Question

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

3, -13, and 5 + 4i

Answer 1

P(x) = (x-3)(x+13)(x-5-4i)(x-5+4i)

Answer 2

f(x) = x^4 - 8x^3 - 12x^2 + 400x - 1599

f(x) = x^4 - 200x^2 + 800x - 1599

f(x) = x^4 - 98x^2 + 800x - 1599

f(x) = x^4 - 8x^3 + 12x^2 - 400x + 1599

Answer 3

@experimentX

Answer 4

Do you understand that for each zero C there must be a factor, (x-C)?

And also - any complex root must also have its complex conjugate as a root.

So since 5+4i is a root, 5-4i is a root.

experimentX used these concepts in setting up the polynomial above, but I just want to make sure you understand them.

Answer 5

So the set-up is:

\[\Large P(x) = (x-3)(x+13)[x-(5-4i)][x-(5+4i)]\]

Answer 6

Start by dealing with the product involving the complex conjugates. use some neat little algebra tricks to simplify:

\(\large [x-(5-4i)][x-(5+4i)]=[x-5+4i][x-5-4i]\) by distribution

\(\large =[(x-5)+4i][(x-5)-4i]\) by associativity

Now look at that product... it should look familiar, as a form (a+b)(a-b), where
a=(x-5)
b=4i

Take that product and you'll have a 2nd degree polynomial.

So you'll be at this point:
\[\Large P(x) = (x-3)(x+13)(........)\]
Where that last .... is the 2nd degree polynomial obtained above. Then foil the first 2 factors, you'll have 2 polynomials, and then just use distribution.

Answer 7

@timo86m

Answer 8

http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem

This states that complex roots always come in pairs

a+bi its pair would be a-bi

in your case

3, -13, and 5 + 4i AND...

5-4i <- the missing pair :)
----

Hold on i will explain more

Answer 9

3, -13, 5 + 4i ,5-4i
a, b , c , d
THe roots can be written like this

(x-a) * (x-b) * (x-c) *(x-d) we just plug in

(x-2) * (x-(-13)) * (x-c) *(x-d) The first 2 are straitforward.

(x-2) * (x-(-13)) * (x-(5 + 4i)) *(x-(5-4i)) c and you just plug in as well.

now you expand

Answer 10

But how do you do that with the i

Answer 11

lets do that first then
(x-(5 + 4i)) *(x-(5-4i))
(x-5-4i)*(x-5+4i) got rid of parenthesis by distributing the - with me so far?