SOLUTION: N =4; -4(⅓), and 2 +3i are zeros; f(1) = 100
1)Since 32+3i are zeros. Conjugate them making it 2-3i.
2)Then set x to each one.
3) x=-3, x=¼, x=3+2i, x=3-2i
x+3=0, 4x=1, x
Algebra.Com
Question 1103671: N =4; -4(⅓), and 2 +3i are zeros; f(1) = 100
1)Since 32+3i are zeros. Conjugate them making it 2-3i.
2)Then set x to each one.
3) x=-3, x=¼, x=3+2i, x=3-2i
x+3=0, 4x=1, x-3-2i=0, x-3+2i=0, x+3=0, 4x=1,x-3-2i=0, x-3+2i=0, 4x-1 =0
4) Use the zero factor property in reverse. When doing that multiply it by a constant “a”
a(x+3)(4x-1)(x-3-2i)=0
5)
F(x) = a(x+3)(4x-1)(x-3-2i)(x-3+2i)
F(x) = a(x+3)(4x-1)[(x-3)-2i][(x-3)+2i]
F(x) = a(4x²+11x-3)[(x-3)²-(2i)²]
F(x) = a(4x²+11x-3)[x²-6x+9-4i²]
F(x) = a(4x²+11x-3)[x²-6x+9-4(-1)]
F(x) = a(4x²+11x-3)[x²-6x+9+4]
F(x) = a(4x²+11x-3)(x²-6x+13)
F(x) = a(4x4-24x³+52x²+11x³-66x²+143-3x²+18x-39)
F(x) = a(4x4-13x³-17x²+161x-39)
F(1) = 100 = a(4∙14-13∙1^3-17∙1^2+161∙1-39)
F(1) = 100 = a(4∙16-13∙8-17∙4+161∙2-39)
F(1) = 100 = a(96)
100/96=a
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!
(1) I can see what you are doing in the work you show; but I don't know how you get to your starting point from the information you show on the first line of your message.
What does the "N=4" mean?
And it appears you are saying that "-4(⅓)" is a root; but I don't know how to interpret that expression. From the work you show, the real roots are -3 and 1/4....
(2) But my real concern is this: is there a question here?
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