## anonymous one year ago f(x) = 2x^3 - 5x^2 + px -5 = 0. I've found the roots: 1+2i and 1-2i. How do I solve the equation? Do I simply use the formula to find the quadratic? (x-(1-2i))(x-(1+2i))?

1. anonymous

I found the quadratic to be: \[x ^{2}-2x+5\] What do I do from here?

2. triciaal

missing a root, x^3 means it will have 3 roots

3. triciaal

how did you find the roots with 2 unknowns?

4. triciaal

given the eqn factor then find what value of p makes the eqn true

5. triciaal

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6. triciaal

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7. triciaal

any questions?

8. triciaal

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9. anonymous

I'm just reading your answer. It game me one root and the other is the complex conjugate.

10. anonymous

I got p = 12. (x-(1-2i))(x-(1+2i)) =x^2 -2x + 5 if you multiply that with 2x-1, you arrive at the original question. so x is also 1/2

11. triciaal

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12. triciaal

I found the value of p the roots would then be the values of x when (2x - 5) = 0 and (x^2 + 1) = 0 x = 5/2 and I

13. triciaal

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14. triciaal

p = 2 review your steps when you have 12 maybe it should be p+2*5 =12 maybe you forgot to multiply by the leading coefficient 2 you did not show your work