Question

form a fifth-degree polynomial function with real coefficients such that 5i,1-2i, and -5 are zeros and f(0)=1875

Answer 1

well its

P(x) = a(x +5)(x -5i)(x + 5i)(x - 1 + 2i)(x - 1 -2i)

a complex root has a congugate pair...

Answer 2

then after distributing, substitute x = 0 and P(x) = 1875 to find a

Answer 3

Roots: 5i,1-2i, and -5

x = 5i
square both sides:
x^2 = -25
x^2 + 25 = 0
Therefore, (x^2+25) is a factor.

x = 1-2i
x - 1 = -2i
(x-1)^2 = 4
x^2 - 2x + 1 - 4 = 0
x^2 - 2x - 3 = 0
Therefore, x^2 - 2x - 3 is a factor

x = -5
Therefore, (x+5) is a factor

Multiply (x^2+25), (x^2 - 2x - 3) and (x+5)

Once you get the polynomial, write it as a constant a * (the polynomial)

f(0)=1875
put x = 0 and f(x) = 1875 to solve for a.

Answer 4

wait whats thw answer

Answer 5

Can't give out answers. I have done most of the work. All you have to do is do a few steps to complete the problem.

Answer 6

ok so this is the polynomial that I got x^5+3x^4+12x^3+60x^2-325x-375

Answer 7

what do I do next?

Answer 8

okay, your polynomial is P(x) = a * ( x^5+3x^4+12x^3+60x^2-325x-375)
put x = 0 and P(x) = 1875 and find 'a'.

Answer 9

ok

Answer 10

I think there is a slight error

x = 1 - 2i

so

x - 1 = -2i

squaring both gives

\[(x - 1)^2 = 4i^2\]

so

(x -1)^2 = -4

because i^2 = -1

Answer 11

yes, sorry about that. The second factor should be (x^2 - 2x + 5)

Answer 12

ok

Answer 13

So the polynomial is: P(x) = a * ( x^5+3x^4+20x^3+100x^2-125x+625 )

Answer 14

1875= a * ( 0^5+3(0)^4+20(0)^3+100(0)^2-125(0)+625 )

Answer 15

Yes. 1875 = 625 * a
a = 1875 / 625 = 3

P(x) = 3(x^5+3x^4+20x^3+100x^2-125x+625)

Answer 16

ok so thats the answer

Answer 17

Yes.

Answer 18

wait but dont I have to multiply three by everything

Answer 19

Either way should be fine. One is in factored form and another is the expanded form but both are the same polynomial.

Answer 20

THANK YOU SO MUCH

Answer 21

You are welcome.