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A polynomial P is given. Find all zeros of P, real and complex, and factor p completely?
A polynomial P is given. P(x) = x^4 + 81x^2. (a) find all zeros of P, real and complex. (Enter answers as a comma-separated list. Enter all answers including repetitions.) and (b) Factor P completely.
here, you should notice that you can factor out x², right?
So if you do that you get: x²(x² + 81)
The zero's are when x² = 0 and x² + 81 = 0.
Can you solve each of these equations?
Answer that, then we will take it from there if need be.
Hopefully no one will spoil you the answer. That would be very irresponsible of them. Don't forget to vote me best answer for being the first to Walk you through without spoiling the answer. That way it gives you a chance to work on it and to get good at it.
(a) 0, 0, 9i, -9i.
- nyphdinmdLv 71 month ago
P(x) = x^4 + 81 x^2
Set to zero --> 0 = x^4 + 81x^2 = x^2*(x^2 + 81)
Right away we see there is a degenerative (repeated) root x = 0 that accounts fro two real roots
then we have x^2 + 81 = 0 --> x = +/- sqrt(-81) = +/- 9 i
so there are two complex (actually pure imaginary) roots x = -9 i, +9 i
so x = 0, 0, -9 i, 9 i
P(x) = x^2(x^2 +81) --> that's all you can do in factoring
- Anonymous1 month ago
X^2(x^2 + 81) = 0
x^2 = 0; x^2 + 81 = 0
x = 0, x = -9, 9
-9, 0, 9
x^2(x^2 + 81)
x^2(x + 9)(x - 9)