(b3⋅(8b)+8)÷(b-2)

Remove parentheses.

b3⋅(8b)+8b-2

Reorder b3 and 8.

8b3b+8b-2

Add 3 and 1.

8b4+8b-2

8b4+8b-2

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 0.

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 |

Divide the highest order term in the dividend 8b4 by the highest order term in divisor b.

8b3 | |||||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 |

Multiply the new quotient term by the divisor.

8b3 | |||||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

+ | 8b4 | – | 16b3 |

The expression needs to be subtracted from the dividend, so change all the signs in 8b4-16b3

8b3 | |||||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 |

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

8b3 | |||||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 |

Pull the next terms from the original dividend down into the current dividend.

8b3 | |||||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 |

Divide the highest order term in the dividend 16b3 by the highest order term in divisor b.

8b3 | + | 16b2 | |||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 |

Multiply the new quotient term by the divisor.

8b3 | + | 16b2 | |||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

+ | 16b3 | – | 32b2 |

The expression needs to be subtracted from the dividend, so change all the signs in 16b3-32b2

8b3 | + | 16b2 | |||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 |

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

8b3 | + | 16b2 | |||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 |

Pull the next terms from the original dividend down into the current dividend.

8b3 | + | 16b2 | |||||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b |

Divide the highest order term in the dividend 32b2 by the highest order term in divisor b.

8b3 | + | 16b2 | + | 32b | |||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b |

Multiply the new quotient term by the divisor.

8b3 | + | 16b2 | + | 32b | |||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b | ||||||||||

+ | 32b2 | – | 64b |

The expression needs to be subtracted from the dividend, so change all the signs in 32b2-64b

8b3 | + | 16b2 | + | 32b | |||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b | ||||||||||

– | 32b2 | + | 64b |

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

8b3 | + | 16b2 | + | 32b | |||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b | ||||||||||

– | 32b2 | + | 64b | ||||||||||

+ | 64b |

Pull the next terms from the original dividend down into the current dividend.

8b3 | + | 16b2 | + | 32b | |||||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b | ||||||||||

– | 32b2 | + | 64b | ||||||||||

+ | 64b | + | 8 |

Divide the highest order term in the dividend 64b by the highest order term in divisor b.

8b3 | + | 16b2 | + | 32b | + | 64 | |||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b | ||||||||||

– | 32b2 | + | 64b | ||||||||||

+ | 64b | + | 8 |

Multiply the new quotient term by the divisor.

8b3 | + | 16b2 | + | 32b | + | 64 | |||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b | ||||||||||

– | 32b2 | + | 64b | ||||||||||

+ | 64b | + | 8 | ||||||||||

+ | 64b | – | 128 |

The expression needs to be subtracted from the dividend, so change all the signs in 64b-128

8b3 | + | 16b2 | + | 32b | + | 64 | |||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b | ||||||||||

– | 32b2 | + | 64b | ||||||||||

+ | 64b | + | 8 | ||||||||||

– | 64b | + | 128 |

8b3 | + | 16b2 | + | 32b | + | 64 | |||||||

b | – | 2 | 8b4 | + | 0b3 | + | 0b2 | + | 0b | + | 8 | ||

– | 8b4 | + | 16b3 | ||||||||||

+ | 16b3 | + | 0b2 | ||||||||||

– | 16b3 | + | 32b2 | ||||||||||

+ | 32b2 | + | 0b | ||||||||||

– | 32b2 | + | 64b | ||||||||||

+ | 64b | + | 8 | ||||||||||

– | 64b | + | 128 | ||||||||||

+ | 136 |

The final answer is the quotient plus the remainder over the divisor.

8b3+16b2+32b+64+136b-2

Divide (b^3*(8b)+8)÷(b-2)