Question

PLEASE HELP
Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

5, -3, and -1 + 3i

Answer 1

\((x-5)(x+3)(x+1)(x-3i)(x+3i)\)

If you know the zero of your polynomial, then you can factor out the polynomial as \(x-x_1)(x-x_2)\cdots\) where \(x_1,x_2, \ldots\) are the zeros.

We multiply though the last term \((x-3i)\) by its conjugate though since you need real coefficients, and \((x-3i)(x+3i)\) is a factorization of a difference of squares: \((x^2+9) = (x^2 -(3i)^2)\)

Answer 2

Just expand this product to get it in standard form

Answer 3

i got x4-4x3+15x2+25+150

Answer 4

It needs to be of degree 5 since you have a product of 5 terms each with x in them.

Answer 5

does that mean that the exponents add up to five? (degree 5)?

Answer 6

it means that the term with the highest exponent is 5, that means you need some term \(x^5\)

Answer 7

i dont think thats right... none of the answers i have to choose from are to the degree of 5 then

Answer 8

f(x) = x4 + 12.5x2 - 50x - 150

f(x) = x4 - 4x3 + 15x2 + 25x + 150

f(x) = x4 - 4x3 - 15x2 - 25x - 150

f(x) = x4 - 9x2 - 50x - 150

Answer 9

Oooh I see why... I am terribly sorry I misread the zeroes... I read it as 5, -3, -1, 3i ... my bad! !
the product is then \((x-5)(x+3)(x-(-1+3i))(x-(-1-3i))=(x-5)(x+3)(x+1-3i)(x+1+3i) \)

Answer 10

then how do you do that?

Answer 11

just multiply it out

Answer 12

i cant see all of what you wrote... :(

Answer 13

\((x-5)(x+3)(x-(-1+3i))(x-(-1-3i))=\\
(x-5)(x+3)(x+1-3i)(x+1+3i)\)

Answer 14

i dont understand how to multiply the 3i

Answer 15

just treat it as any number that you usually use.