Question 1100423: Form the fifth-degree polynomial function with real coefficients,
given that 1+i is a zero, and that the graph passes through (-2,0),
(0,8)(1,0)
Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388)   (Show Source):
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
I assumed you meant 1+1 instead of 11+i, since 11+1 made it so complicated.
-2 and 1 are zeros because the graph passes through (-2,0) and (1,0).
1+i is a zero, and since it has real coefficients,
its conjugate 1-i is also a zero.
Therefore when the polynomial, f(x) is set equal to 0, we get
x = -2, x = 1, x = 1+i, x = 1-i
But there must be one more real zero, so that it will have degree 5.
Let that fifth zero be "a", making the polynomial have 5 zeros
x = -2, x = 1, x = 1+i, x = 1-i, x = a
Getting 0 on each side we have
x+2 = 0, x-1 = 0, x-1-i = 0, x-1+i = 0, x-a = 0
Multiplying all left sides and right sides:
(x+2)(x-1)(x-1-i)(x-1+i)(x-a) = (0)(0)(0)(0)(0)
(x+2)(x-1)[(x-1)-i][(x-1)+i](x-a) = 0
So
f(x) = (x+2)(x-1)(x-1-i)(x-1+i)(x-a)
We are also given that the graph passes through (0,8), so
f(0) = (0+2)(0-1)(0-1-i)(0-1+i)(0-a) = 8
(2)(-1)(-1+i)(-1-i)(-a) = 8
(2a)(-1+i)(-1-i) = 8
2a(1-iČ) = 8
2a[1-(-1)] = 8
2a(1+1) = 8
2a(2) = 8
4a = 8
a = 2
Therefore
f(x) = (x+2)(x-1)(x-1-i)(x-1+i)(x-2)
Now you hav to go to all the troubl to muiltiply
that out and collect terms. It may help to group
the first two terms in the factors with the i's as
f(x) = (x+2)(x-1)[(x-1)-i][(x-1)+i](x-2)
So I'll leave the multiplying up to you.
Edwin
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