Question

Find the unique polynomial in factored form with real coefficients that meets these cond:
degree:4
zeros:-5,12-3i,f(0)=130

Answer 1

fx= (x+5)^2(x-1)(x-(2-3i))+130
I got this

Answer 2

alternate forms assuming f and x are real.
fx = x4 + 7 x3 - 3 x2 + i ( 3 x3 + 27 x2 + 45 x - 75) - 55 x + 180

Answer 3

not quite ... with complex zeros come in pairs
2+3i is also a zero

and dont add 130 at end...only if x=0, f = 130
there is an "a" coefficient in fron of factors dictated by point f(0) = 130

\[f(x) = a(x+5)(x-1)(x-(2-3i))(x-(2+3i))\]
you have to expand to get rid of "i" and solve for "a"

Answer 4

wait but where are the exponents and 130.. ? you don't add that in? @dumbcow

Answer 5

no for example ... say zeros are -5,1 f(0) = 130
you would say...
f(x) = (x+5)(x-1) +130

but if you plug in x=0, it wont work
f(0) = (5)(-1) +130 = 125

make sense?

Answer 6

oh I see!

Answer 7

So how can I approach this problem

Answer 8

write it in factored form and put a coefficient in front
...in prev example

--> f(x) = a(x+5)(x-1)
f(0) = a(5)(-1) = 130
-5a = 130
a = -26

Answer 9

ohhhh wait can we totally disregard the imaginary numbers

Answer 10

now for complex zeros ...you can write it out as factors then expand
but i suggest this approach
\[x = 2-3i\]
work backwards until right side is 0
\[x-2 = -3i\]
\[(x-2)^{2} = (-3i)^{2}\]
\[x^{2} -4x +4 = -9\]
\[x^{2} -4x +13 = 0\]
that is the factor for both complex zeros
\[f(x) = a(x+5)(x-1)(x^{2} -4x+13)\]

Answer 11

iohhh

Answer 12

a=-26

Answer 13

haha no i only did that to make it easier to understand the "a" concept

no a is not -26 ... just for my example

Answer 14

lol oh ok

Answer 15

its not hard solving for a though ... just plug in x=0

Answer 16

ahhh

Answer 17

wait and y=130?

Answer 18

yes

Answer 19

oh wow so

Answer 20

y=2(x+5)(x-1)(x^2-4x+13)

Answer 21

yep...well it should be -2 in front

Answer 22

YESYES lol sorry that's what I got too ahaha

Answer 23

THNX!

Answer 24

yw :)