Question

I'll give a medal if you help me :D
Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form. 2, -4, and 1 + 3i

Answer 1

from the zeros, make the factors
then calculate the product of these factors and you will have your polynomial
finally arrange the terms in decreasing power of variable.

Answer 2

I tried but I can't seem to factor the 1+3i and the 1-3i

Answer 3

(x-2)(x+4)[x-(1+3i)][x-(1-3i)]
(x-2)(x+4)(x-1-3i)(x-1+3i)
calculate this product

Answer 4

how do you factor (x-1-3i)(x-1+3i)?

Answer 5

you don't have to factor......you have to multiply them

Answer 6

I meant multiply, sorry, do you multiply everything by the 2 1's or is 1-3i a single term? I keep getting confused on that part

Answer 7

x, -1 and -3i are to be treated as independent of each other....

Answer 8

(x-2)(x+4)[x-(1+3i)][x-(1-3i)]
(x-2)(x+4)(x-1-3i)(x-1+3i)
for first two brackets
(x-2)(x+4) = (x^2 - 2x + 4x - 8) = (x^2 + 2x -8)

Answer 9

is it x^4 + x^2 + 34x - 72?

Answer 10

Have not calculated complete answer....hold on

Answer 11

for next two brackets
(x-1-3i)(x-1+3i) = (x^2 - x - 3ix - x + 1 + 3i + 3ix - 3i - 9 i^2)
= (x^2 - 2x + 1 - 9*-1) = (x^2 - 2x + 10)

Answer 12

So,
(x-2)(x+4)[x-(1+3i)][x-(1-3i)] = (x^2 + 2x -8)(x^2 - 2x + 10)
= (x^4 + 2x^3 - 8x^2 - 2x^3 - 4x^2 + 16x + 10x^2 + 20x - 80)
= (x^4 - 2x^2 + 36x - 80)

Answer 13

My answer is different from yours...............pls check if my calculations are correct .......

Answer 14

I think you're right, I probably made a mistake in the multiplying :P

Answer 15

ok.....so now you know how to handle imaginary roots.....

Answer 16

yeah, thank you so much!

Answer 17

u r welcome....☺