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jaramillotina

  • 3 years ago

can you please show me how to solve x^3-6x^2=16x-96

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  1. Maegan
    • 3 years ago
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    First bring everything over to the same side of the =. This will make 16 become a negtive and 96 a positive. x^3 - 6x^2-16X+96

  2. Maegan
    • 3 years ago
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    Then treat them as 2 different equation soo (x^3 - 6x^2) and (-16x+96)

  3. Maegan
    • 3 years ago
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    Then simplfy each: (x^3 - 6x^2) = x^2(x - 6) and (-16x+96) = -16(x-6) So your answer is (x-6) & (X^2-16)

  4. Maegan
    • 3 years ago
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    Then simplfy them so (x-6) becomes x-6=0 Bring the 6 over x=6 And (x^2-16)=0 x^2=16 \[\sqrt{x}=\sqrt{16}\] x=4

  5. joemath314159
    • 3 years ago
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    youre just missing one more solution. instead of taking the x^2-16=0 and solving, you could have factored even further:\[x^2-16=0\iff (x-4)(x+4)=0\iff x=4,x=-4\]so overall you should have x= 6, 4, -4 as solutions.

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