Question 1000165: Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 4; zeros: -2-5i; -4 multiplicity 2
Found 2 solutions by stanbon, Boreal:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 4; zeros: -2-5i; -4 multiplicity 2
Comment:: Since the coefficients are Real, -2+5i is also a zero.
-----
f(x) = (x-(-2-5i))(x-(-2+5i)(x+4)^2
-----
f(x) = ((x+2)+5i)((x+2)-5i)(x+4)^2
-----
f(x) = [(x+2)^2+5^2](x+4)^2
-----
f(x) = (x^2+4x+29)(x^2+8x+16)
----------
Cheers,
Stan H.
--------------
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! (x+4)^2 deals with root of -4 with multiplicity 2.
(1/2a)(-b+/- sqrt(b^2-4ac) has to be -2 +5i and -2 -5i, because complex roots are conjugates. The square root has to be -25 or some multiple of it. b^2-4ac=-25*constant. b=2 or some multiple of it. Let's make a 1, (1/2)(4+/- sqrt(16-4c)). Here, b is 4. To get 5i, I need the square root of 100, because I want 10i, which divided by 2 gives 5i. So 4c=116, and c=29.
x^2-4x+29 would be that polynomial.
(x+4)^2(x^-4x+29)
x^4+4x^3+13x^2+168x+464
|
|
|
