Introduction to Creating a Generator Polynomial

Below, you will find the steps to create a generator polynomial for 13 error correction code words.

However, since we are working with the galois field GF256, we need to use special methods to make sure our numbers don't get too big or too small to stay in the galois field.

When we multiply two alpha values, normally we add the two exponents together. But if the result is bigger than 255, we have a problem.

For example, consider ɑ²⁵¹ * ɑ¹⁰. Multiplying these two gives us ɑ²⁶¹, but that is too big.

To prevent that from happening, we put the resulting exponent through the following formula:
(exponent % 256) + floor(exponent / 256)

Using the exponent of 261 from the above example, this gives us:
(261 % 256) + floor(261 / 256) = (5) + floor(1.01953125) = (5 + 1) = 6.

So when we multiply ɑ²⁵¹ * ɑ¹⁰, the result is ɑ⁶.

Another problem is combining like terms. Because of the nature of the galois field, we can't simply add the numbers together. We have to XOR them.

This will cause some unexpected results. For example:
ɑ²⁰¹x² + ɑ¹⁹⁹x² = 56x² + 14x²

If we add those together normally, we get 70x². This number is not larger than 255, so it would seem acceptable. However, if we perform an XOR instead of addition, we get:
56 ⊕ 14 = 54.

This is the correct result. When adding like terms during the creation of a generator polynomial, you must always XOR the integers rather than adding them.

How to Create a Generator Polynomial

In each step of creating a generator polynomial, you multiply a polynomial by a polynomial. The very first polynomial that you start with in the first step is always (ɑ⁰x¹ + ɑ⁰x⁰). For each multiplication step, you multiply the current polynomial by (ɑ⁰x¹ + ɑ^jx⁰) where j is 1 for the first multiplication, 2 for the second multiplication, 3 for the third, and so on.

Multiplication Step #1

We will multiply (ɑ⁰x¹ + ɑ⁰x⁰) by (ɑ⁰x¹ + ɑ¹x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x¹ * ɑ⁰x¹) + (ɑ⁰x⁰ * ɑ⁰x¹) + (ɑ⁰x¹ * ɑ¹x⁰) + (ɑ⁰x⁰ * ɑ¹x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽¹⁺¹⁾) + (ɑ⁽⁰⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺¹⁾x⁽¹⁺⁰⁾) + (ɑ⁽⁰⁺¹⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x² + ɑ⁰x¹ + ɑ¹x¹ + ɑ¹x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x¹ term.

We will now combine the x¹ terms. they are:
ɑ⁰x¹ + ɑ¹x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
1x¹ + 2x¹

After XORing the integers, we convert the result back to an alpha term.
(1 ⊕ 2)x¹ = 3x¹ = ɑ²⁵x¹

Below is the final polynomial from this step after combining the like terms. If we needed 2 error correction code words, this would be the generator polynomial required.
ɑ⁰x² + ɑ²⁵x¹ + ɑ¹x⁰

Multiplication Step #2

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x² + ɑ²⁵x¹ + ɑ¹x⁰) by (ɑ⁰x¹ + ɑ²x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x² * ɑ⁰x¹) + (ɑ²⁵x¹ * ɑ⁰x¹) + (ɑ¹x⁰ * ɑ⁰x¹) + (ɑ⁰x² * ɑ²x⁰) + (ɑ²⁵x¹ * ɑ²x⁰) + (ɑ¹x⁰ * ɑ²x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽²⁺¹⁾) + (ɑ⁽²⁵⁺⁰⁾x⁽¹⁺¹⁾) + (ɑ⁽¹⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺²⁾x⁽²⁺⁰⁾) + (ɑ⁽²⁵⁺²⁾x⁽¹⁺⁰⁾) + (ɑ⁽¹⁺²⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x³ + ɑ²⁵x² + ɑ¹x¹ + ɑ²x² + ɑ²⁷x¹ + ɑ³x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x² and x¹ term.

We will now combine the x² terms. they are:
ɑ²⁵x² + ɑ²x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
3x² + 4x²

After XORing the integers, we convert the result back to an alpha term.
(3 ⊕ 4)x² = 7x² = ɑ¹⁹⁸x²

We will now combine the x¹ terms. they are:
ɑ¹x¹ + ɑ²⁷x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
2x¹ + 12x¹

After XORing the integers, we convert the result back to an alpha term.
(2 ⊕ 12)x¹ = 14x¹ = ɑ¹⁹⁹x¹

Below is the final polynomial from this step after combining the like terms. If we needed 3 error correction code words, this would be the generator polynomial required.
ɑ⁰x³ + ɑ¹⁹⁸x² + ɑ¹⁹⁹x¹ + ɑ³x⁰

Multiplication Step #3

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x³ + ɑ¹⁹⁸x² + ɑ¹⁹⁹x¹ + ɑ³x⁰) by (ɑ⁰x¹ + ɑ³x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x³ * ɑ⁰x¹) + (ɑ¹⁹⁸x² * ɑ⁰x¹) + (ɑ¹⁹⁹x¹ * ɑ⁰x¹) + (ɑ³x⁰ * ɑ⁰x¹) + (ɑ⁰x³ * ɑ³x⁰) + (ɑ¹⁹⁸x² * ɑ³x⁰) + (ɑ¹⁹⁹x¹ * ɑ³x⁰) + (ɑ³x⁰ * ɑ³x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽³⁺¹⁾) + (ɑ^(198+0)x⁽²⁺¹⁾) + (ɑ^(199+0)x⁽¹⁺¹⁾) + (ɑ⁽³⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺³⁾x⁽³⁺⁰⁾) + (ɑ^(198+3)x⁽²⁺⁰⁾) + (ɑ^(199+3)x⁽¹⁺⁰⁾) + (ɑ⁽³⁺³⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x⁴ + ɑ¹⁹⁸x³ + ɑ¹⁹⁹x² + ɑ³x¹ + ɑ³x³ + ɑ²⁰¹x² + ɑ²⁰²x¹ + ɑ⁶x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x³, x² and x¹ term.

We will now combine the x³ terms. they are:
ɑ¹⁹⁸x³ + ɑ³x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
7x³ + 8x³

After XORing the integers, we convert the result back to an alpha term.
(7 ⊕ 8)x³ = 15x³ = ɑ⁷⁵x³

We will now combine the x² terms. they are:
ɑ¹⁹⁹x² + ɑ²⁰¹x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
14x² + 56x²

After XORing the integers, we convert the result back to an alpha term.
(14 ⊕ 56)x² = 54x² = ɑ²⁴⁹x²

We will now combine the x¹ terms. they are:
ɑ³x¹ + ɑ²⁰²x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
8x¹ + 112x¹

After XORing the integers, we convert the result back to an alpha term.
(8 ⊕ 112)x¹ = 120x¹ = ɑ⁷⁸x¹

Below is the final polynomial from this step after combining the like terms. If we needed 4 error correction code words, this would be the generator polynomial required.
ɑ⁰x⁴ + ɑ⁷⁵x³ + ɑ²⁴⁹x² + ɑ⁷⁸x¹ + ɑ⁶x⁰

Multiplication Step #4

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x⁴ + ɑ⁷⁵x³ + ɑ²⁴⁹x² + ɑ⁷⁸x¹ + ɑ⁶x⁰) by (ɑ⁰x¹ + ɑ⁴x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x⁴ * ɑ⁰x¹) + (ɑ⁷⁵x³ * ɑ⁰x¹) + (ɑ²⁴⁹x² * ɑ⁰x¹) + (ɑ⁷⁸x¹ * ɑ⁰x¹) + (ɑ⁶x⁰ * ɑ⁰x¹) + (ɑ⁰x⁴ * ɑ⁴x⁰) + (ɑ⁷⁵x³ * ɑ⁴x⁰) + (ɑ²⁴⁹x² * ɑ⁴x⁰) + (ɑ⁷⁸x¹ * ɑ⁴x⁰) + (ɑ⁶x⁰ * ɑ⁴x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽⁴⁺¹⁾) + (ɑ⁽⁷⁵⁺⁰⁾x⁽³⁺¹⁾) + (ɑ^(249+0)x⁽²⁺¹⁾) + (ɑ⁽⁷⁸⁺⁰⁾x⁽¹⁺¹⁾) + (ɑ⁽⁶⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺⁴⁾x⁽⁴⁺⁰⁾) + (ɑ⁽⁷⁵⁺⁴⁾x⁽³⁺⁰⁾) + (ɑ^(249+4)x⁽²⁺⁰⁾) + (ɑ⁽⁷⁸⁺⁴⁾x⁽¹⁺⁰⁾) + (ɑ⁽⁶⁺⁴⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x⁵ + ɑ⁷⁵x⁴ + ɑ²⁴⁹x³ + ɑ⁷⁸x² + ɑ⁶x¹ + ɑ⁴x⁴ + ɑ⁷⁹x³ + ɑ²⁵³x² + ɑ⁸²x¹ + ɑ¹⁰x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x⁴, x³, x² and x¹ term.

We will now combine the x⁴ terms. they are:
ɑ⁷⁵x⁴ + ɑ⁴x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
15x⁴ + 16x⁴

After XORing the integers, we convert the result back to an alpha term.
(15 ⊕ 16)x⁴ = 31x⁴ = ɑ¹¹³x⁴

We will now combine the x³ terms. they are:
ɑ²⁴⁹x³ + ɑ⁷⁹x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
54x³ + 240x³

After XORing the integers, we convert the result back to an alpha term.
(54 ⊕ 240)x³ = 198x³ = ɑ¹⁶⁴x³

We will now combine the x² terms. they are:
ɑ⁷⁸x² + ɑ²⁵³x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
120x² + 71x²

After XORing the integers, we convert the result back to an alpha term.
(120 ⊕ 71)x² = 63x² = ɑ¹⁶⁶x²

We will now combine the x¹ terms. they are:
ɑ⁶x¹ + ɑ⁸²x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
64x¹ + 211x¹

After XORing the integers, we convert the result back to an alpha term.
(64 ⊕ 211)x¹ = 147x¹ = ɑ¹¹⁹x¹

Below is the final polynomial from this step after combining the like terms. If we needed 5 error correction code words, this would be the generator polynomial required.
ɑ⁰x⁵ + ɑ¹¹³x⁴ + ɑ¹⁶⁴x³ + ɑ¹⁶⁶x² + ɑ¹¹⁹x¹ + ɑ¹⁰x⁰

Multiplication Step #5

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x⁵ + ɑ¹¹³x⁴ + ɑ¹⁶⁴x³ + ɑ¹⁶⁶x² + ɑ¹¹⁹x¹ + ɑ¹⁰x⁰) by (ɑ⁰x¹ + ɑ⁵x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x⁵ * ɑ⁰x¹) + (ɑ¹¹³x⁴ * ɑ⁰x¹) + (ɑ¹⁶⁴x³ * ɑ⁰x¹) + (ɑ¹⁶⁶x² * ɑ⁰x¹) + (ɑ¹¹⁹x¹ * ɑ⁰x¹) + (ɑ¹⁰x⁰ * ɑ⁰x¹) + (ɑ⁰x⁵ * ɑ⁵x⁰) + (ɑ¹¹³x⁴ * ɑ⁵x⁰) + (ɑ¹⁶⁴x³ * ɑ⁵x⁰) + (ɑ¹⁶⁶x² * ɑ⁵x⁰) + (ɑ¹¹⁹x¹ * ɑ⁵x⁰) + (ɑ¹⁰x⁰ * ɑ⁵x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽⁵⁺¹⁾) + (ɑ^(113+0)x⁽⁴⁺¹⁾) + (ɑ^(164+0)x⁽³⁺¹⁾) + (ɑ^(166+0)x⁽²⁺¹⁾) + (ɑ^(119+0)x⁽¹⁺¹⁾) + (ɑ⁽¹⁰⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺⁵⁾x⁽⁵⁺⁰⁾) + (ɑ^(113+5)x⁽⁴⁺⁰⁾) + (ɑ^(164+5)x⁽³⁺⁰⁾) + (ɑ^(166+5)x⁽²⁺⁰⁾) + (ɑ^(119+5)x⁽¹⁺⁰⁾) + (ɑ⁽¹⁰⁺⁵⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x⁶ + ɑ¹¹³x⁵ + ɑ¹⁶⁴x⁴ + ɑ¹⁶⁶x³ + ɑ¹¹⁹x² + ɑ¹⁰x¹ + ɑ⁵x⁵ + ɑ¹¹⁸x⁴ + ɑ¹⁶⁹x³ + ɑ¹⁷¹x² + ɑ¹²⁴x¹ + ɑ¹⁵x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x⁵, x⁴, x³, x² and x¹ term.

We will now combine the x⁵ terms. they are:
ɑ¹¹³x⁵ + ɑ⁵x⁵

We will convert the above alphas to integers using the log antilog table to make this step easier:
31x⁵ + 32x⁵

After XORing the integers, we convert the result back to an alpha term.
(31 ⊕ 32)x⁵ = 63x⁵ = ɑ¹⁶⁶x⁵

We will now combine the x⁴ terms. they are:
ɑ¹⁶⁴x⁴ + ɑ¹¹⁸x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
198x⁴ + 199x⁴

After XORing the integers, we convert the result back to an alpha term.
(198 ⊕ 199)x⁴ = 1x⁴ = ɑ⁰x⁴

We will now combine the x³ terms. they are:
ɑ¹⁶⁶x³ + ɑ¹⁶⁹x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
63x³ + 229x³

After XORing the integers, we convert the result back to an alpha term.
(63 ⊕ 229)x³ = 218x³ = ɑ¹³⁴x³

We will now combine the x² terms. they are:
ɑ¹¹⁹x² + ɑ¹⁷¹x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
147x² + 179x²

After XORing the integers, we convert the result back to an alpha term.
(147 ⊕ 179)x² = 32x² = ɑ⁵x²

We will now combine the x¹ terms. they are:
ɑ¹⁰x¹ + ɑ¹²⁴x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
116x¹ + 151x¹

After XORing the integers, we convert the result back to an alpha term.
(116 ⊕ 151)x¹ = 227x¹ = ɑ¹⁷⁶x¹

Below is the final polynomial from this step after combining the like terms. If we needed 6 error correction code words, this would be the generator polynomial required.
ɑ⁰x⁶ + ɑ¹⁶⁶x⁵ + ɑ⁰x⁴ + ɑ¹³⁴x³ + ɑ⁵x² + ɑ¹⁷⁶x¹ + ɑ¹⁵x⁰

Multiplication Step #6

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x⁶ + ɑ¹⁶⁶x⁵ + ɑ⁰x⁴ + ɑ¹³⁴x³ + ɑ⁵x² + ɑ¹⁷⁶x¹ + ɑ¹⁵x⁰) by (ɑ⁰x¹ + ɑ⁶x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x⁶ * ɑ⁰x¹) + (ɑ¹⁶⁶x⁵ * ɑ⁰x¹) + (ɑ⁰x⁴ * ɑ⁰x¹) + (ɑ¹³⁴x³ * ɑ⁰x¹) + (ɑ⁵x² * ɑ⁰x¹) + (ɑ¹⁷⁶x¹ * ɑ⁰x¹) + (ɑ¹⁵x⁰ * ɑ⁰x¹) + (ɑ⁰x⁶ * ɑ⁶x⁰) + (ɑ¹⁶⁶x⁵ * ɑ⁶x⁰) + (ɑ⁰x⁴ * ɑ⁶x⁰) + (ɑ¹³⁴x³ * ɑ⁶x⁰) + (ɑ⁵x² * ɑ⁶x⁰) + (ɑ¹⁷⁶x¹ * ɑ⁶x⁰) + (ɑ¹⁵x⁰ * ɑ⁶x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽⁶⁺¹⁾) + (ɑ^(166+0)x⁽⁵⁺¹⁾) + (ɑ⁽⁰⁺⁰⁾x⁽⁴⁺¹⁾) + (ɑ^(134+0)x⁽³⁺¹⁾) + (ɑ⁽⁵⁺⁰⁾x⁽²⁺¹⁾) + (ɑ^(176+0)x⁽¹⁺¹⁾) + (ɑ⁽¹⁵⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺⁶⁾x⁽⁶⁺⁰⁾) + (ɑ^(166+6)x⁽⁵⁺⁰⁾) + (ɑ⁽⁰⁺⁶⁾x⁽⁴⁺⁰⁾) + (ɑ^(134+6)x⁽³⁺⁰⁾) + (ɑ⁽⁵⁺⁶⁾x⁽²⁺⁰⁾) + (ɑ^(176+6)x⁽¹⁺⁰⁾) + (ɑ⁽¹⁵⁺⁶⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x⁷ + ɑ¹⁶⁶x⁶ + ɑ⁰x⁵ + ɑ¹³⁴x⁴ + ɑ⁵x³ + ɑ¹⁷⁶x² + ɑ¹⁵x¹ + ɑ⁶x⁶ + ɑ¹⁷²x⁵ + ɑ⁶x⁴ + ɑ¹⁴⁰x³ + ɑ¹¹x² + ɑ¹⁸²x¹ + ɑ²¹x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x⁶, x⁵, x⁴, x³, x² and x¹ term.

We will now combine the x⁶ terms. they are:
ɑ¹⁶⁶x⁶ + ɑ⁶x⁶

We will convert the above alphas to integers using the log antilog table to make this step easier:
63x⁶ + 64x⁶

After XORing the integers, we convert the result back to an alpha term.
(63 ⊕ 64)x⁶ = 127x⁶ = ɑ⁸⁷x⁶

We will now combine the x⁵ terms. they are:
ɑ⁰x⁵ + ɑ¹⁷²x⁵

We will convert the above alphas to integers using the log antilog table to make this step easier:
1x⁵ + 123x⁵

After XORing the integers, we convert the result back to an alpha term.
(1 ⊕ 123)x⁵ = 122x⁵ = ɑ²²⁹x⁵

We will now combine the x⁴ terms. they are:
ɑ¹³⁴x⁴ + ɑ⁶x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
218x⁴ + 64x⁴

After XORing the integers, we convert the result back to an alpha term.
(218 ⊕ 64)x⁴ = 154x⁴ = ɑ¹⁴⁶x⁴

We will now combine the x³ terms. they are:
ɑ⁵x³ + ɑ¹⁴⁰x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
32x³ + 132x³

After XORing the integers, we convert the result back to an alpha term.
(32 ⊕ 132)x³ = 164x³ = ɑ¹⁴⁹x³

We will now combine the x² terms. they are:
ɑ¹⁷⁶x² + ɑ¹¹x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
227x² + 232x²

After XORing the integers, we convert the result back to an alpha term.
(227 ⊕ 232)x² = 11x² = ɑ²³⁸x²

We will now combine the x¹ terms. they are:
ɑ¹⁵x¹ + ɑ¹⁸²x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
38x¹ + 98x¹

After XORing the integers, we convert the result back to an alpha term.
(38 ⊕ 98)x¹ = 68x¹ = ɑ¹⁰²x¹

Below is the final polynomial from this step after combining the like terms. If we needed 7 error correction code words, this would be the generator polynomial required.
ɑ⁰x⁷ + ɑ⁸⁷x⁶ + ɑ²²⁹x⁵ + ɑ¹⁴⁶x⁴ + ɑ¹⁴⁹x³ + ɑ²³⁸x² + ɑ¹⁰²x¹ + ɑ²¹x⁰

Multiplication Step #7

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x⁷ + ɑ⁸⁷x⁶ + ɑ²²⁹x⁵ + ɑ¹⁴⁶x⁴ + ɑ¹⁴⁹x³ + ɑ²³⁸x² + ɑ¹⁰²x¹ + ɑ²¹x⁰) by (ɑ⁰x¹ + ɑ⁷x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x⁷ * ɑ⁰x¹) + (ɑ⁸⁷x⁶ * ɑ⁰x¹) + (ɑ²²⁹x⁵ * ɑ⁰x¹) + (ɑ¹⁴⁶x⁴ * ɑ⁰x¹) + (ɑ¹⁴⁹x³ * ɑ⁰x¹) + (ɑ²³⁸x² * ɑ⁰x¹) + (ɑ¹⁰²x¹ * ɑ⁰x¹) + (ɑ²¹x⁰ * ɑ⁰x¹) + (ɑ⁰x⁷ * ɑ⁷x⁰) + (ɑ⁸⁷x⁶ * ɑ⁷x⁰) + (ɑ²²⁹x⁵ * ɑ⁷x⁰) + (ɑ¹⁴⁶x⁴ * ɑ⁷x⁰) + (ɑ¹⁴⁹x³ * ɑ⁷x⁰) + (ɑ²³⁸x² * ɑ⁷x⁰) + (ɑ¹⁰²x¹ * ɑ⁷x⁰) + (ɑ²¹x⁰ * ɑ⁷x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽⁷⁺¹⁾) + (ɑ⁽⁸⁷⁺⁰⁾x⁽⁶⁺¹⁾) + (ɑ^(229+0)x⁽⁵⁺¹⁾) + (ɑ^(146+0)x⁽⁴⁺¹⁾) + (ɑ^(149+0)x⁽³⁺¹⁾) + (ɑ^(238+0)x⁽²⁺¹⁾) + (ɑ^(102+0)x⁽¹⁺¹⁾) + (ɑ⁽²¹⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺⁷⁾x⁽⁷⁺⁰⁾) + (ɑ⁽⁸⁷⁺⁷⁾x⁽⁶⁺⁰⁾) + (ɑ^(229+7)x⁽⁵⁺⁰⁾) + (ɑ^(146+7)x⁽⁴⁺⁰⁾) + (ɑ^(149+7)x⁽³⁺⁰⁾) + (ɑ^(238+7)x⁽²⁺⁰⁾) + (ɑ^(102+7)x⁽¹⁺⁰⁾) + (ɑ⁽²¹⁺⁷⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x⁸ + ɑ⁸⁷x⁷ + ɑ²²⁹x⁶ + ɑ¹⁴⁶x⁵ + ɑ¹⁴⁹x⁴ + ɑ²³⁸x³ + ɑ¹⁰²x² + ɑ²¹x¹ + ɑ⁷x⁷ + ɑ⁹⁴x⁶ + ɑ²³⁶x⁵ + ɑ¹⁵³x⁴ + ɑ¹⁵⁶x³ + ɑ²⁴⁵x² + ɑ¹⁰⁹x¹ + ɑ²⁸x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x⁷, x⁶, x⁵, x⁴, x³, x² and x¹ term.

We will now combine the x⁷ terms. they are:
ɑ⁸⁷x⁷ + ɑ⁷x⁷

We will convert the above alphas to integers using the log antilog table to make this step easier:
127x⁷ + 128x⁷

After XORing the integers, we convert the result back to an alpha term.
(127 ⊕ 128)x⁷ = 255x⁷ = ɑ¹⁷⁵x⁷

We will now combine the x⁶ terms. they are:
ɑ²²⁹x⁶ + ɑ⁹⁴x⁶

We will convert the above alphas to integers using the log antilog table to make this step easier:
122x⁶ + 113x⁶

After XORing the integers, we convert the result back to an alpha term.
(122 ⊕ 113)x⁶ = 11x⁶ = ɑ²³⁸x⁶

We will now combine the x⁵ terms. they are:
ɑ¹⁴⁶x⁵ + ɑ²³⁶x⁵

We will convert the above alphas to integers using the log antilog table to make this step easier:
154x⁵ + 203x⁵

After XORing the integers, we convert the result back to an alpha term.
(154 ⊕ 203)x⁵ = 81x⁵ = ɑ²⁰⁸x⁵

We will now combine the x⁴ terms. they are:
ɑ¹⁴⁹x⁴ + ɑ¹⁵³x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
164x⁴ + 146x⁴

After XORing the integers, we convert the result back to an alpha term.
(164 ⊕ 146)x⁴ = 54x⁴ = ɑ²⁴⁹x⁴

We will now combine the x³ terms. they are:
ɑ²³⁸x³ + ɑ¹⁵⁶x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
11x³ + 228x³

After XORing the integers, we convert the result back to an alpha term.
(11 ⊕ 228)x³ = 239x³ = ɑ²¹⁵x³

We will now combine the x² terms. they are:
ɑ¹⁰²x² + ɑ²⁴⁵x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
68x² + 233x²

After XORing the integers, we convert the result back to an alpha term.
(68 ⊕ 233)x² = 173x² = ɑ²⁵²x²

We will now combine the x¹ terms. they are:
ɑ²¹x¹ + ɑ¹⁰⁹x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
117x¹ + 189x¹

After XORing the integers, we convert the result back to an alpha term.
(117 ⊕ 189)x¹ = 200x¹ = ɑ¹⁹⁶x¹

Below is the final polynomial from this step after combining the like terms. If we needed 8 error correction code words, this would be the generator polynomial required.
ɑ⁰x⁸ + ɑ¹⁷⁵x⁷ + ɑ²³⁸x⁶ + ɑ²⁰⁸x⁵ + ɑ²⁴⁹x⁴ + ɑ²¹⁵x³ + ɑ²⁵²x² + ɑ¹⁹⁶x¹ + ɑ²⁸x⁰

Multiplication Step #8

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x⁸ + ɑ¹⁷⁵x⁷ + ɑ²³⁸x⁶ + ɑ²⁰⁸x⁵ + ɑ²⁴⁹x⁴ + ɑ²¹⁵x³ + ɑ²⁵²x² + ɑ¹⁹⁶x¹ + ɑ²⁸x⁰) by (ɑ⁰x¹ + ɑ⁸x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x⁸ * ɑ⁰x¹) + (ɑ¹⁷⁵x⁷ * ɑ⁰x¹) + (ɑ²³⁸x⁶ * ɑ⁰x¹) + (ɑ²⁰⁸x⁵ * ɑ⁰x¹) + (ɑ²⁴⁹x⁴ * ɑ⁰x¹) + (ɑ²¹⁵x³ * ɑ⁰x¹) + (ɑ²⁵²x² * ɑ⁰x¹) + (ɑ¹⁹⁶x¹ * ɑ⁰x¹) + (ɑ²⁸x⁰ * ɑ⁰x¹) + (ɑ⁰x⁸ * ɑ⁸x⁰) + (ɑ¹⁷⁵x⁷ * ɑ⁸x⁰) + (ɑ²³⁸x⁶ * ɑ⁸x⁰) + (ɑ²⁰⁸x⁵ * ɑ⁸x⁰) + (ɑ²⁴⁹x⁴ * ɑ⁸x⁰) + (ɑ²¹⁵x³ * ɑ⁸x⁰) + (ɑ²⁵²x² * ɑ⁸x⁰) + (ɑ¹⁹⁶x¹ * ɑ⁸x⁰) + (ɑ²⁸x⁰ * ɑ⁸x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽⁸⁺¹⁾) + (ɑ^(175+0)x⁽⁷⁺¹⁾) + (ɑ^(238+0)x⁽⁶⁺¹⁾) + (ɑ^(208+0)x⁽⁵⁺¹⁾) + (ɑ^(249+0)x⁽⁴⁺¹⁾) + (ɑ^(215+0)x⁽³⁺¹⁾) + (ɑ^(252+0)x⁽²⁺¹⁾) + (ɑ^(196+0)x⁽¹⁺¹⁾) + (ɑ⁽²⁸⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺⁸⁾x⁽⁸⁺⁰⁾) + (ɑ^(175+8)x⁽⁷⁺⁰⁾) + (ɑ^(238+8)x⁽⁶⁺⁰⁾) + (ɑ^(208+8)x⁽⁵⁺⁰⁾) + (ɑ^(249+8)x⁽⁴⁺⁰⁾) + (ɑ^(215+8)x⁽³⁺⁰⁾) + (ɑ^(252+8)x⁽²⁺⁰⁾) + (ɑ^(196+8)x⁽¹⁺⁰⁾) + (ɑ⁽²⁸⁺⁸⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x⁹ + ɑ¹⁷⁵x⁸ + ɑ²³⁸x⁷ + ɑ²⁰⁸x⁶ + ɑ²⁴⁹x⁵ + ɑ²¹⁵x⁴ + ɑ²⁵²x³ + ɑ¹⁹⁶x² + ɑ²⁸x¹ + ɑ⁸x⁸ + ɑ¹⁸³x⁷ + ɑ²⁴⁶x⁶ + ɑ²¹⁶x⁵ + ɑ²⁵⁷x⁴ + ɑ²²³x³ + ɑ²⁶⁰x² + ɑ²⁰⁴x¹ + ɑ³⁶x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256)
ɑ⁰x⁹ + ɑ¹⁷⁵x⁸ + ɑ²³⁸x⁷ + ɑ²⁰⁸x⁶ + ɑ²⁴⁹x⁵ + ɑ²¹⁵x⁴ + ɑ²⁵²x³ + ɑ¹⁹⁶x² + ɑ²⁸x¹ + ɑ⁸x⁸ + ɑ¹⁸³x⁷ + ɑ²⁴⁶x⁶ + ɑ²¹⁶x⁵ + ɑ²x⁴ + ɑ²²³x³ + ɑ⁵x² + ɑ²⁰⁴x¹ + ɑ³⁶x⁰

Now you have to combine terms that have the same exponent. In this case, there is more than one x⁸, x⁷, x⁶, x⁵, x⁴, x³, x² and x¹ term.

We will now combine the x⁸ terms. they are:
ɑ¹⁷⁵x⁸ + ɑ⁸x⁸

We will convert the above alphas to integers using the log antilog table to make this step easier:
255x⁸ + 29x⁸

After XORing the integers, we convert the result back to an alpha term.
(255 ⊕ 29)x⁸ = 226x⁸ = ɑ⁹⁵x⁸

We will now combine the x⁷ terms. they are:
ɑ²³⁸x⁷ + ɑ¹⁸³x⁷

We will convert the above alphas to integers using the log antilog table to make this step easier:
11x⁷ + 196x⁷

After XORing the integers, we convert the result back to an alpha term.
(11 ⊕ 196)x⁷ = 207x⁷ = ɑ²⁴⁶x⁷

We will now combine the x⁶ terms. they are:
ɑ²⁰⁸x⁶ + ɑ²⁴⁶x⁶

We will convert the above alphas to integers using the log antilog table to make this step easier:
81x⁶ + 207x⁶

After XORing the integers, we convert the result back to an alpha term.
(81 ⊕ 207)x⁶ = 158x⁶ = ɑ¹³⁷x⁶

We will now combine the x⁵ terms. they are:
ɑ²⁴⁹x⁵ + ɑ²¹⁶x⁵

We will convert the above alphas to integers using the log antilog table to make this step easier:
54x⁵ + 195x⁵

After XORing the integers, we convert the result back to an alpha term.
(54 ⊕ 195)x⁵ = 245x⁵ = ɑ²³¹x⁵

We will now combine the x⁴ terms. they are:
ɑ²¹⁵x⁴ + ɑ²x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
239x⁴ + 4x⁴

After XORing the integers, we convert the result back to an alpha term.
(239 ⊕ 4)x⁴ = 235x⁴ = ɑ²³⁵x⁴

We will now combine the x³ terms. they are:
ɑ²⁵²x³ + ɑ²²³x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
173x³ + 9x³

After XORing the integers, we convert the result back to an alpha term.
(173 ⊕ 9)x³ = 164x³ = ɑ¹⁴⁹x³

We will now combine the x² terms. they are:
ɑ¹⁹⁶x² + ɑ⁵x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
200x² + 32x²

After XORing the integers, we convert the result back to an alpha term.
(200 ⊕ 32)x² = 232x² = ɑ¹¹x²

We will now combine the x¹ terms. they are:
ɑ²⁸x¹ + ɑ²⁰⁴x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
24x¹ + 221x¹

After XORing the integers, we convert the result back to an alpha term.
(24 ⊕ 221)x¹ = 197x¹ = ɑ¹²³x¹

Below is the final polynomial from this step after combining the like terms. If we needed 9 error correction code words, this would be the generator polynomial required.
ɑ⁰x⁹ + ɑ⁹⁵x⁸ + ɑ²⁴⁶x⁷ + ɑ¹³⁷x⁶ + ɑ²³¹x⁵ + ɑ²³⁵x⁴ + ɑ¹⁴⁹x³ + ɑ¹¹x² + ɑ¹²³x¹ + ɑ³⁶x⁰

Multiplication Step #9

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x⁹ + ɑ⁹⁵x⁸ + ɑ²⁴⁶x⁷ + ɑ¹³⁷x⁶ + ɑ²³¹x⁵ + ɑ²³⁵x⁴ + ɑ¹⁴⁹x³ + ɑ¹¹x² + ɑ¹²³x¹ + ɑ³⁶x⁰) by (ɑ⁰x¹ + ɑ⁹x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x⁹ * ɑ⁰x¹) + (ɑ⁹⁵x⁸ * ɑ⁰x¹) + (ɑ²⁴⁶x⁷ * ɑ⁰x¹) + (ɑ¹³⁷x⁶ * ɑ⁰x¹) + (ɑ²³¹x⁵ * ɑ⁰x¹) + (ɑ²³⁵x⁴ * ɑ⁰x¹) + (ɑ¹⁴⁹x³ * ɑ⁰x¹) + (ɑ¹¹x² * ɑ⁰x¹) + (ɑ¹²³x¹ * ɑ⁰x¹) + (ɑ³⁶x⁰ * ɑ⁰x¹) + (ɑ⁰x⁹ * ɑ⁹x⁰) + (ɑ⁹⁵x⁸ * ɑ⁹x⁰) + (ɑ²⁴⁶x⁷ * ɑ⁹x⁰) + (ɑ¹³⁷x⁶ * ɑ⁹x⁰) + (ɑ²³¹x⁵ * ɑ⁹x⁰) + (ɑ²³⁵x⁴ * ɑ⁹x⁰) + (ɑ¹⁴⁹x³ * ɑ⁹x⁰) + (ɑ¹¹x² * ɑ⁹x⁰) + (ɑ¹²³x¹ * ɑ⁹x⁰) + (ɑ³⁶x⁰ * ɑ⁹x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽⁹⁺¹⁾) + (ɑ⁽⁹⁵⁺⁰⁾x⁽⁸⁺¹⁾) + (ɑ^(246+0)x⁽⁷⁺¹⁾) + (ɑ^(137+0)x⁽⁶⁺¹⁾) + (ɑ^(231+0)x⁽⁵⁺¹⁾) + (ɑ^(235+0)x⁽⁴⁺¹⁾) + (ɑ^(149+0)x⁽³⁺¹⁾) + (ɑ⁽¹¹⁺⁰⁾x⁽²⁺¹⁾) + (ɑ^(123+0)x⁽¹⁺¹⁾) + (ɑ⁽³⁶⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺⁹⁾x⁽⁹⁺⁰⁾) + (ɑ⁽⁹⁵⁺⁹⁾x⁽⁸⁺⁰⁾) + (ɑ^(246+9)x⁽⁷⁺⁰⁾) + (ɑ^(137+9)x⁽⁶⁺⁰⁾) + (ɑ^(231+9)x⁽⁵⁺⁰⁾) + (ɑ^(235+9)x⁽⁴⁺⁰⁾) + (ɑ^(149+9)x⁽³⁺⁰⁾) + (ɑ⁽¹¹⁺⁹⁾x⁽²⁺⁰⁾) + (ɑ^(123+9)x⁽¹⁺⁰⁾) + (ɑ⁽³⁶⁺⁹⁾x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x¹⁰ + ɑ⁹⁵x⁹ + ɑ²⁴⁶x⁸ + ɑ¹³⁷x⁷ + ɑ²³¹x⁶ + ɑ²³⁵x⁵ + ɑ¹⁴⁹x⁴ + ɑ¹¹x³ + ɑ¹²³x² + ɑ³⁶x¹ + ɑ⁹x⁹ + ɑ¹⁰⁴x⁸ + ɑ²⁵⁵x⁷ + ɑ¹⁴⁶x⁶ + ɑ²⁴⁰x⁵ + ɑ²⁴⁴x⁴ + ɑ¹⁵⁸x³ + ɑ²⁰x² + ɑ¹³²x¹ + ɑ⁴⁵x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x⁹, x⁸, x⁷, x⁶, x⁵, x⁴, x³, x² and x¹ term.

We will now combine the x⁹ terms. they are:
ɑ⁹⁵x⁹ + ɑ⁹x⁹

We will convert the above alphas to integers using the log antilog table to make this step easier:
226x⁹ + 58x⁹

After XORing the integers, we convert the result back to an alpha term.
(226 ⊕ 58)x⁹ = 216x⁹ = ɑ²⁵¹x⁹

We will now combine the x⁸ terms. they are:
ɑ²⁴⁶x⁸ + ɑ¹⁰⁴x⁸

We will convert the above alphas to integers using the log antilog table to make this step easier:
207x⁸ + 13x⁸

After XORing the integers, we convert the result back to an alpha term.
(207 ⊕ 13)x⁸ = 194x⁸ = ɑ⁶⁷x⁸

We will now combine the x⁷ terms. they are:
ɑ¹³⁷x⁷ + ɑ²⁵⁵x⁷

We will convert the above alphas to integers using the log antilog table to make this step easier:
158x⁷ + 1x⁷

After XORing the integers, we convert the result back to an alpha term.
(158 ⊕ 1)x⁷ = 159x⁷ = ɑ⁴⁶x⁷

We will now combine the x⁶ terms. they are:
ɑ²³¹x⁶ + ɑ¹⁴⁶x⁶

We will convert the above alphas to integers using the log antilog table to make this step easier:
245x⁶ + 154x⁶

After XORing the integers, we convert the result back to an alpha term.
(245 ⊕ 154)x⁶ = 111x⁶ = ɑ⁶¹x⁶

We will now combine the x⁵ terms. they are:
ɑ²³⁵x⁵ + ɑ²⁴⁰x⁵

We will convert the above alphas to integers using the log antilog table to make this step easier:
235x⁵ + 44x⁵

After XORing the integers, we convert the result back to an alpha term.
(235 ⊕ 44)x⁵ = 199x⁵ = ɑ¹¹⁸x⁵

We will now combine the x⁴ terms. they are:
ɑ¹⁴⁹x⁴ + ɑ²⁴⁴x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
164x⁴ + 250x⁴

After XORing the integers, we convert the result back to an alpha term.
(164 ⊕ 250)x⁴ = 94x⁴ = ɑ⁷⁰x⁴

We will now combine the x³ terms. they are:
ɑ¹¹x³ + ɑ¹⁵⁸x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
232x³ + 183x³

After XORing the integers, we convert the result back to an alpha term.
(232 ⊕ 183)x³ = 95x³ = ɑ⁶⁴x³

We will now combine the x² terms. they are:
ɑ¹²³x² + ɑ²⁰x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
197x² + 180x²

After XORing the integers, we convert the result back to an alpha term.
(197 ⊕ 180)x² = 113x² = ɑ⁹⁴x²

We will now combine the x¹ terms. they are:
ɑ³⁶x¹ + ɑ¹³²x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
37x¹ + 184x¹

After XORing the integers, we convert the result back to an alpha term.
(37 ⊕ 184)x¹ = 157x¹ = ɑ³²x¹

Below is the final polynomial from this step after combining the like terms. If we needed 10 error correction code words, this would be the generator polynomial required.
ɑ⁰x¹⁰ + ɑ²⁵¹x⁹ + ɑ⁶⁷x⁸ + ɑ⁴⁶x⁷ + ɑ⁶¹x⁶ + ɑ¹¹⁸x⁵ + ɑ⁷⁰x⁴ + ɑ⁶⁴x³ + ɑ⁹⁴x² + ɑ³²x¹ + ɑ⁴⁵x⁰

Multiplication Step #10

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x¹⁰ + ɑ²⁵¹x⁹ + ɑ⁶⁷x⁸ + ɑ⁴⁶x⁷ + ɑ⁶¹x⁶ + ɑ¹¹⁸x⁵ + ɑ⁷⁰x⁴ + ɑ⁶⁴x³ + ɑ⁹⁴x² + ɑ³²x¹ + ɑ⁴⁵x⁰) by (ɑ⁰x¹ + ɑ¹⁰x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x¹⁰ * ɑ⁰x¹) + (ɑ²⁵¹x⁹ * ɑ⁰x¹) + (ɑ⁶⁷x⁸ * ɑ⁰x¹) + (ɑ⁴⁶x⁷ * ɑ⁰x¹) + (ɑ⁶¹x⁶ * ɑ⁰x¹) + (ɑ¹¹⁸x⁵ * ɑ⁰x¹) + (ɑ⁷⁰x⁴ * ɑ⁰x¹) + (ɑ⁶⁴x³ * ɑ⁰x¹) + (ɑ⁹⁴x² * ɑ⁰x¹) + (ɑ³²x¹ * ɑ⁰x¹) + (ɑ⁴⁵x⁰ * ɑ⁰x¹) + (ɑ⁰x¹⁰ * ɑ¹⁰x⁰) + (ɑ²⁵¹x⁹ * ɑ¹⁰x⁰) + (ɑ⁶⁷x⁸ * ɑ¹⁰x⁰) + (ɑ⁴⁶x⁷ * ɑ¹⁰x⁰) + (ɑ⁶¹x⁶ * ɑ¹⁰x⁰) + (ɑ¹¹⁸x⁵ * ɑ¹⁰x⁰) + (ɑ⁷⁰x⁴ * ɑ¹⁰x⁰) + (ɑ⁶⁴x³ * ɑ¹⁰x⁰) + (ɑ⁹⁴x² * ɑ¹⁰x⁰) + (ɑ³²x¹ * ɑ¹⁰x⁰) + (ɑ⁴⁵x⁰ * ɑ¹⁰x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽¹⁰⁺¹⁾) + (ɑ^(251+0)x⁽⁹⁺¹⁾) + (ɑ⁽⁶⁷⁺⁰⁾x⁽⁸⁺¹⁾) + (ɑ⁽⁴⁶⁺⁰⁾x⁽⁷⁺¹⁾) + (ɑ⁽⁶¹⁺⁰⁾x⁽⁶⁺¹⁾) + (ɑ^(118+0)x⁽⁵⁺¹⁾) + (ɑ⁽⁷⁰⁺⁰⁾x⁽⁴⁺¹⁾) + (ɑ⁽⁶⁴⁺⁰⁾x⁽³⁺¹⁾) + (ɑ⁽⁹⁴⁺⁰⁾x⁽²⁺¹⁾) + (ɑ⁽³²⁺⁰⁾x⁽¹⁺¹⁾) + (ɑ⁽⁴⁵⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺¹⁰⁾x⁽¹⁰⁺⁰⁾) + (ɑ^(251+10)x⁽⁹⁺⁰⁾) + (ɑ^(67+10)x⁽⁸⁺⁰⁾) + (ɑ^(46+10)x⁽⁷⁺⁰⁾) + (ɑ^(61+10)x⁽⁶⁺⁰⁾) + (ɑ^(118+10)x⁽⁵⁺⁰⁾) + (ɑ^(70+10)x⁽⁴⁺⁰⁾) + (ɑ^(64+10)x⁽³⁺⁰⁾) + (ɑ^(94+10)x⁽²⁺⁰⁾) + (ɑ^(32+10)x⁽¹⁺⁰⁾) + (ɑ^(45+10)x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x¹¹ + ɑ²⁵¹x¹⁰ + ɑ⁶⁷x⁹ + ɑ⁴⁶x⁸ + ɑ⁶¹x⁷ + ɑ¹¹⁸x⁶ + ɑ⁷⁰x⁵ + ɑ⁶⁴x⁴ + ɑ⁹⁴x³ + ɑ³²x² + ɑ⁴⁵x¹ + ɑ¹⁰x¹⁰ + ɑ²⁶¹x⁹ + ɑ⁷⁷x⁸ + ɑ⁵⁶x⁷ + ɑ⁷¹x⁶ + ɑ¹²⁸x⁵ + ɑ⁸⁰x⁴ + ɑ⁷⁴x³ + ɑ¹⁰⁴x² + ɑ⁴²x¹ + ɑ⁵⁵x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256)
ɑ⁰x¹¹ + ɑ²⁵¹x¹⁰ + ɑ⁶⁷x⁹ + ɑ⁴⁶x⁸ + ɑ⁶¹x⁷ + ɑ¹¹⁸x⁶ + ɑ⁷⁰x⁵ + ɑ⁶⁴x⁴ + ɑ⁹⁴x³ + ɑ³²x² + ɑ⁴⁵x¹ + ɑ¹⁰x¹⁰ + ɑ⁶x⁹ + ɑ⁷⁷x⁸ + ɑ⁵⁶x⁷ + ɑ⁷¹x⁶ + ɑ¹²⁸x⁵ + ɑ⁸⁰x⁴ + ɑ⁷⁴x³ + ɑ¹⁰⁴x² + ɑ⁴²x¹ + ɑ⁵⁵x⁰

Now you have to combine terms that have the same exponent. In this case, there is more than one x¹⁰, x⁹, x⁸, x⁷, x⁶, x⁵, x⁴, x³, x² and x¹ term.

We will now combine the x¹⁰ terms. they are:
ɑ²⁵¹x¹⁰ + ɑ¹⁰x¹⁰

We will convert the above alphas to integers using the log antilog table to make this step easier:
216x¹⁰ + 116x¹⁰

After XORing the integers, we convert the result back to an alpha term.
(216 ⊕ 116)x¹⁰ = 172x¹⁰ = ɑ²²⁰x¹⁰

We will now combine the x⁹ terms. they are:
ɑ⁶⁷x⁹ + ɑ⁶x⁹

We will convert the above alphas to integers using the log antilog table to make this step easier:
194x⁹ + 64x⁹

After XORing the integers, we convert the result back to an alpha term.
(194 ⊕ 64)x⁹ = 130x⁹ = ɑ¹⁹²x⁹

We will now combine the x⁸ terms. they are:
ɑ⁴⁶x⁸ + ɑ⁷⁷x⁸

We will convert the above alphas to integers using the log antilog table to make this step easier:
159x⁸ + 60x⁸

After XORing the integers, we convert the result back to an alpha term.
(159 ⊕ 60)x⁸ = 163x⁸ = ɑ⁹¹x⁸

We will now combine the x⁷ terms. they are:
ɑ⁶¹x⁷ + ɑ⁵⁶x⁷

We will convert the above alphas to integers using the log antilog table to make this step easier:
111x⁷ + 93x⁷

After XORing the integers, we convert the result back to an alpha term.
(111 ⊕ 93)x⁷ = 50x⁷ = ɑ¹⁹⁴x⁷

We will now combine the x⁶ terms. they are:
ɑ¹¹⁸x⁶ + ɑ⁷¹x⁶

We will convert the above alphas to integers using the log antilog table to make this step easier:
199x⁶ + 188x⁶

After XORing the integers, we convert the result back to an alpha term.
(199 ⊕ 188)x⁶ = 123x⁶ = ɑ¹⁷²x⁶

We will now combine the x⁵ terms. they are:
ɑ⁷⁰x⁵ + ɑ¹²⁸x⁵

We will convert the above alphas to integers using the log antilog table to make this step easier:
94x⁵ + 133x⁵

After XORing the integers, we convert the result back to an alpha term.
(94 ⊕ 133)x⁵ = 219x⁵ = ɑ¹⁷⁷x⁵

We will now combine the x⁴ terms. they are:
ɑ⁶⁴x⁴ + ɑ⁸⁰x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
95x⁴ + 253x⁴

After XORing the integers, we convert the result back to an alpha term.
(95 ⊕ 253)x⁴ = 162x⁴ = ɑ²⁰⁹x⁴

We will now combine the x³ terms. they are:
ɑ⁹⁴x³ + ɑ⁷⁴x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
113x³ + 137x³

After XORing the integers, we convert the result back to an alpha term.
(113 ⊕ 137)x³ = 248x³ = ɑ¹¹⁶x³

We will now combine the x² terms. they are:
ɑ³²x² + ɑ¹⁰⁴x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
157x² + 13x²

After XORing the integers, we convert the result back to an alpha term.
(157 ⊕ 13)x² = 144x² = ɑ²²⁷x²

We will now combine the x¹ terms. they are:
ɑ⁴⁵x¹ + ɑ⁴²x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
193x¹ + 181x¹

After XORing the integers, we convert the result back to an alpha term.
(193 ⊕ 181)x¹ = 116x¹ = ɑ¹⁰x¹

Below is the final polynomial from this step after combining the like terms. If we needed 11 error correction code words, this would be the generator polynomial required.
ɑ⁰x¹¹ + ɑ²²⁰x¹⁰ + ɑ¹⁹²x⁹ + ɑ⁹¹x⁸ + ɑ¹⁹⁴x⁷ + ɑ¹⁷²x⁶ + ɑ¹⁷⁷x⁵ + ɑ²⁰⁹x⁴ + ɑ¹¹⁶x³ + ɑ²²⁷x² + ɑ¹⁰x¹ + ɑ⁵⁵x⁰

Multiplication Step #11

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x¹¹ + ɑ²²⁰x¹⁰ + ɑ¹⁹²x⁹ + ɑ⁹¹x⁸ + ɑ¹⁹⁴x⁷ + ɑ¹⁷²x⁶ + ɑ¹⁷⁷x⁵ + ɑ²⁰⁹x⁴ + ɑ¹¹⁶x³ + ɑ²²⁷x² + ɑ¹⁰x¹ + ɑ⁵⁵x⁰) by (ɑ⁰x¹ + ɑ¹¹x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x¹¹ * ɑ⁰x¹) + (ɑ²²⁰x¹⁰ * ɑ⁰x¹) + (ɑ¹⁹²x⁹ * ɑ⁰x¹) + (ɑ⁹¹x⁸ * ɑ⁰x¹) + (ɑ¹⁹⁴x⁷ * ɑ⁰x¹) + (ɑ¹⁷²x⁶ * ɑ⁰x¹) + (ɑ¹⁷⁷x⁵ * ɑ⁰x¹) + (ɑ²⁰⁹x⁴ * ɑ⁰x¹) + (ɑ¹¹⁶x³ * ɑ⁰x¹) + (ɑ²²⁷x² * ɑ⁰x¹) + (ɑ¹⁰x¹ * ɑ⁰x¹) + (ɑ⁵⁵x⁰ * ɑ⁰x¹) + (ɑ⁰x¹¹ * ɑ¹¹x⁰) + (ɑ²²⁰x¹⁰ * ɑ¹¹x⁰) + (ɑ¹⁹²x⁹ * ɑ¹¹x⁰) + (ɑ⁹¹x⁸ * ɑ¹¹x⁰) + (ɑ¹⁹⁴x⁷ * ɑ¹¹x⁰) + (ɑ¹⁷²x⁶ * ɑ¹¹x⁰) + (ɑ¹⁷⁷x⁵ * ɑ¹¹x⁰) + (ɑ²⁰⁹x⁴ * ɑ¹¹x⁰) + (ɑ¹¹⁶x³ * ɑ¹¹x⁰) + (ɑ²²⁷x² * ɑ¹¹x⁰) + (ɑ¹⁰x¹ * ɑ¹¹x⁰) + (ɑ⁵⁵x⁰ * ɑ¹¹x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽¹¹⁺¹⁾) + (ɑ^(220+0)x⁽¹⁰⁺¹⁾) + (ɑ^(192+0)x⁽⁹⁺¹⁾) + (ɑ⁽⁹¹⁺⁰⁾x⁽⁸⁺¹⁾) + (ɑ^(194+0)x⁽⁷⁺¹⁾) + (ɑ^(172+0)x⁽⁶⁺¹⁾) + (ɑ^(177+0)x⁽⁵⁺¹⁾) + (ɑ^(209+0)x⁽⁴⁺¹⁾) + (ɑ^(116+0)x⁽³⁺¹⁾) + (ɑ^(227+0)x⁽²⁺¹⁾) + (ɑ⁽¹⁰⁺⁰⁾x⁽¹⁺¹⁾) + (ɑ⁽⁵⁵⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺¹¹⁾x⁽¹¹⁺⁰⁾) + (ɑ^(220+11)x⁽¹⁰⁺⁰⁾) + (ɑ^(192+11)x⁽⁹⁺⁰⁾) + (ɑ^(91+11)x⁽⁸⁺⁰⁾) + (ɑ^(194+11)x⁽⁷⁺⁰⁾) + (ɑ^(172+11)x⁽⁶⁺⁰⁾) + (ɑ^(177+11)x⁽⁵⁺⁰⁾) + (ɑ^(209+11)x⁽⁴⁺⁰⁾) + (ɑ^(116+11)x⁽³⁺⁰⁾) + (ɑ^(227+11)x⁽²⁺⁰⁾) + (ɑ^(10+11)x⁽¹⁺⁰⁾) + (ɑ^(55+11)x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x¹² + ɑ²²⁰x¹¹ + ɑ¹⁹²x¹⁰ + ɑ⁹¹x⁹ + ɑ¹⁹⁴x⁸ + ɑ¹⁷²x⁷ + ɑ¹⁷⁷x⁶ + ɑ²⁰⁹x⁵ + ɑ¹¹⁶x⁴ + ɑ²²⁷x³ + ɑ¹⁰x² + ɑ⁵⁵x¹ + ɑ¹¹x¹¹ + ɑ²³¹x¹⁰ + ɑ²⁰³x⁹ + ɑ¹⁰²x⁸ + ɑ²⁰⁵x⁷ + ɑ¹⁸³x⁶ + ɑ¹⁸⁸x⁵ + ɑ²²⁰x⁴ + ɑ¹²⁷x³ + ɑ²³⁸x² + ɑ²¹x¹ + ɑ⁶⁶x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x¹¹, x¹⁰, x⁹, x⁸, x⁷, x⁶, x⁵, x⁴, x³, x² and x¹ term.

We will now combine the x¹¹ terms. they are:
ɑ²²⁰x¹¹ + ɑ¹¹x¹¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
172x¹¹ + 232x¹¹

After XORing the integers, we convert the result back to an alpha term.
(172 ⊕ 232)x¹¹ = 68x¹¹ = ɑ¹⁰²x¹¹

We will now combine the x¹⁰ terms. they are:
ɑ¹⁹²x¹⁰ + ɑ²³¹x¹⁰

We will convert the above alphas to integers using the log antilog table to make this step easier:
130x¹⁰ + 245x¹⁰

After XORing the integers, we convert the result back to an alpha term.
(130 ⊕ 245)x¹⁰ = 119x¹⁰ = ɑ⁴³x¹⁰

We will now combine the x⁹ terms. they are:
ɑ⁹¹x⁹ + ɑ²⁰³x⁹

We will convert the above alphas to integers using the log antilog table to make this step easier:
163x⁹ + 224x⁹

After XORing the integers, we convert the result back to an alpha term.
(163 ⊕ 224)x⁹ = 67x⁹ = ɑ⁹⁸x⁹

We will now combine the x⁸ terms. they are:
ɑ¹⁹⁴x⁸ + ɑ¹⁰²x⁸

We will convert the above alphas to integers using the log antilog table to make this step easier:
50x⁸ + 68x⁸

After XORing the integers, we convert the result back to an alpha term.
(50 ⊕ 68)x⁸ = 118x⁸ = ɑ¹²¹x⁸

We will now combine the x⁷ terms. they are:
ɑ¹⁷²x⁷ + ɑ²⁰⁵x⁷

We will convert the above alphas to integers using the log antilog table to make this step easier:
123x⁷ + 167x⁷

After XORing the integers, we convert the result back to an alpha term.
(123 ⊕ 167)x⁷ = 220x⁷ = ɑ¹⁸⁷x⁷

We will now combine the x⁶ terms. they are:
ɑ¹⁷⁷x⁶ + ɑ¹⁸³x⁶

We will convert the above alphas to integers using the log antilog table to make this step easier:
219x⁶ + 196x⁶

After XORing the integers, we convert the result back to an alpha term.
(219 ⊕ 196)x⁶ = 31x⁶ = ɑ¹¹³x⁶

We will now combine the x⁵ terms. they are:
ɑ²⁰⁹x⁵ + ɑ¹⁸⁸x⁵

We will convert the above alphas to integers using the log antilog table to make this step easier:
162x⁵ + 165x⁵

After XORing the integers, we convert the result back to an alpha term.
(162 ⊕ 165)x⁵ = 7x⁵ = ɑ¹⁹⁸x⁵

We will now combine the x⁴ terms. they are:
ɑ¹¹⁶x⁴ + ɑ²²⁰x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
248x⁴ + 172x⁴

After XORing the integers, we convert the result back to an alpha term.
(248 ⊕ 172)x⁴ = 84x⁴ = ɑ¹⁴³x⁴

We will now combine the x³ terms. they are:
ɑ²²⁷x³ + ɑ¹²⁷x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
144x³ + 204x³

After XORing the integers, we convert the result back to an alpha term.
(144 ⊕ 204)x³ = 92x³ = ɑ¹³¹x³

We will now combine the x² terms. they are:
ɑ¹⁰x² + ɑ²³⁸x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
116x² + 11x²

After XORing the integers, we convert the result back to an alpha term.
(116 ⊕ 11)x² = 127x² = ɑ⁸⁷x²

We will now combine the x¹ terms. they are:
ɑ⁵⁵x¹ + ɑ²¹x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
160x¹ + 117x¹

After XORing the integers, we convert the result back to an alpha term.
(160 ⊕ 117)x¹ = 213x¹ = ɑ¹⁵⁷x¹

Below is the final polynomial from this step after combining the like terms. If we needed 12 error correction code words, this would be the generator polynomial required.
ɑ⁰x¹² + ɑ¹⁰²x¹¹ + ɑ⁴³x¹⁰ + ɑ⁹⁸x⁹ + ɑ¹²¹x⁸ + ɑ¹⁸⁷x⁷ + ɑ¹¹³x⁶ + ɑ¹⁹⁸x⁵ + ɑ¹⁴³x⁴ + ɑ¹³¹x³ + ɑ⁸⁷x² + ɑ¹⁵⁷x¹ + ɑ⁶⁶x⁰

Multiplication Step #12

Using the polynomial that we got from the previous step, we will multiply (ɑ⁰x¹² + ɑ¹⁰²x¹¹ + ɑ⁴³x¹⁰ + ɑ⁹⁸x⁹ + ɑ¹²¹x⁸ + ɑ¹⁸⁷x⁷ + ɑ¹¹³x⁶ + ɑ¹⁹⁸x⁵ + ɑ¹⁴³x⁴ + ɑ¹³¹x³ + ɑ⁸⁷x² + ɑ¹⁵⁷x¹ + ɑ⁶⁶x⁰) by (ɑ⁰x¹ + ɑ¹²x⁰)

After multiplying each term from the first part with each term from the second part, we get:
(ɑ⁰x¹² * ɑ⁰x¹) + (ɑ¹⁰²x¹¹ * ɑ⁰x¹) + (ɑ⁴³x¹⁰ * ɑ⁰x¹) + (ɑ⁹⁸x⁹ * ɑ⁰x¹) + (ɑ¹²¹x⁸ * ɑ⁰x¹) + (ɑ¹⁸⁷x⁷ * ɑ⁰x¹) + (ɑ¹¹³x⁶ * ɑ⁰x¹) + (ɑ¹⁹⁸x⁵ * ɑ⁰x¹) + (ɑ¹⁴³x⁴ * ɑ⁰x¹) + (ɑ¹³¹x³ * ɑ⁰x¹) + (ɑ⁸⁷x² * ɑ⁰x¹) + (ɑ¹⁵⁷x¹ * ɑ⁰x¹) + (ɑ⁶⁶x⁰ * ɑ⁰x¹) + (ɑ⁰x¹² * ɑ¹²x⁰) + (ɑ¹⁰²x¹¹ * ɑ¹²x⁰) + (ɑ⁴³x¹⁰ * ɑ¹²x⁰) + (ɑ⁹⁸x⁹ * ɑ¹²x⁰) + (ɑ¹²¹x⁸ * ɑ¹²x⁰) + (ɑ¹⁸⁷x⁷ * ɑ¹²x⁰) + (ɑ¹¹³x⁶ * ɑ¹²x⁰) + (ɑ¹⁹⁸x⁵ * ɑ¹²x⁰) + (ɑ¹⁴³x⁴ * ɑ¹²x⁰) + (ɑ¹³¹x³ * ɑ¹²x⁰) + (ɑ⁸⁷x² * ɑ¹²x⁰) + (ɑ¹⁵⁷x¹ * ɑ¹²x⁰) + (ɑ⁶⁶x⁰ * ɑ¹²x⁰)

To multiply, you add the exponents together like this:
(ɑ⁽⁰⁺⁰⁾x⁽¹²⁺¹⁾) + (ɑ^(102+0)x⁽¹¹⁺¹⁾) + (ɑ⁽⁴³⁺⁰⁾x⁽¹⁰⁺¹⁾) + (ɑ⁽⁹⁸⁺⁰⁾x⁽⁹⁺¹⁾) + (ɑ^(121+0)x⁽⁸⁺¹⁾) + (ɑ^(187+0)x⁽⁷⁺¹⁾) + (ɑ^(113+0)x⁽⁶⁺¹⁾) + (ɑ^(198+0)x⁽⁵⁺¹⁾) + (ɑ^(143+0)x⁽⁴⁺¹⁾) + (ɑ^(131+0)x⁽³⁺¹⁾) + (ɑ⁽⁸⁷⁺⁰⁾x⁽²⁺¹⁾) + (ɑ^(157+0)x⁽¹⁺¹⁾) + (ɑ⁽⁶⁶⁺⁰⁾x⁽⁰⁺¹⁾) + (ɑ⁽⁰⁺¹²⁾x⁽¹²⁺⁰⁾) + (ɑ^(102+12)x⁽¹¹⁺⁰⁾) + (ɑ^(43+12)x⁽¹⁰⁺⁰⁾) + (ɑ^(98+12)x⁽⁹⁺⁰⁾) + (ɑ^(121+12)x⁽⁸⁺⁰⁾) + (ɑ^(187+12)x⁽⁷⁺⁰⁾) + (ɑ^(113+12)x⁽⁶⁺⁰⁾) + (ɑ^(198+12)x⁽⁵⁺⁰⁾) + (ɑ^(143+12)x⁽⁴⁺⁰⁾) + (ɑ^(131+12)x⁽³⁺⁰⁾) + (ɑ^(87+12)x⁽²⁺⁰⁾) + (ɑ^(157+12)x⁽¹⁺⁰⁾) + (ɑ^(66+12)x⁽⁰⁺⁰⁾)

After adding the exponents, this is the result:
ɑ⁰x¹³ + ɑ¹⁰²x¹² + ɑ⁴³x¹¹ + ɑ⁹⁸x¹⁰ + ɑ¹²¹x⁹ + ɑ¹⁸⁷x⁸ + ɑ¹¹³x⁷ + ɑ¹⁹⁸x⁶ + ɑ¹⁴³x⁵ + ɑ¹³¹x⁴ + ɑ⁸⁷x³ + ɑ¹⁵⁷x² + ɑ⁶⁶x¹ + ɑ¹²x¹² + ɑ¹¹⁴x¹¹ + ɑ⁵⁵x¹⁰ + ɑ¹¹⁰x⁹ + ɑ¹³³x⁸ + ɑ¹⁹⁹x⁷ + ɑ¹²⁵x⁶ + ɑ²¹⁰x⁵ + ɑ¹⁵⁵x⁴ + ɑ¹⁴³x³ + ɑ⁹⁹x² + ɑ¹⁶⁹x¹ + ɑ⁷⁸x⁰

If there are any exponents that are larger than 255, fix them by putting them through the formula
(exponent % 256) + floor(exponent / 256). In this case, all the exponents are less than or equal to 255, so we can continue.

Now you have to combine terms that have the same exponent. In this case, there is more than one x¹², x¹¹, x¹⁰, x⁹, x⁸, x⁷, x⁶, x⁵, x⁴, x³, x² and x¹ term.

We will now combine the x¹² terms. they are:
ɑ¹⁰²x¹² + ɑ¹²x¹²

We will convert the above alphas to integers using the log antilog table to make this step easier:
68x¹² + 205x¹²

After XORing the integers, we convert the result back to an alpha term.
(68 ⊕ 205)x¹² = 137x¹² = ɑ⁷⁴x¹²

We will now combine the x¹¹ terms. they are:
ɑ⁴³x¹¹ + ɑ¹¹⁴x¹¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
119x¹¹ + 62x¹¹

After XORing the integers, we convert the result back to an alpha term.
(119 ⊕ 62)x¹¹ = 73x¹¹ = ɑ¹⁵²x¹¹

We will now combine the x¹⁰ terms. they are:
ɑ⁹⁸x¹⁰ + ɑ⁵⁵x¹⁰

We will convert the above alphas to integers using the log antilog table to make this step easier:
67x¹⁰ + 160x¹⁰

After XORing the integers, we convert the result back to an alpha term.
(67 ⊕ 160)x¹⁰ = 227x¹⁰ = ɑ¹⁷⁶x¹⁰

We will now combine the x⁹ terms. they are:
ɑ¹²¹x⁹ + ɑ¹¹⁰x⁹

We will convert the above alphas to integers using the log antilog table to make this step easier:
118x⁹ + 103x⁹

After XORing the integers, we convert the result back to an alpha term.
(118 ⊕ 103)x⁹ = 17x⁹ = ɑ¹⁰⁰x⁹

We will now combine the x⁸ terms. they are:
ɑ¹⁸⁷x⁸ + ɑ¹³³x⁸

We will convert the above alphas to integers using the log antilog table to make this step easier:
220x⁸ + 109x⁸

After XORing the integers, we convert the result back to an alpha term.
(220 ⊕ 109)x⁸ = 177x⁸ = ɑ⁸⁶x⁸

We will now combine the x⁷ terms. they are:
ɑ¹¹³x⁷ + ɑ¹⁹⁹x⁷

We will convert the above alphas to integers using the log antilog table to make this step easier:
31x⁷ + 14x⁷

After XORing the integers, we convert the result back to an alpha term.
(31 ⊕ 14)x⁷ = 17x⁷ = ɑ¹⁰⁰x⁷

We will now combine the x⁶ terms. they are:
ɑ¹⁹⁸x⁶ + ɑ¹²⁵x⁶

We will convert the above alphas to integers using the log antilog table to make this step easier:
7x⁶ + 51x⁶

After XORing the integers, we convert the result back to an alpha term.
(7 ⊕ 51)x⁶ = 52x⁶ = ɑ¹⁰⁶x⁶

We will now combine the x⁵ terms. they are:
ɑ¹⁴³x⁵ + ɑ²¹⁰x⁵

We will convert the above alphas to integers using the log antilog table to make this step easier:
84x⁵ + 89x⁵

After XORing the integers, we convert the result back to an alpha term.
(84 ⊕ 89)x⁵ = 13x⁵ = ɑ¹⁰⁴x⁵

We will now combine the x⁴ terms. they are:
ɑ¹³¹x⁴ + ɑ¹⁵⁵x⁴

We will convert the above alphas to integers using the log antilog table to make this step easier:
92x⁴ + 114x⁴

After XORing the integers, we convert the result back to an alpha term.
(92 ⊕ 114)x⁴ = 46x⁴ = ɑ¹³⁰x⁴

We will now combine the x³ terms. they are:
ɑ⁸⁷x³ + ɑ¹⁴³x³

We will convert the above alphas to integers using the log antilog table to make this step easier:
127x³ + 84x³

After XORing the integers, we convert the result back to an alpha term.
(127 ⊕ 84)x³ = 43x³ = ɑ²¹⁸x³

We will now combine the x² terms. they are:
ɑ¹⁵⁷x² + ɑ⁹⁹x²

We will convert the above alphas to integers using the log antilog table to make this step easier:
213x² + 134x²

After XORing the integers, we convert the result back to an alpha term.
(213 ⊕ 134)x² = 83x² = ɑ²⁰⁶x²

We will now combine the x¹ terms. they are:
ɑ⁶⁶x¹ + ɑ¹⁶⁹x¹

We will convert the above alphas to integers using the log antilog table to make this step easier:
97x¹ + 229x¹

After XORing the integers, we convert the result back to an alpha term.
(97 ⊕ 229)x¹ = 132x¹ = ɑ¹⁴⁰x¹

Below is the final polynomial from this step after combining the like terms. If we needed 13 error correction code words, this would be the generator polynomial required.
ɑ⁰x¹³ + ɑ⁷⁴x¹² + ɑ¹⁵²x¹¹ + ɑ¹⁷⁶x¹⁰ + ɑ¹⁰⁰x⁹ + ɑ⁸⁶x⁸ + ɑ¹⁰⁰x⁷ + ɑ¹⁰⁶x⁶ + ɑ¹⁰⁴x⁵ + ɑ¹³⁰x⁴ + ɑ²¹⁸x³ + ɑ²⁰⁶x² + ɑ¹⁴⁰x¹ + ɑ⁷⁸x⁰

Final result:
(ɑ⁰x¹³ + ɑ⁷⁴x¹² + ɑ¹⁵²x¹¹ + ɑ¹⁷⁶x¹⁰ + ɑ¹⁰⁰x⁹ + ɑ⁸⁶x⁸ + ɑ¹⁰⁰x⁷ + ɑ¹⁰⁶x⁶ + ɑ¹⁰⁴x⁵ + ɑ¹³⁰x⁴ + ɑ²¹⁸x³ + ɑ²⁰⁶x² + ɑ¹⁴⁰x¹ + ɑ⁷⁸x⁰)