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Product of Linear FactorsProblem: Rewrite (4x^2+5x)^2 - 5(4x^2+5x) - 6 as a product of linear factors. I have no idea where to start. I tried using the substitution method to get (4x^2 + 5x - 6)(4x^2 + 5x + 1) and using the quadratic formula to get x = 3/4, -2, -1, and -1/4. |

That's a good approach. If you've got x = 3/4, then 4x - 3 = 0. (4x - 3)(x + 2)(4x + 1)(x + 1) is equivalent to the original expression. |

Quote:
I overlooked 4x - 3 = 0 and used x = 3/4 instead. |

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