Question

how to put -5 and 4+3i as zeros?
Ex. -3 is a zero. (x+3)
Ex. 5i is a zero. (x+5i)(x-5i)

Answer 1

Both in one equation or two separate equations?

Answer 2

3 separate

Answer 3

But you have only two zeros. How three separate equations?

Answer 4

i comes in pairs

Answer 5

But that is not what I am asking. Are the given zeros to the same equation or separate equations.

Answer 6

given to the same polynomial

Answer 7

Exactly that was my original question.
Proceed as follows:

Answer 8

n=3
-5 and 4+3i are zeroes
f(2)=91

Answer 9

find the nth-degree polynomial function with real coefficients satisfying the given conditions

Answer 10

-5 and 4+3i are zeros

x = -5
x + 5 = 0. Therefore (x+5) is one factor.

x = 4 + 3i
x - 4 = 3i
square both sides to get rid of the "i""
(x-4)^2 = (3i)^2 = -9
x^2 - 8x + 16 = 9
x^2 - 8x + 7 = 0. Therefore, (x^2 - 8x + 7) is another factor.

Multiplying them together and introduce a constant in front of the polynomial for the more general case: The polynomial f(x) = a(x + 5)(x^2 - 8x + 7)
f(2)=91
put x = 2 and f(x) = 91:
91 = a(2+5)(2^2 -8*2 + 7)
91 = a(7)(4-16+7)
91 = a(7)(-5)
a = -91/35

f(x) =-91/35(x + 5)(x^2 - 8x + 7)
multiply the (x + 5)(x^2 - 8x + 7) part.

just double check my arithmetic to see if they are correct.

Answer 11

thank you that helped a lot

Answer 12

You are welcome.

Answer 13

Just noticed a mistake above. Correcting it here:
-5 and 4+3i are zeros

x = -5
x + 5 = 0. Therefore (x+5) is one factor.

x = 4 + 3i
x - 4 = 3i Square both sides to get rid of the "i"
(x-4)^2 = (3i)^2 = -9
x^2 - 8x + 16 = -9
x^2 - 8x + 25 = 0. Therefore, (x^2 - 8x + 25) is another factor.

Multiplying the two factors together and introduce a constant in front of the polynomial for the more general case: The polynomial f(x) = a(x + 5)(x^2 - 8x + 25)
f(2)=91 put x = 2 and f(x) = 91:
91 = a(2+5)(2^2 -8*2 + 25)
91 = a(7)(4-16+25)
91 = a(7)(13)
91 = 91a
a = 1

f(x) = (x + 5)(x^2 - 8x + 25)
f(x) = x^3 - 8x^2 + 25x + 5x^2 - 40x + 125
f(x) = x^3 - 3x^2 - 15x + 125 is the polynomial.