x¹² + (79 ⊕ 13)x¹¹ + (11 ⊕ 68)x¹⁰ + (9 ⊕ 133)x⁹ + (119 ⊕ 195)x⁸ + (207 ⊕ 221)x⁷ + (217 ⊕ 33)x⁶ + (0 ⊕ 30)x⁵

The result is:

0x¹⁸ + 196x¹⁷ + 98x¹⁶ + 203x¹⁵ + 208x¹⁴ + 197x¹³ + 145x¹² + 66x¹¹ + 79x¹⁰ + 140x⁹ + 180x⁸ + 18x⁷ + 248x⁶ + 30x⁵

Discard the lead 0 term to get:

196x¹⁷ + 98x¹⁶ + 203x¹⁵ + 208x¹⁴ + 197x¹³ + 145x¹² + 66x¹¹ + 79x¹⁰ + 140x⁹ + 180x⁸ + 18x⁷ + 248x⁶ + 30x⁵

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 196x¹⁷. Convert 196x¹⁷ to alpha notation. According to the log antilog table, for the integer value 196, the alpha exponent is 183. Therefore 196 = α¹⁸³. Multiply the generator polynomial by α¹⁸³:

(α¹⁸³ * α⁰)x¹⁷ + (α¹⁸³ * α⁷⁴)x¹⁶ + (α¹⁸³ * α¹⁵²)x¹⁵ + (α¹⁸³ * α¹⁷⁶)x¹⁴ + (α¹⁸³ * α¹⁰⁰)x¹³ + (α¹⁸³ * α⁸⁶)x¹² + (α¹⁸³ * α¹⁰⁰)x¹¹ + (α¹⁸³ * α¹⁰⁶)x¹⁰ + (α¹⁸³ * α¹⁰⁴)x⁹ + (α¹⁸³ * α¹³⁰)x⁸ + (α¹⁸³ * α²¹⁸)x⁷ + (α¹⁸³ * α²⁰⁶)x⁶ + (α¹⁸³ * α¹⁴⁰)x⁵ + (α¹⁸³ * α⁷⁸)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:

α¹⁸³x¹⁷ + α^{(257 % 255)}x¹⁶ + α^{(335 % 255)}x¹⁵ + α^{(359 % 255)}x¹⁴ + α^{(283 % 255)}x¹³ + α^{(269 % 255)}x¹² + α^{(283 % 255)}x¹¹ + α^{(289 % 255)}x¹⁰ + α^{(287 % 255)}x⁹ + α^{(313 % 255)}x⁸ + α^{(401 % 255)}x⁷ + α^{(389 % 255)}x⁶ + α^{(323 % 255)}x⁵ + α^{(261 % 255)}x⁴

The result is:

α¹⁸³x¹⁷ + α²x¹⁶ + α⁸⁰x¹⁵ + α¹⁰⁴x¹⁴ + α²⁸x¹³ + α¹⁴x¹² + α²⁸x¹¹ + α³⁴x¹⁰ + α³²x⁹ + α⁵⁸x⁸ + α¹⁴⁶x⁷ + α¹³⁴x⁶ + α⁶⁸x⁵ + α⁶x⁴

Now, convert this to integer notation:

196x¹⁷ + 4x¹⁶ + 253x¹⁵ + 13x¹⁴ + 24x¹³ + 19x¹² + 24x¹¹ + 78x¹⁰ + 157x⁹ + 105x⁸ + 154x⁷ + 218x⁶ + 153x⁵ + 64x⁴

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

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

(196 ⊕ 196)x¹⁷ + (98 ⊕ 4)x¹⁶ + (203 ⊕ 253)x¹⁵ + (208 ⊕ 13)x¹⁴ + (197 ⊕ 24)x¹³ + (145 ⊕ 19)x¹² + (66 ⊕ 24)x¹¹ + (79 ⊕ 78)x¹⁰ + (140 ⊕ 157)x⁹ + (180 ⊕ 105)x⁸ + (18 ⊕ 154)x⁷ + (248 ⊕ 218)x⁶ + (30 ⊕ 153)x⁵ + (0 ⊕ 64)x⁴

The result is:

0x¹⁷ + 102x¹⁶ + 54x¹⁵ + 221x¹⁴ + 221x¹³ + 130x¹² + 90x¹¹ + 1x¹⁰ + 17x⁹ + 221x⁸ + 136x⁷ + 34x⁶ + 135x⁵ + 64x⁴

Discard the lead 0 term to get:

102x¹⁶ + 54x¹⁵ + 221x¹⁴ + 221x¹³ + 130x¹² + 90x¹¹ + 1x¹⁰ + 17x⁹ + 221x⁸ + 136x⁷ + 34x⁶ + 135x⁵ + 64x⁴

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 102x¹⁶. Convert 102x¹⁶ to alpha notation. According to the log antilog table, for the integer value 102, the alpha exponent is 126. Therefore 102 = α¹²⁶. Multiply the generator polynomial by α¹²⁶:

(α¹²⁶ * α⁰)x¹⁶ + (α¹²⁶ * α⁷⁴)x¹⁵ + (α¹²⁶ * α¹⁵²)x¹⁴ + (α¹²⁶ * α¹⁷⁶)x¹³ + (α¹²⁶ * α¹⁰⁰)x¹² + (α¹²⁶ * α⁸⁶)x¹¹ + (α¹²⁶ * α¹⁰⁰)x¹⁰ + (α¹²⁶ * α¹⁰⁶)x⁹ + (α¹²⁶ * α¹⁰⁴)x⁸ + (α¹²⁶ * α¹³⁰)x⁷ + (α¹²⁶ * α²¹⁸)x⁶ + (α¹²⁶ * α²⁰⁶)x⁵ + (α¹²⁶ * α¹⁴⁰)x⁴ + (α¹²⁶ * α⁷⁸)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:

α¹²⁶x¹⁶ + α²⁰⁰x¹⁵ + α^{(278 % 255)}x¹⁴ + α^{(302 % 255)}x¹³ + α²²⁶x¹² + α²¹²x¹¹ + α²²⁶x¹⁰ + α²³²x⁹ + α²³⁰x⁸ + α^{(256 % 255)}x⁷ + α^{(344 % 255)}x⁶ + α^{(332 % 255)}x⁵ + α^{(266 % 255)}x⁴ + α²⁰⁴x³

The result is:

α¹²⁶x¹⁶ + α²⁰⁰x¹⁵ + α²³x¹⁴ + α⁴⁷x¹³ + α²²⁶x¹² + α²¹²x¹¹ + α²²⁶x¹⁰ + α²³²x⁹ + α²³⁰x⁸ + α¹x⁷ + α⁸⁹x⁶ + α⁷⁷x⁵ + α¹¹x⁴ + α²⁰⁴x³

Now, convert this to integer notation:

102x¹⁶ + 28x¹⁵ + 201x¹⁴ + 35x¹³ + 72x¹² + 121x¹¹ + 72x¹⁰ + 247x⁹ + 244x⁸ + 2x⁷ + 225x⁶ + 60x⁵ + 232x⁴ + 221x³

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

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

(102 ⊕ 102)x¹⁶ + (54 ⊕ 28)x¹⁵ + (221 ⊕ 201)x¹⁴ + (221 ⊕ 35)x¹³ + (130 ⊕ 72)x¹² + (90 ⊕ 121)x¹¹ + (1 ⊕ 72)x¹⁰ + (17 ⊕ 247)x⁹ + (221 ⊕ 244)x⁸ + (136 ⊕ 2)x⁷ + (34 ⊕ 225)x⁶ + (135 ⊕ 60)x⁵ + (64 ⊕ 232)x⁴ + (0 ⊕ 221)x³

The result is:

0x¹⁶ + 42x¹⁵ + 20x¹⁴ + 254x¹³ + 202x¹² + 35x¹¹ + 73x¹⁰ + 230x⁹ + 41x⁸ + 138x⁷ + 195x⁶ + 187x⁵ + 168x⁴ + 221x³

Discard the lead 0 term to get:

42x¹⁵ + 20x¹⁴ + 254x¹³ + 202x¹² + 35x¹¹ + 73x¹⁰ + 230x⁹ + 41x⁸ + 138x⁷ + 195x⁶ + 187x⁵ + 168x⁴ + 221x³

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 42x¹⁵. Convert 42x¹⁵ to alpha notation. According to the log antilog table, for the integer value 42, the alpha exponent is 142. Therefore 42 = α¹⁴². Multiply the generator polynomial by α¹⁴²:

(α¹⁴² * α⁰)x¹⁵ + (α¹⁴² * α⁷⁴)x¹⁴ + (α¹⁴² * α¹⁵²)x¹³ + (α¹⁴² * α¹⁷⁶)x¹² + (α¹⁴² * α¹⁰⁰)x¹¹ + (α¹⁴² * α⁸⁶)x¹⁰ + (α¹⁴² * α¹⁰⁰)x⁹ + (α¹⁴² * α¹⁰⁶)x⁸ + (α¹⁴² * α¹⁰⁴)x⁷ + (α¹⁴² * α¹³⁰)x⁶ + (α¹⁴² * α²¹⁸)x⁵ + (α¹⁴² * α²⁰⁶)x⁴ + (α¹⁴² * α¹⁴⁰)x³ + (α¹⁴² * α⁷⁸)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:

α¹⁴²x¹⁵ + α²¹⁶x¹⁴ + α^{(294 % 255)}x¹³ + α^{(318 % 255)}x¹² + α²⁴²x¹¹ + α²²⁸x¹⁰ + α²⁴²x⁹ + α²⁴⁸x⁸ + α²⁴⁶x⁷ + α^{(272 % 255)}x⁶ + α^{(360 % 255)}x⁵ + α^{(348 % 255)}x⁴ + α^{(282 % 255)}x³ + α²²⁰x²

The result is:

α¹⁴²x¹⁵ + α²¹⁶x¹⁴ + α³⁹x¹³ + α⁶³x¹² + α²⁴²x¹¹ + α²²⁸x¹⁰ + α²⁴²x⁹ + α²⁴⁸x⁸ + α²⁴⁶x⁷ + α¹⁷x⁶ + α¹⁰⁵x⁵ + α⁹³x⁴ + α²⁷x³ + α²²⁰x²

Now, convert this to integer notation:

42x¹⁵ + 195x¹⁴ + 53x¹³ + 161x¹² + 176x¹¹ + 61x¹⁰ + 176x⁹ + 27x⁸ + 207x⁷ + 152x⁶ + 26x⁵ + 182x⁴ + 12x³ + 172x²

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

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

(42 ⊕ 42)x¹⁵ + (20 ⊕ 195)x¹⁴ + (254 ⊕ 53)x¹³ + (202 ⊕ 161)x¹² + (35 ⊕ 176)x¹¹ + (73 ⊕ 61)x¹⁰ + (230 ⊕ 176)x⁹ + (41 ⊕ 27)x⁸ + (138 ⊕ 207)x⁷ + (195 ⊕ 152)x⁶ + (187 ⊕ 26)x⁵ + (168 ⊕ 182)x⁴ + (221 ⊕ 12)x³ + (0 ⊕ 172)x²

The result is:

0x¹⁵ + 215x¹⁴ + 203x¹³ + 107x¹² + 147x¹¹ + 116x¹⁰ + 86x⁹ + 50x⁸ + 69x⁷ + 91x⁶ + 161x⁵ + 30x⁴ + 209x³ + 172x²

Discard the lead 0 term to get:

215x¹⁴ + 203x¹³ + 107x¹² + 147x¹¹ + 116x¹⁰ + 86x⁹ + 50x⁸ + 69x⁷ + 91x⁶ + 161x⁵ + 30x⁴ + 209x³ + 172x²

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 215x¹⁴. Convert 215x¹⁴ to alpha notation. According to the log antilog table, for the integer value 215, the alpha exponent is 170. Therefore 215 = α¹⁷⁰. Multiply the generator polynomial by α¹⁷⁰:

(α¹⁷⁰ * α⁰)x¹⁴ + (α¹⁷⁰ * α⁷⁴)x¹³ + (α¹⁷⁰ * α¹⁵²)x¹² + (α¹⁷⁰ * α¹⁷⁶)x¹¹ + (α¹⁷⁰ * α¹⁰⁰)x¹⁰ + (α¹⁷⁰ * α⁸⁶)x⁹ + (α¹⁷⁰ * α¹⁰⁰)x⁸ + (α¹⁷⁰ * α¹⁰⁶)x⁷ + (α¹⁷⁰ * α¹⁰⁴)x⁶ + (α¹⁷⁰ * α¹³⁰)x⁵ + (α¹⁷⁰ * α²¹⁸)x⁴ + (α¹⁷⁰ * α²⁰⁶)x³ + (α¹⁷⁰ * α¹⁴⁰)x² + (α¹⁷⁰ * α⁷⁸)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:

α¹⁷⁰x¹⁴ + α²⁴⁴x¹³ + α^{(322 % 255)}x¹² + α^{(346 % 255)}x¹¹ + α^{(270 % 255)}x¹⁰ + α^{(256 % 255)}x⁹ + α^{(270 % 255)}x⁸ + α^{(276 % 255)}x⁷ + α^{(274 % 255)}x⁶ + α^{(300 % 255)}x⁵ + α^{(388 % 255)}x⁴ + α^{(376 % 255)}x³ + α^{(310 % 255)}x² + α²⁴⁸x¹

The result is:

α¹⁷⁰x¹⁴ + α²⁴⁴x¹³ + α⁶⁷x¹² + α⁹¹x¹¹ + α¹⁵x¹⁰ + α¹x⁹ + α¹⁵x⁸ + α²¹x⁷ + α¹⁹x⁶ + α⁴⁵x⁵ + α¹³³x⁴ + α¹²¹x³ + α⁵⁵x² + α²⁴⁸x¹

Now, convert this to integer notation:

215x¹⁴ + 250x¹³ + 194x¹² + 163x¹¹ + 38x¹⁰ + 2x⁹ + 38x⁸ + 117x⁷ + 90x⁶ + 193x⁵ + 109x⁴ + 118x³ + 160x² + 27x¹

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

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

(215 ⊕ 215)x¹⁴ + (203 ⊕ 250)x¹³ + (107 ⊕ 194)x¹² + (147 ⊕ 163)x¹¹ + (116 ⊕ 38)x¹⁰ + (86 ⊕ 2)x⁹ + (50 ⊕ 38)x⁸ + (69 ⊕ 117)x⁷ + (91 ⊕ 90)x⁶ + (161 ⊕ 193)x⁵ + (30 ⊕ 109)x⁴ + (209 ⊕ 118)x³ + (172 ⊕ 160)x² + (0 ⊕ 27)x¹

The result is:

0x¹⁴ + 49x¹³ + 169x¹² + 48x¹¹ + 82x¹⁰ + 84x⁹ + 20x⁸ + 48x⁷ + 1x⁶ + 96x⁵ + 115x⁴ + 167x³ + 12x² + 27x¹

Discard the lead 0 term to get:

49x¹³ + 169x¹² + 48x¹¹ + 82x¹⁰ + 84x⁹ + 20x⁸ + 48x⁷ + 1x⁶ + 96x⁵ + 115x⁴ + 167x³ + 12x² + 27x¹

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 49x¹³. Convert 49x¹³ to alpha notation. According to the log antilog table, for the integer value 49, the alpha exponent is 181. Therefore 49 = α¹⁸¹. Multiply the generator polynomial by α¹⁸¹:

(α¹⁸¹ * α⁰)x¹³ + (α¹⁸¹ * α⁷⁴)x¹² + (α¹⁸¹ * α¹⁵²)x¹¹ + (α¹⁸¹ * α¹⁷⁶)x¹⁰ + (α¹⁸¹ * α¹⁰⁰)x⁹ + (α¹⁸¹ * α⁸⁶)x⁸ + (α¹⁸¹ * α¹⁰⁰)x⁷ + (α¹⁸¹ * α¹⁰⁶)x⁶ + (α¹⁸¹ * α¹⁰⁴)x⁵ + (α¹⁸¹ * α¹³⁰)x⁴ + (α¹⁸¹ * α²¹⁸)x³ + (α¹⁸¹ * α²⁰⁶)x² + (α¹⁸¹ * α¹⁴⁰)x¹ + (α¹⁸¹ * α⁷⁸)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:

α¹⁸¹x¹³ + α²⁵⁵x¹² + α^{(333 % 255)}x¹¹ + α^{(357 % 255)}x¹⁰ + α^{(281 % 255)}x⁹ + α^{(267 % 255)}x⁸ + α^{(281 % 255)}x⁷ + α^{(287 % 255)}x⁶ + α^{(285 % 255)}x⁵ + α^{(311 % 255)}x⁴ + α^{(399 % 255)}x³ + α^{(387 % 255)}x² + α^{(321 % 255)}x¹ + α^{(259 % 255)}x⁰

The result is:

α¹⁸¹x¹³ + α⁰x¹² + α⁷⁸x¹¹ + α¹⁰²x¹⁰ + α²⁶x⁹ + α¹²x⁸ + α²⁶x⁷ + α³²x⁶ + α³⁰x⁵ + α⁵⁶x⁴ + α¹⁴⁴x³ + α¹³²x² + α⁶⁶x¹ + α⁴x⁰

Now, convert this to integer notation:

49x¹³ + 1x¹² + 120x¹¹ + 68x¹⁰ + 6x⁹ + 205x⁸ + 6x⁷ + 157x⁶ + 96x⁵ + 93x⁴ + 168x³ + 184x² + 97x¹ + 16

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

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

(49 ⊕ 49)x¹³ + (169 ⊕ 1)x¹² + (48 ⊕ 120)x¹¹ + (82 ⊕ 68)x¹⁰ + (84 ⊕ 6)x⁹ + (20 ⊕ 205)x⁸ + (48 ⊕ 6)x⁷ + (1 ⊕ 157)x⁶ + (96 ⊕ 96)x⁵ + (115 ⊕ 93)x⁴ + (167 ⊕ 168)x³ + (12 ⊕ 184)x² + (27 ⊕ 97)x¹ + (0 ⊕ 16)x⁰

The result is:

0x¹³ + 168x¹² + 72x¹¹ + 22x¹⁰ + 82x⁹ + 217x⁸ + 54x⁷ + 156x⁶ + 0x⁵ + 46x⁴ + 15x³ + 180x² + 122x¹ + 16

Discard the lead 0 term to get:

168x¹² + 72x¹¹ + 22x¹⁰ + 82x⁹ + 217x⁸ + 54x⁷ + 156x⁶ + 0x⁵ + 46x⁴ + 15x³ + 180x² + 122x¹ + 16

Use the terms of the remainder as the error correction codewords

The division has been performed 13 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:

168  72  22  82  217  54  156  0  46  15  180  122  16