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

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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 ? \[2x^2-7x-15=0\]
Do you get this part @odehye ?
yes, i did

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Other answers:

Can you solve the quadratic to find the other two roots?
i'm lost
using P/Q theorem
The easiest way is to us the quadratic formual to solve for zeros.
x=5 and x=-3/2
@ash2326 x=x=5 and x=-3/2

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