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fari321

  • 3 years ago

One factor of 4x^3+15x^2-31x-30 is x-2. Find the other factors

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  1. freewilly922
    • 3 years ago
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    You could use long division or synthetic division. Your choice.

  2. fari321
    • 3 years ago
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    how

  3. fari321
    • 3 years ago
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    with long division

  4. freewilly922
    • 3 years ago
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    Well to start with you know that \[4x^2 \times (x-2) = 4x^3-8x^2\] so just like you would do long division with numbers you can also do it with variables like this. so \[ 4x^3-15x^2 -\bigg[4x^3-8x^2 \bigg]=-7x^2\] you drop the -31x like long division to get \[-7x^2 -31x\] and then divide this by (x-2) again \[-7x\times (x-2) = -7x^2 +14x\] and so on

  5. freewilly922
    • 3 years ago
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    sorry the \[-15x^2 \] should be positive 15

  6. freewilly922
    • 3 years ago
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    so \[4x^3+15x^2 -\bigg[4x^3-8x^2 \bigg]=23x^2\] \[23x\times (x-2) = 23x^2-46x\] \[23x^2-31x -\bigg[23x^2-46x\bigg]=+15x\] \[15\times(x-2)=15x-30\] \[+15x-30 -\bigg[15x-30 \bigg]=0\] so \[4x^3+15x^2-31x-30 \to (x-2)(4x^2+23x+15)\] and you can factor the rest... hopefully

  7. freewilly922
    • 3 years ago
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    Explaining how to do long division or synthetic division is a real pain on open study, I would suggest you take the time outside of this to study it from a book with examples. Or go online and search for a youtube video. It isn't hard.

  8. fari321
    • 3 years ago
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    thank you

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