#1**+1 **

Just like in the previous problem you posted, factor out the GCF of each term in the parentheses.

\((8x^4+8x^3)+(27x+27)\) | Now, factor out the GCF. |

\(8x^3(x+1)+27(x+1)\) | (x+1) is common to both factors, so we can rewrite this. |

\((8x^3+27)(x+1)\) | (8x^3+27) happens to be a sum of cubes and must be dealth with as such. |

\(a=\sqrt[3]{8x^3}=\sqrt[3]{(2x)^3}=2x\\ b=\sqrt[3]{27}=3\) | Now that a and b have been identified expand the sum of cubes. |

\((2x+3)((2x)^2-(3)(2x)+3^2)(x+1)\) | Now, simplify. |

\((2x+3)(4x^2-6x+9)(x+1)\) | 4x^2-6x+9 is an irreducible trinomial and nothing else can be reduced further. |

TheXSquaredFactor
Nov 28, 2017

#1**+1 **

Best Answer

Just like in the previous problem you posted, factor out the GCF of each term in the parentheses.

\((8x^4+8x^3)+(27x+27)\) | Now, factor out the GCF. |

\(8x^3(x+1)+27(x+1)\) | (x+1) is common to both factors, so we can rewrite this. |

\((8x^3+27)(x+1)\) | (8x^3+27) happens to be a sum of cubes and must be dealth with as such. |

\(a=\sqrt[3]{8x^3}=\sqrt[3]{(2x)^3}=2x\\ b=\sqrt[3]{27}=3\) | Now that a and b have been identified expand the sum of cubes. |

\((2x+3)((2x)^2-(3)(2x)+3^2)(x+1)\) | Now, simplify. |

\((2x+3)(4x^2-6x+9)(x+1)\) | 4x^2-6x+9 is an irreducible trinomial and nothing else can be reduced further. |

TheXSquaredFactor
Nov 28, 2017