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odehye

  • 3 years ago

Find the zeros for P(x) = 2x^3 - 7x^2 - 15x

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  1. ash2326
    • 3 years ago
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    @odehye We have to find the solution of this \[2x^3-7x^2-15x=0\] Take x common \[x(2x^2-7x-15)=0\] Now we have \[x=0\ \text{or}\ (2x^2-7x-15)=0\] So one root is zero Can you solve this quadratic equation to find the other two roots ? \[2x^2-7x-15=0\]

  2. ash2326
    • 3 years ago
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    Do you get this part @odehye ?

  3. odehye
    • 3 years ago
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    yes, i did

  4. ash2326
    • 3 years ago
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    Can you solve the quadratic to find the other two roots?

  5. odehye
    • 3 years ago
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    i'm lost

  6. odehye
    • 3 years ago
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    using P/Q theorem

  7. rob1525
    • 3 years ago
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    The easiest way is to us the quadratic formual to solve for zeros.

  8. odehye
    • 3 years ago
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    x=5 and x=-3/2

  9. odehye
    • 3 years ago
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    @ash2326 x=x=5 and x=-3/2

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