|
Question 102465: the degree three polynomial f(x) with real coefficients and leading coefficient 1, has -3 and +4i among its roots. Express f(x) as the productof of linear and quadriatic polynomials with real coefficients
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! the degree three polynomial f(x) with real coefficients and leading coefficient 1, has -3 and +4i among its roots. Express f(x) as the productof of linear and quadriatic polynomials with real coefficients
If a polynomial has all real coefficients and has A+Bi as a root,
then its conjugate A-Bi is also a root of the the polynomial.
So since the polynomial we want has +4i as a root, then it also
has -4i as a root.
That's because +4i is really 0+4i and so its conjugate is 0-4i,
which is just -4i.
So we work the problem backwards:
We are to end up with:
x = -3; x = 4i x = -4i
Before that we must have had this:
x + 3 = 0; x - 4i = 0; x + 4i = 0
Before that we must have had this:
(x + 3)(x - 4i)(x + 4i) = 0
Multiplying the second and third parentheses together
using FOIL:
(x + 3)(x² + 4ix - 4ix - 16i²) = 0
Canceling the +4ix and the -4ix
(x + 3)(x² - 16i²) = 0
Replacing i² by -1
(x + 3)[x² -16(-1)] = 0
Simplifying:
(x + 3)[x² + 16] = 0
(x + 3)(x² + 16) = 0
Multiplying those using FOIL:
x³ + 16x + 3x² + 48 = 0
Arranging in descending order:
x³ + 3x² + 16x + 48 = 0
Edwin
|
|
|
| |
