odehye
Find the zeros for P(x) = 2x^3 - 7x^2 - 15x
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ash2326
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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 ?
\[2x^2-7x-15=0\]
ash2326
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Do you get this part @odehye ?
odehye
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yes, i did
ash2326
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Can you solve the quadratic to find the other two roots?
odehye
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i'm lost
odehye
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using P/Q theorem
rob1525
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The easiest way is to us the quadratic formual to solve for zeros.
odehye
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x=5 and x=-3/2
odehye
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@ash2326 x=x=5 and x=-3/2