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:

64x¹⁵ + 149x¹⁴ + 6x¹³ + 22x¹² + 118x¹¹ + 86x¹⁰ + 68x⁹ + 247x⁸ + 87x⁷ + 66x⁶ + 16x⁵ + 236x⁴ + 17x³ + 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 10, for 10 error correction codewords, so multiply the message polynomial by x¹⁰, which gives us:

64x²⁵ + 149x²⁴ + 6x²³ + 22x²² + 118x²¹ + 86x²⁰ + 68x¹⁹ + 247x¹⁸ + 87x¹⁷ + 66x¹⁶ + 16x¹⁵ + 236x¹⁴ + 17x¹³ + 236x¹² + 17x¹¹ + 236x¹⁰

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

α⁰x²⁵ + α²⁵¹x²⁴ + α⁶⁷x²³ + α⁴⁶x²² + α⁶¹x²¹ + α¹¹⁸x²⁰ + α⁷⁰x¹⁹ + α⁶⁴x¹⁸ + α⁹⁴x¹⁷ + α³²x¹⁶ + α⁴⁵x¹⁵

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 10 terms. These terms will be the 10 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 64x²⁵. Since alpha notation makes it easier to perform the multiplication, it is recommended to convert 64x²⁵ to alpha notation. According to the log antilog table, for the integer value 64, the alpha exponent is 6. Therefore 64 = α⁶. Multiply the generator polynomial by α⁶:

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:

α⁶x²⁵ + α^{(257 % 255)}x²⁴ + α⁷³x²³ + α⁵²x²² + α⁶⁷x²¹ + α¹²⁴x²⁰ + α⁷⁶x¹⁹ + α⁷⁰x¹⁸ + α¹⁰⁰x¹⁷ + α³⁸x¹⁶ + α⁵¹x¹⁵

The result is:

α⁶x²⁵ + α²x²⁴ + α⁷³x²³ + α⁵²x²² + α⁶⁷x²¹ + α¹²⁴x²⁰ + α⁷⁶x¹⁹ + α⁷⁰x¹⁸ + α¹⁰⁰x¹⁷ + α³⁸x¹⁶ + α⁵¹x¹⁵

Now, convert this to integer notation:

64x²⁵ + 4x²⁴ + 202x²³ + 20x²² + 194x²¹ + 151x²⁰ + 30x¹⁹ + 94x¹⁸ + 17x¹⁷ + 148x¹⁶ + 10x¹⁵

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.

(64 ⊕ 64)x²⁵ + (149 ⊕ 4)x²⁴ + (6 ⊕ 202)x²³ + (22 ⊕ 20)x²² + (118 ⊕ 194)x²¹ + (86 ⊕ 151)x²⁰ + (68 ⊕ 30)x¹⁹ + (247 ⊕ 94)x¹⁸ + (87 ⊕ 17)x¹⁷ + (66 ⊕ 148)x¹⁶ + (16 ⊕ 10)x¹⁵ + (236 ⊕ 0)x¹⁴ + (17 ⊕ 0)x¹³ + (236 ⊕ 0)x¹² + (17 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²⁵ + 145x²⁴ + 204x²³ + 2x²² + 180x²¹ + 193x²⁰ + 90x¹⁹ + 169x¹⁸ + 70x¹⁷ + 214x¹⁶ + 26x¹⁵ + 236x¹⁴ + 17x¹³ + 236x¹² + 17x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

145x²⁴ + 204x²³ + 2x²² + 180x²¹ + 193x²⁰ + 90x¹⁹ + 169x¹⁸ + 70x¹⁷ + 214x¹⁶ + 26x¹⁵ + 236x¹⁴ + 17x¹³ + 236x¹² + 17x¹¹ + 236x¹⁰

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 145x²⁴. Convert 145x²⁴ 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)x²⁴ + (α¹⁶⁵ * α²⁵¹)x²³ + (α¹⁶⁵ * α⁶⁷)x²² + (α¹⁶⁵ * α⁴⁶)x²¹ + (α¹⁶⁵ * α⁶¹)x²⁰ + (α¹⁶⁵ * α¹¹⁸)x¹⁹ + (α¹⁶⁵ * α⁷⁰)x¹⁸ + (α¹⁶⁵ * α⁶⁴)x¹⁷ + (α¹⁶⁵ * α⁹⁴)x¹⁶ + (α¹⁶⁵ * α³²)x¹⁵ + (α¹⁶⁵ * α⁴⁵)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:

α¹⁶⁵x²⁴ + α^{(416 % 255)}x²³ + α²³²x²² + α²¹¹x²¹ + α²²⁶x²⁰ + α^{(283 % 255)}x¹⁹ + α²³⁵x¹⁸ + α²²⁹x¹⁷ + α^{(259 % 255)}x¹⁶ + α¹⁹⁷x¹⁵ + α²¹⁰x¹⁴

The result is:

α¹⁶⁵x²⁴ + α¹⁶¹x²³ + α²³²x²² + α²¹¹x²¹ + α²²⁶x²⁰ + α²⁸x¹⁹ + α²³⁵x¹⁸ + α²²⁹x¹⁷ + α⁴x¹⁶ + α¹⁹⁷x¹⁵ + α²¹⁰x¹⁴

Now, convert this to integer notation:

145x²⁴ + 209x²³ + 247x²² + 178x²¹ + 72x²⁰ + 24x¹⁹ + 235x¹⁸ + 122x¹⁷ + 16x¹⁶ + 141x¹⁵ + 89x¹⁴

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

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

(145 ⊕ 145)x²⁴ + (204 ⊕ 209)x²³ + (2 ⊕ 247)x²² + (180 ⊕ 178)x²¹ + (193 ⊕ 72)x²⁰ + (90 ⊕ 24)x¹⁹ + (169 ⊕ 235)x¹⁸ + (70 ⊕ 122)x¹⁷ + (214 ⊕ 16)x¹⁶ + (26 ⊕ 141)x¹⁵ + (236 ⊕ 89)x¹⁴ + (17 ⊕ 0)x¹³ + (236 ⊕ 0)x¹² + (17 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²⁴ + 29x²³ + 245x²² + 6x²¹ + 137x²⁰ + 66x¹⁹ + 66x¹⁸ + 60x¹⁷ + 198x¹⁶ + 151x¹⁵ + 181x¹⁴ + 17x¹³ + 236x¹² + 17x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

29x²³ + 245x²² + 6x²¹ + 137x²⁰ + 66x¹⁹ + 66x¹⁸ + 60x¹⁷ + 198x¹⁶ + 151x¹⁵ + 181x¹⁴ + 17x¹³ + 236x¹² + 17x¹¹ + 236x¹⁰

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 29x²³. Convert 29x²³ to alpha notation. According to the log antilog table, for the integer value 29, the alpha exponent is 8. Therefore 29 = α⁸. Multiply the generator polynomial by α⁸:

(α⁸ * α⁰)x²³ + (α⁸ * α²⁵¹)x²² + (α⁸ * α⁶⁷)x²¹ + (α⁸ * α⁴⁶)x²⁰ + (α⁸ * α⁶¹)x¹⁹ + (α⁸ * α¹¹⁸)x¹⁸ + (α⁸ * α⁷⁰)x¹⁷ + (α⁸ * α⁶⁴)x¹⁶ + (α⁸ * α⁹⁴)x¹⁵ + (α⁸ * α³²)x¹⁴ + (α⁸ * α⁴⁵)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:

α⁸x²³ + α^{(259 % 255)}x²² + α⁷⁵x²¹ + α⁵⁴x²⁰ + α⁶⁹x¹⁹ + α¹²⁶x¹⁸ + α⁷⁸x¹⁷ + α⁷²x¹⁶ + α¹⁰²x¹⁵ + α⁴⁰x¹⁴ + α⁵³x¹³

The result is:

α⁸x²³ + α⁴x²² + α⁷⁵x²¹ + α⁵⁴x²⁰ + α⁶⁹x¹⁹ + α¹²⁶x¹⁸ + α⁷⁸x¹⁷ + α⁷²x¹⁶ + α¹⁰²x¹⁵ + α⁴⁰x¹⁴ + α⁵³x¹³

Now, convert this to integer notation:

29x²³ + 16x²² + 15x²¹ + 80x²⁰ + 47x¹⁹ + 102x¹⁸ + 120x¹⁷ + 101x¹⁶ + 68x¹⁵ + 106x¹⁴ + 40x¹³

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

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

(29 ⊕ 29)x²³ + (245 ⊕ 16)x²² + (6 ⊕ 15)x²¹ + (137 ⊕ 80)x²⁰ + (66 ⊕ 47)x¹⁹ + (66 ⊕ 102)x¹⁸ + (60 ⊕ 120)x¹⁷ + (198 ⊕ 101)x¹⁶ + (151 ⊕ 68)x¹⁵ + (181 ⊕ 106)x¹⁴ + (17 ⊕ 40)x¹³ + (236 ⊕ 0)x¹² + (17 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²³ + 229x²² + 9x²¹ + 217x²⁰ + 109x¹⁹ + 36x¹⁸ + 68x¹⁷ + 163x¹⁶ + 211x¹⁵ + 223x¹⁴ + 57x¹³ + 236x¹² + 17x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

229x²² + 9x²¹ + 217x²⁰ + 109x¹⁹ + 36x¹⁸ + 68x¹⁷ + 163x¹⁶ + 211x¹⁵ + 223x¹⁴ + 57x¹³ + 236x¹² + 17x¹¹ + 236x¹⁰

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 229x²². Convert 229x²² to alpha notation. According to the log antilog table, for the integer value 229, the alpha exponent is 169. Therefore 229 = α¹⁶⁹. Multiply the generator polynomial by α¹⁶⁹:

(α¹⁶⁹ * α⁰)x²² + (α¹⁶⁹ * α²⁵¹)x²¹ + (α¹⁶⁹ * α⁶⁷)x²⁰ + (α¹⁶⁹ * α⁴⁶)x¹⁹ + (α¹⁶⁹ * α⁶¹)x¹⁸ + (α¹⁶⁹ * α¹¹⁸)x¹⁷ + (α¹⁶⁹ * α⁷⁰)x¹⁶ + (α¹⁶⁹ * α⁶⁴)x¹⁵ + (α¹⁶⁹ * α⁹⁴)x¹⁴ + (α¹⁶⁹ * α³²)x¹³ + (α¹⁶⁹ * α⁴⁵)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:

α¹⁶⁹x²² + α^{(420 % 255)}x²¹ + α²³⁶x²⁰ + α²¹⁵x¹⁹ + α²³⁰x¹⁸ + α^{(287 % 255)}x¹⁷ + α²³⁹x¹⁶ + α²³³x¹⁵ + α^{(263 % 255)}x¹⁴ + α²⁰¹x¹³ + α²¹⁴x¹²

The result is:

α¹⁶⁹x²² + α¹⁶⁵x²¹ + α²³⁶x²⁰ + α²¹⁵x¹⁹ + α²³⁰x¹⁸ + α³²x¹⁷ + α²³⁹x¹⁶ + α²³³x¹⁵ + α⁸x¹⁴ + α²⁰¹x¹³ + α²¹⁴x¹²

Now, convert this to integer notation:

229x²² + 145x²¹ + 203x²⁰ + 239x¹⁹ + 244x¹⁸ + 157x¹⁷ + 22x¹⁶ + 243x¹⁵ + 29x¹⁴ + 56x¹³ + 249x¹²

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

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

(229 ⊕ 229)x²² + (9 ⊕ 145)x²¹ + (217 ⊕ 203)x²⁰ + (109 ⊕ 239)x¹⁹ + (36 ⊕ 244)x¹⁸ + (68 ⊕ 157)x¹⁷ + (163 ⊕ 22)x¹⁶ + (211 ⊕ 243)x¹⁵ + (223 ⊕ 29)x¹⁴ + (57 ⊕ 56)x¹³ + (236 ⊕ 249)x¹² + (17 ⊕ 0)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²² + 152x²¹ + 18x²⁰ + 130x¹⁹ + 208x¹⁸ + 217x¹⁷ + 181x¹⁶ + 32x¹⁵ + 194x¹⁴ + 1x¹³ + 21x¹² + 17x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

152x²¹ + 18x²⁰ + 130x¹⁹ + 208x¹⁸ + 217x¹⁷ + 181x¹⁶ + 32x¹⁵ + 194x¹⁴ + 1x¹³ + 21x¹² + 17x¹¹ + 236x¹⁰

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 152x²¹. Convert 152x²¹ to alpha notation. According to the log antilog table, for the integer value 152, the alpha exponent is 17. Therefore 152 = α¹⁷. Multiply the generator polynomial by α¹⁷:

(α¹⁷ * α⁰)x²¹ + (α¹⁷ * α²⁵¹)x²⁰ + (α¹⁷ * α⁶⁷)x¹⁹ + (α¹⁷ * α⁴⁶)x¹⁸ + (α¹⁷ * α⁶¹)x¹⁷ + (α¹⁷ * α¹¹⁸)x¹⁶ + (α¹⁷ * α⁷⁰)x¹⁵ + (α¹⁷ * α⁶⁴)x¹⁴ + (α¹⁷ * α⁹⁴)x¹³ + (α¹⁷ * α³²)x¹² + (α¹⁷ * α⁴⁵)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:

α¹⁷x²¹ + α^{(268 % 255)}x²⁰ + α⁸⁴x¹⁹ + α⁶³x¹⁸ + α⁷⁸x¹⁷ + α¹³⁵x¹⁶ + α⁸⁷x¹⁵ + α⁸¹x¹⁴ + α¹¹¹x¹³ + α⁴⁹x¹² + α⁶²x¹¹

The result is:

α¹⁷x²¹ + α¹³x²⁰ + α⁸⁴x¹⁹ + α⁶³x¹⁸ + α⁷⁸x¹⁷ + α¹³⁵x¹⁶ + α⁸⁷x¹⁵ + α⁸¹x¹⁴ + α¹¹¹x¹³ + α⁴⁹x¹² + α⁶²x¹¹

Now, convert this to integer notation:

152x²¹ + 135x²⁰ + 107x¹⁹ + 161x¹⁸ + 120x¹⁷ + 169x¹⁶ + 127x¹⁵ + 231x¹⁴ + 206x¹³ + 140x¹² + 222x¹¹

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

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

(152 ⊕ 152)x²¹ + (18 ⊕ 135)x²⁰ + (130 ⊕ 107)x¹⁹ + (208 ⊕ 161)x¹⁸ + (217 ⊕ 120)x¹⁷ + (181 ⊕ 169)x¹⁶ + (32 ⊕ 127)x¹⁵ + (194 ⊕ 231)x¹⁴ + (1 ⊕ 206)x¹³ + (21 ⊕ 140)x¹² + (17 ⊕ 222)x¹¹ + (236 ⊕ 0)x¹⁰

The result is:

0x²¹ + 149x²⁰ + 233x¹⁹ + 113x¹⁸ + 161x¹⁷ + 28x¹⁶ + 95x¹⁵ + 37x¹⁴ + 207x¹³ + 153x¹² + 207x¹¹ + 236x¹⁰

Discard the lead 0 term to get:

149x²⁰ + 233x¹⁹ + 113x¹⁸ + 161x¹⁷ + 28x¹⁶ + 95x¹⁵ + 37x¹⁴ + 207x¹³ + 153x¹² + 207x¹¹ + 236x¹⁰

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 149x²⁰. Convert 149x²⁰ to alpha notation. According to the log antilog table, for the integer value 149, the alpha exponent is 184. Therefore 149 = α¹⁸⁴. Multiply the generator polynomial by α¹⁸⁴:

(α¹⁸⁴ * α⁰)x²⁰ + (α¹⁸⁴ * α²⁵¹)x¹⁹ + (α¹⁸⁴ * α⁶⁷)x¹⁸ + (α¹⁸⁴ * α⁴⁶)x¹⁷ + (α¹⁸⁴ * α⁶¹)x¹⁶ + (α¹⁸⁴ * α¹¹⁸)x¹⁵ + (α¹⁸⁴ * α⁷⁰)x¹⁴ + (α¹⁸⁴ * α⁶⁴)x¹³ + (α¹⁸⁴ * α⁹⁴)x¹² + (α¹⁸⁴ * α³²)x¹¹ + (α¹⁸⁴ * α⁴⁵)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:

α¹⁸⁴x²⁰ + α^{(435 % 255)}x¹⁹ + α²⁵¹x¹⁸ + α²³⁰x¹⁷ + α²⁴⁵x¹⁶ + α^{(302 % 255)}x¹⁵ + α²⁵⁴x¹⁴ + α²⁴⁸x¹³ + α^{(278 % 255)}x¹² + α²¹⁶x¹¹ + α²²⁹x¹⁰

The result is:

α¹⁸⁴x²⁰ + α¹⁸⁰x¹⁹ + α²⁵¹x¹⁸ + α²³⁰x¹⁷ + α²⁴⁵x¹⁶ + α⁴⁷x¹⁵ + α²⁵⁴x¹⁴ + α²⁴⁸x¹³ + α²³x¹² + α²¹⁶x¹¹ + α²²⁹x¹⁰

Now, convert this to integer notation:

149x²⁰ + 150x¹⁹ + 216x¹⁸ + 244x¹⁷ + 233x¹⁶ + 35x¹⁵ + 142x¹⁴ + 27x¹³ + 201x¹² + 195x¹¹ + 122x¹⁰

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

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

(149 ⊕ 149)x²⁰ + (233 ⊕ 150)x¹⁹ + (113 ⊕ 216)x¹⁸ + (161 ⊕ 244)x¹⁷ + (28 ⊕ 233)x¹⁶ + (95 ⊕ 35)x¹⁵ + (37 ⊕ 142)x¹⁴ + (207 ⊕ 27)x¹³ + (153 ⊕ 201)x¹² + (207 ⊕ 195)x¹¹ + (236 ⊕ 122)x¹⁰

The result is:

0x²⁰ + 127x¹⁹ + 169x¹⁸ + 85x¹⁷ + 245x¹⁶ + 124x¹⁵ + 171x¹⁴ + 212x¹³ + 80x¹² + 12x¹¹ + 150x¹⁰

Discard the lead 0 term to get:

127x¹⁹ + 169x¹⁸ + 85x¹⁷ + 245x¹⁶ + 124x¹⁵ + 171x¹⁴ + 212x¹³ + 80x¹² + 12x¹¹ + 150x¹⁰

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 127x¹⁹. Convert 127x¹⁹ to alpha notation. According to the log antilog table, for the integer value 127, the alpha exponent is 87. Therefore 127 = α⁸⁷. Multiply the generator polynomial by α⁸⁷:

(α⁸⁷ * α⁰)x¹⁹ + (α⁸⁷ * α²⁵¹)x¹⁸ + (α⁸⁷ * α⁶⁷)x¹⁷ + (α⁸⁷ * α⁴⁶)x¹⁶ + (α⁸⁷ * α⁶¹)x¹⁵ + (α⁸⁷ * α¹¹⁸)x¹⁴ + (α⁸⁷ * α⁷⁰)x¹³ + (α⁸⁷ * α⁶⁴)x¹² + (α⁸⁷ * α⁹⁴)x¹¹ + (α⁸⁷ * α³²)x¹⁰ + (α⁸⁷ * α⁴⁵)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:

α⁸⁷x¹⁹ + α^{(338 % 255)}x¹⁸ + α¹⁵⁴x¹⁷ + α¹³³x¹⁶ + α¹⁴⁸x¹⁵ + α²⁰⁵x¹⁴ + α¹⁵⁷x¹³ + α¹⁵¹x¹² + α¹⁸¹x¹¹ + α¹¹⁹x¹⁰ + α¹³²x⁹

The result is:

α⁸⁷x¹⁹ + α⁸³x¹⁸ + α¹⁵⁴x¹⁷ + α¹³³x¹⁶ + α¹⁴⁸x¹⁵ + α²⁰⁵x¹⁴ + α¹⁵⁷x¹³ + α¹⁵¹x¹² + α¹⁸¹x¹¹ + α¹¹⁹x¹⁰ + α¹³²x⁹

Now, convert this to integer notation:

127x¹⁹ + 187x¹⁸ + 57x¹⁷ + 109x¹⁶ + 82x¹⁵ + 167x¹⁴ + 213x¹³ + 170x¹² + 49x¹¹ + 147x¹⁰ + 184x⁹

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

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

(127 ⊕ 127)x¹⁹ + (169 ⊕ 187)x¹⁸ + (85 ⊕ 57)x¹⁷ + (245 ⊕ 109)x¹⁶ + (124 ⊕ 82)x¹⁵ + (171 ⊕ 167)x¹⁴ + (212 ⊕ 213)x¹³ + (80 ⊕ 170)x¹² + (12 ⊕ 49)x¹¹ + (150 ⊕ 147)x¹⁰ + (0 ⊕ 184)x⁹

The result is:

0x¹⁹ + 18x¹⁸ + 108x¹⁷ + 152x¹⁶ + 46x¹⁵ + 12x¹⁴ + 1x¹³ + 250x¹² + 61x¹¹ + 5x¹⁰ + 184x⁹

Discard the lead 0 term to get:

18x¹⁸ + 108x¹⁷ + 152x¹⁶ + 46x¹⁵ + 12x¹⁴ + 1x¹³ + 250x¹² + 61x¹¹ + 5x¹⁰ + 184x⁹

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 18x¹⁸. Convert 18x¹⁸ to alpha notation. According to the log antilog table, for the integer value 18, the alpha exponent is 224. Therefore 18 = α²²⁴. Multiply the generator polynomial by α²²⁴:

(α²²⁴ * α⁰)x¹⁸ + (α²²⁴ * α²⁵¹)x¹⁷ + (α²²⁴ * α⁶⁷)x¹⁶ + (α²²⁴ * α⁴⁶)x¹⁵ + (α²²⁴ * α⁶¹)x¹⁴ + (α²²⁴ * α¹¹⁸)x¹³ + (α²²⁴ * α⁷⁰)x¹² + (α²²⁴ * α⁶⁴)x¹¹ + (α²²⁴ * α⁹⁴)x¹⁰ + (α²²⁴ * α³²)x⁹ + (α²²⁴ * α⁴⁵)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:

α²²⁴x¹⁸ + α^{(475 % 255)}x¹⁷ + α^{(291 % 255)}x¹⁶ + α^{(270 % 255)}x¹⁵ + α^{(285 % 255)}x¹⁴ + α^{(342 % 255)}x¹³ + α^{(294 % 255)}x¹² + α^{(288 % 255)}x¹¹ + α^{(318 % 255)}x¹⁰ + α^{(256 % 255)}x⁹ + α^{(269 % 255)}x⁸

The result is:

α²²⁴x¹⁸ + α²²⁰x¹⁷ + α³⁶x¹⁶ + α¹⁵x¹⁵ + α³⁰x¹⁴ + α⁸⁷x¹³ + α³⁹x¹² + α³³x¹¹ + α⁶³x¹⁰ + α¹x⁹ + α¹⁴x⁸

Now, convert this to integer notation:

18x¹⁸ + 172x¹⁷ + 37x¹⁶ + 38x¹⁵ + 96x¹⁴ + 127x¹³ + 53x¹² + 39x¹¹ + 161x¹⁰ + 2x⁹ + 19x⁸

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

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

(18 ⊕ 18)x¹⁸ + (108 ⊕ 172)x¹⁷ + (152 ⊕ 37)x¹⁶ + (46 ⊕ 38)x¹⁵ + (12 ⊕ 96)x¹⁴ + (1 ⊕ 127)x¹³ + (250 ⊕ 53)x¹² + (61 ⊕ 39)x¹¹ + (5 ⊕ 161)x¹⁰ + (184 ⊕ 2)x⁹ + (0 ⊕ 19)x⁸

The result is:

0x¹⁸ + 192x¹⁷ + 189x¹⁶ + 8x¹⁵ + 108x¹⁴ + 126x¹³ + 207x¹² + 26x¹¹ + 164x¹⁰ + 186x⁹ + 19x⁸

Discard the lead 0 term to get:

192x¹⁷ + 189x¹⁶ + 8x¹⁵ + 108x¹⁴ + 126x¹³ + 207x¹² + 26x¹¹ + 164x¹⁰ + 186x⁹ + 19x⁸

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 192x¹⁷. Convert 192x¹⁷ to alpha notation. According to the log antilog table, for the integer value 192, the alpha exponent is 31. Therefore 192 = α³¹. Multiply the generator polynomial by α³¹:

(α³¹ * α⁰)x¹⁷ + (α³¹ * α²⁵¹)x¹⁶ + (α³¹ * α⁶⁷)x¹⁵ + (α³¹ * α⁴⁶)x¹⁴ + (α³¹ * α⁶¹)x¹³ + (α³¹ * α¹¹⁸)x¹² + (α³¹ * α⁷⁰)x¹¹ + (α³¹ * α⁶⁴)x¹⁰ + (α³¹ * α⁹⁴)x⁹ + (α³¹ * α³²)x⁸ + (α³¹ * α⁴⁵)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:

α³¹x¹⁷ + α^{(282 % 255)}x¹⁶ + α⁹⁸x¹⁵ + α⁷⁷x¹⁴ + α⁹²x¹³ + α¹⁴⁹x¹² + α¹⁰¹x¹¹ + α⁹⁵x¹⁰ + α¹²⁵x⁹ + α⁶³x⁸ + α⁷⁶x⁷

The result is:

α³¹x¹⁷ + α²⁷x¹⁶ + α⁹⁸x¹⁵ + α⁷⁷x¹⁴ + α⁹²x¹³ + α¹⁴⁹x¹² + α¹⁰¹x¹¹ + α⁹⁵x¹⁰ + α¹²⁵x⁹ + α⁶³x⁸ + α⁷⁶x⁷

Now, convert this to integer notation:

192x¹⁷ + 12x¹⁶ + 67x¹⁵ + 60x¹⁴ + 91x¹³ + 164x¹² + 34x¹¹ + 226x¹⁰ + 51x⁹ + 161x⁸ + 30x⁷

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

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

(192 ⊕ 192)x¹⁷ + (189 ⊕ 12)x¹⁶ + (8 ⊕ 67)x¹⁵ + (108 ⊕ 60)x¹⁴ + (126 ⊕ 91)x¹³ + (207 ⊕ 164)x¹² + (26 ⊕ 34)x¹¹ + (164 ⊕ 226)x¹⁰ + (186 ⊕ 51)x⁹ + (19 ⊕ 161)x⁸ + (0 ⊕ 30)x⁷

The result is:

0x¹⁷ + 177x¹⁶ + 75x¹⁵ + 80x¹⁴ + 37x¹³ + 107x¹² + 56x¹¹ + 70x¹⁰ + 137x⁹ + 178x⁸ + 30x⁷

Discard the lead 0 term to get:

177x¹⁶ + 75x¹⁵ + 80x¹⁴ + 37x¹³ + 107x¹² + 56x¹¹ + 70x¹⁰ + 137x⁹ + 178x⁸ + 30x⁷

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 177x¹⁶. Convert 177x¹⁶ to alpha notation. According to the log antilog table, for the integer value 177, the alpha exponent is 86. Therefore 177 = α⁸⁶. Multiply the generator polynomial by α⁸⁶:

(α⁸⁶ * α⁰)x¹⁶ + (α⁸⁶ * α²⁵¹)x¹⁵ + (α⁸⁶ * α⁶⁷)x¹⁴ + (α⁸⁶ * α⁴⁶)x¹³ + (α⁸⁶ * α⁶¹)x¹² + (α⁸⁶ * α¹¹⁸)x¹¹ + (α⁸⁶ * α⁷⁰)x¹⁰ + (α⁸⁶ * α⁶⁴)x⁹ + (α⁸⁶ * α⁹⁴)x⁸ + (α⁸⁶ * α³²)x⁷ + (α⁸⁶ * α⁴⁵)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:

α⁸⁶x¹⁶ + α^{(337 % 255)}x¹⁵ + α¹⁵³x¹⁴ + α¹³²x¹³ + α¹⁴⁷x¹² + α²⁰⁴x¹¹ + α¹⁵⁶x¹⁰ + α¹⁵⁰x⁹ + α¹⁸⁰x⁸ + α¹¹⁸x⁷ + α¹³¹x⁶

The result is:

α⁸⁶x¹⁶ + α⁸²x¹⁵ + α¹⁵³x¹⁴ + α¹³²x¹³ + α¹⁴⁷x¹² + α²⁰⁴x¹¹ + α¹⁵⁶x¹⁰ + α¹⁵⁰x⁹ + α¹⁸⁰x⁸ + α¹¹⁸x⁷ + α¹³¹x⁶

Now, convert this to integer notation:

177x¹⁶ + 211x¹⁵ + 146x¹⁴ + 184x¹³ + 41x¹² + 221x¹¹ + 228x¹⁰ + 85x⁹ + 150x⁸ + 199x⁷ + 92x⁶

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

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

(177 ⊕ 177)x¹⁶ + (75 ⊕ 211)x¹⁵ + (80 ⊕ 146)x¹⁴ + (37 ⊕ 184)x¹³ + (107 ⊕ 41)x¹² + (56 ⊕ 221)x¹¹ + (70 ⊕ 228)x¹⁰ + (137 ⊕ 85)x⁹ + (178 ⊕ 150)x⁸ + (30 ⊕ 199)x⁷ + (0 ⊕ 92)x⁶

The result is:

0x¹⁶ + 152x¹⁵ + 194x¹⁴ + 157x¹³ + 66x¹² + 229x¹¹ + 162x¹⁰ + 220x⁹ + 36x⁸ + 217x⁷ + 92x⁶

Discard the lead 0 term to get:

152x¹⁵ + 194x¹⁴ + 157x¹³ + 66x¹² + 229x¹¹ + 162x¹⁰ + 220x⁹ + 36x⁸ + 217x⁷ + 92x⁶

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 152x¹⁵. Convert 152x¹⁵ to alpha notation. According to the log antilog table, for the integer value 152, the alpha exponent is 17. Therefore 152 = α¹⁷. Multiply the generator polynomial by α¹⁷:

(α¹⁷ * α⁰)x¹⁵ + (α¹⁷ * α²⁵¹)x¹⁴ + (α¹⁷ * α⁶⁷)x¹³ + (α¹⁷ * α⁴⁶)x¹² + (α¹⁷ * α⁶¹)x¹¹ + (α¹⁷ * α¹¹⁸)x¹⁰ + (α¹⁷ * α⁷⁰)x⁹ + (α¹⁷ * α⁶⁴)x⁸ + (α¹⁷ * α⁹⁴)x⁷ + (α¹⁷ * α³²)x⁶ + (α¹⁷ * α⁴⁵)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:

α¹⁷x¹⁵ + α^{(268 % 255)}x¹⁴ + α⁸⁴x¹³ + α⁶³x¹² + α⁷⁸x¹¹ + α¹³⁵x¹⁰ + α⁸⁷x⁹ + α⁸¹x⁸ + α¹¹¹x⁷ + α⁴⁹x⁶ + α⁶²x⁵

The result is:

α¹⁷x¹⁵ + α¹³x¹⁴ + α⁸⁴x¹³ + α⁶³x¹² + α⁷⁸x¹¹ + α¹³⁵x¹⁰ + α⁸⁷x⁹ + α⁸¹x⁸ + α¹¹¹x⁷ + α⁴⁹x⁶ + α⁶²x⁵

Now, convert this to integer notation:

152x¹⁵ + 135x¹⁴ + 107x¹³ + 161x¹² + 120x¹¹ + 169x¹⁰ + 127x⁹ + 231x⁸ + 206x⁷ + 140x⁶ + 222x⁵

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

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

(152 ⊕ 152)x¹⁵ + (194 ⊕ 135)x¹⁴ + (157 ⊕ 107)x¹³ + (66 ⊕ 161)x¹² + (229 ⊕ 120)x¹¹ + (162 ⊕ 169)x¹⁰ + (220 ⊕ 127)x⁹ + (36 ⊕ 231)x⁸ + (217 ⊕ 206)x⁷ + (92 ⊕ 140)x⁶ + (0 ⊕ 222)x⁵

The result is:

0x¹⁵ + 69x¹⁴ + 246x¹³ + 227x¹² + 157x¹¹ + 11x¹⁰ + 163x⁹ + 195x⁸ + 23x⁷ + 208x⁶ + 222x⁵

Discard the lead 0 term to get:

69x¹⁴ + 246x¹³ + 227x¹² + 157x¹¹ + 11x¹⁰ + 163x⁹ + 195x⁸ + 23x⁷ + 208x⁶ + 222x⁵

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 69x¹⁴. Convert 69x¹⁴ to alpha notation. According to the log antilog table, for the integer value 69, the alpha exponent is 221. Therefore 69 = α²²¹. Multiply the generator polynomial by α²²¹:

(α²²¹ * α⁰)x¹⁴ + (α²²¹ * α²⁵¹)x¹³ + (α²²¹ * α⁶⁷)x¹² + (α²²¹ * α⁴⁶)x¹¹ + (α²²¹ * α⁶¹)x¹⁰ + (α²²¹ * α¹¹⁸)x⁹ + (α²²¹ * α⁷⁰)x⁸ + (α²²¹ * α⁶⁴)x⁷ + (α²²¹ * α⁹⁴)x⁶ + (α²²¹ * α³²)x⁵ + (α²²¹ * α⁴⁵)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:

α²²¹x¹⁴ + α^{(472 % 255)}x¹³ + α^{(288 % 255)}x¹² + α^{(267 % 255)}x¹¹ + α^{(282 % 255)}x¹⁰ + α^{(339 % 255)}x⁹ + α^{(291 % 255)}x⁸ + α^{(285 % 255)}x⁷ + α^{(315 % 255)}x⁶ + α²⁵³x⁵ + α^{(266 % 255)}x⁴

The result is:

α²²¹x¹⁴ + α²¹⁷x¹³ + α³³x¹² + α¹²x¹¹ + α²⁷x¹⁰ + α⁸⁴x⁹ + α³⁶x⁸ + α³⁰x⁷ + α⁶⁰x⁶ + α²⁵³x⁵ + α¹¹x⁴

Now, convert this to integer notation:

69x¹⁴ + 155x¹³ + 39x¹² + 205x¹¹ + 12x¹⁰ + 107x⁹ + 37x⁸ + 96x⁷ + 185x⁶ + 71x⁵ + 232x⁴

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

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

(69 ⊕ 69)x¹⁴ + (246 ⊕ 155)x¹³ + (227 ⊕ 39)x¹² + (157 ⊕ 205)x¹¹ + (11 ⊕ 12)x¹⁰ + (163 ⊕ 107)x⁹ + (195 ⊕ 37)x⁸ + (23 ⊕ 96)x⁷ + (208 ⊕ 185)x⁶ + (222 ⊕ 71)x⁵ + (0 ⊕ 232)x⁴

The result is:

0x¹⁴ + 109x¹³ + 196x¹² + 80x¹¹ + 7x¹⁰ + 200x⁹ + 230x⁸ + 119x⁷ + 105x⁶ + 153x⁵ + 232x⁴

Discard the lead 0 term to get:

109x¹³ + 196x¹² + 80x¹¹ + 7x¹⁰ + 200x⁹ + 230x⁸ + 119x⁷ + 105x⁶ + 153x⁵ + 232x⁴

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 109x¹³. Convert 109x¹³ to alpha notation. According to the log antilog table, for the integer value 109, the alpha exponent is 133. Therefore 109 = α¹³³. Multiply the generator polynomial by α¹³³:

(α¹³³ * α⁰)x¹³ + (α¹³³ * α²⁵¹)x¹² + (α¹³³ * α⁶⁷)x¹¹ + (α¹³³ * α⁴⁶)x¹⁰ + (α¹³³ * α⁶¹)x⁹ + (α¹³³ * α¹¹⁸)x⁸ + (α¹³³ * α⁷⁰)x⁷ + (α¹³³ * α⁶⁴)x⁶ + (α¹³³ * α⁹⁴)x⁵ + (α¹³³ * α³²)x⁴ + (α¹³³ * α⁴⁵)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:

α¹³³x¹³ + α^{(384 % 255)}x¹² + α²⁰⁰x¹¹ + α¹⁷⁹x¹⁰ + α¹⁹⁴x⁹ + α²⁵¹x⁸ + α²⁰³x⁷ + α¹⁹⁷x⁶ + α²²⁷x⁵ + α¹⁶⁵x⁴ + α¹⁷⁸x³

The result is:

α¹³³x¹³ + α¹²⁹x¹² + α²⁰⁰x¹¹ + α¹⁷⁹x¹⁰ + α¹⁹⁴x⁹ + α²⁵¹x⁸ + α²⁰³x⁷ + α¹⁹⁷x⁶ + α²²⁷x⁵ + α¹⁶⁵x⁴ + α¹⁷⁸x³

Now, convert this to integer notation:

109x¹³ + 23x¹² + 28x¹¹ + 75x¹⁰ + 50x⁹ + 216x⁸ + 224x⁷ + 141x⁶ + 144x⁵ + 145x⁴ + 171x³

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

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

(109 ⊕ 109)x¹³ + (196 ⊕ 23)x¹² + (80 ⊕ 28)x¹¹ + (7 ⊕ 75)x¹⁰ + (200 ⊕ 50)x⁹ + (230 ⊕ 216)x⁸ + (119 ⊕ 224)x⁷ + (105 ⊕ 141)x⁶ + (153 ⊕ 144)x⁵ + (232 ⊕ 145)x⁴ + (0 ⊕ 171)x³

The result is:

0x¹³ + 211x¹² + 76x¹¹ + 76x¹⁰ + 250x⁹ + 62x⁸ + 151x⁷ + 228x⁶ + 9x⁵ + 121x⁴ + 171x³

Discard the lead 0 term to get:

211x¹² + 76x¹¹ + 76x¹⁰ + 250x⁹ + 62x⁸ + 151x⁷ + 228x⁶ + 9x⁵ + 121x⁴ + 171x³

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 211x¹². Convert 211x¹² to alpha notation. According to the log antilog table, for the integer value 211, the alpha exponent is 82. Therefore 211 = α⁸². Multiply the generator polynomial by α⁸²:

(α⁸² * α⁰)x¹² + (α⁸² * α²⁵¹)x¹¹ + (α⁸² * α⁶⁷)x¹⁰ + (α⁸² * α⁴⁶)x⁹ + (α⁸² * α⁶¹)x⁸ + (α⁸² * α¹¹⁸)x⁷ + (α⁸² * α⁷⁰)x⁶ + (α⁸² * α⁶⁴)x⁵ + (α⁸² * α⁹⁴)x⁴ + (α⁸² * α³²)x³ + (α⁸² * α⁴⁵)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:

α⁸²x¹² + α^{(333 % 255)}x¹¹ + α¹⁴⁹x¹⁰ + α¹²⁸x⁹ + α¹⁴³x⁸ + α²⁰⁰x⁷ + α¹⁵²x⁶ + α¹⁴⁶x⁵ + α¹⁷⁶x⁴ + α¹¹⁴x³ + α¹²⁷x²

The result is:

α⁸²x¹² + α⁷⁸x¹¹ + α¹⁴⁹x¹⁰ + α¹²⁸x⁹ + α¹⁴³x⁸ + α²⁰⁰x⁷ + α¹⁵²x⁶ + α¹⁴⁶x⁵ + α¹⁷⁶x⁴ + α¹¹⁴x³ + α¹²⁷x²

Now, convert this to integer notation:

211x¹² + 120x¹¹ + 164x¹⁰ + 133x⁹ + 84x⁸ + 28x⁷ + 73x⁶ + 154x⁵ + 227x⁴ + 62x³ + 204x²

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

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

(211 ⊕ 211)x¹² + (76 ⊕ 120)x¹¹ + (76 ⊕ 164)x¹⁰ + (250 ⊕ 133)x⁹ + (62 ⊕ 84)x⁸ + (151 ⊕ 28)x⁷ + (228 ⊕ 73)x⁶ + (9 ⊕ 154)x⁵ + (121 ⊕ 227)x⁴ + (171 ⊕ 62)x³ + (0 ⊕ 204)x²

The result is:

0x¹² + 52x¹¹ + 232x¹⁰ + 127x⁹ + 106x⁸ + 139x⁷ + 173x⁶ + 147x⁵ + 154x⁴ + 149x³ + 204x²

Discard the lead 0 term to get:

52x¹¹ + 232x¹⁰ + 127x⁹ + 106x⁸ + 139x⁷ + 173x⁶ + 147x⁵ + 154x⁴ + 149x³ + 204x²

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 52x¹¹. Convert 52x¹¹ to alpha notation. According to the log antilog table, for the integer value 52, the alpha exponent is 106. Therefore 52 = α¹⁰⁶. Multiply the generator polynomial by α¹⁰⁶:

(α¹⁰⁶ * α⁰)x¹¹ + (α¹⁰⁶ * α²⁵¹)x¹⁰ + (α¹⁰⁶ * α⁶⁷)x⁹ + (α¹⁰⁶ * α⁴⁶)x⁸ + (α¹⁰⁶ * α⁶¹)x⁷ + (α¹⁰⁶ * α¹¹⁸)x⁶ + (α¹⁰⁶ * α⁷⁰)x⁵ + (α¹⁰⁶ * α⁶⁴)x⁴ + (α¹⁰⁶ * α⁹⁴)x³ + (α¹⁰⁶ * α³²)x² + (α¹⁰⁶ * α⁴⁵)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:

α¹⁰⁶x¹¹ + α^{(357 % 255)}x¹⁰ + α¹⁷³x⁹ + α¹⁵²x⁸ + α¹⁶⁷x⁷ + α²²⁴x⁶ + α¹⁷⁶x⁵ + α¹⁷⁰x⁴ + α²⁰⁰x³ + α¹³⁸x² + α¹⁵¹x¹

The result is:

α¹⁰⁶x¹¹ + α¹⁰²x¹⁰ + α¹⁷³x⁹ + α¹⁵²x⁸ + α¹⁶⁷x⁷ + α²²⁴x⁶ + α¹⁷⁶x⁵ + α¹⁷⁰x⁴ + α²⁰⁰x³ + α¹³⁸x² + α¹⁵¹x¹

Now, convert this to integer notation:

52x¹¹ + 68x¹⁰ + 246x⁹ + 73x⁸ + 126x⁷ + 18x⁶ + 227x⁵ + 215x⁴ + 28x³ + 33x² + 170x¹

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

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

(52 ⊕ 52)x¹¹ + (232 ⊕ 68)x¹⁰ + (127 ⊕ 246)x⁹ + (106 ⊕ 73)x⁸ + (139 ⊕ 126)x⁷ + (173 ⊕ 18)x⁶ + (147 ⊕ 227)x⁵ + (154 ⊕ 215)x⁴ + (149 ⊕ 28)x³ + (204 ⊕ 33)x² + (0 ⊕ 170)x¹

The result is:

0x¹¹ + 172x¹⁰ + 137x⁹ + 35x⁸ + 245x⁷ + 191x⁶ + 112x⁵ + 77x⁴ + 137x³ + 237x² + 170x¹

Discard the lead 0 term to get:

172x¹⁰ + 137x⁹ + 35x⁸ + 245x⁷ + 191x⁶ + 112x⁵ + 77x⁴ + 137x³ + 237x² + 170x¹

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 172x¹⁰. Convert 172x¹⁰ to alpha notation. According to the log antilog table, for the integer value 172, the alpha exponent is 220. Therefore 172 = α²²⁰. Multiply the generator polynomial by α²²⁰:

(α²²⁰ * α⁰)x¹⁰ + (α²²⁰ * α²⁵¹)x⁹ + (α²²⁰ * α⁶⁷)x⁸ + (α²²⁰ * α⁴⁶)x⁷ + (α²²⁰ * α⁶¹)x⁶ + (α²²⁰ * α¹¹⁸)x⁵ + (α²²⁰ * α⁷⁰)x⁴ + (α²²⁰ * α⁶⁴)x³ + (α²²⁰ * α⁹⁴)x² + (α²²⁰ * α³²)x¹ + (α²²⁰ * α⁴⁵)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:

α²²⁰x¹⁰ + α^{(471 % 255)}x⁹ + α^{(287 % 255)}x⁸ + α^{(266 % 255)}x⁷ + α^{(281 % 255)}x⁶ + α^{(338 % 255)}x⁵ + α^{(290 % 255)}x⁴ + α^{(284 % 255)}x³ + α^{(314 % 255)}x² + α²⁵²x¹ + α^{(265 % 255)}x⁰

The result is:

α²²⁰x¹⁰ + α²¹⁶x⁹ + α³²x⁸ + α¹¹x⁷ + α²⁶x⁶ + α⁸³x⁵ + α³⁵x⁴ + α²⁹x³ + α⁵⁹x² + α²⁵²x¹ + α¹⁰x⁰

Now, convert this to integer notation:

172x¹⁰ + 195x⁹ + 157x⁸ + 232x⁷ + 6x⁶ + 187x⁵ + 156x⁴ + 48x³ + 210x² + 173x¹ + 116

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

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

(172 ⊕ 172)x¹⁰ + (137 ⊕ 195)x⁹ + (35 ⊕ 157)x⁸ + (245 ⊕ 232)x⁷ + (191 ⊕ 6)x⁶ + (112 ⊕ 187)x⁵ + (77 ⊕ 156)x⁴ + (137 ⊕ 48)x³ + (237 ⊕ 210)x² + (170 ⊕ 173)x¹ + (0 ⊕ 116)x⁰

The result is:

0x¹⁰ + 74x⁹ + 190x⁸ + 29x⁷ + 185x⁶ + 203x⁵ + 209x⁴ + 185x³ + 63x² + 7x¹ + 116

Discard the lead 0 term to get:

74x⁹ + 190x⁸ + 29x⁷ + 185x⁶ + 203x⁵ + 209x⁴ + 185x³ + 63x² + 7x¹ + 116

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:

74  190  29  185  203  209  185  63  7  116