Question

polynomials in standard form

Answer 1

i have absolutely no idea how to do these :P
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 2

nice one.

Answer 3

standard form means: \(a_nx^n + a_{n-1}x^{n-1} + \dots + a_0\) (degree from highest to lowest --> which is 0 ).

you'll construct the polynomial like this (an example to explain):

if the question was: the zeros include 0 and 2, we would say that
"a polynomial p(x) having 2 as zero can be factorized as \((x-2)q(x)\)".
0 is also zero so, q(x) can be decomposed as \(x h(x)\).
nothing more is asked. let's put \(h(x)=1\) , to keep the degree low.
We constructed the polynomial \(x(x-2)\).



In this exercise, 2 and -4 are zeros. So the polynomial you will construct (call is P(x)) is going to be factoriz-able as (x-2)(x+4)Q(x). --at least that.--
Q(x) will be such that the other conditions are met (1+3i is a zero).

Answer 4

do you understand the approach?

Answer 5

so it will be (x-2)(x+4)(1+3i)?

Answer 6

almost , nope.
The factorization of P(x) is \((x-2)(x+4) \times\)something
something is a polynomial with real coefficients, therefore the following polynomial is not accepted as answer: \( (x-2)(x+4)(x- (1+3i))\).

you must find a "trick". Do you know of polynomial equations who have complex numbers as solution, and yet there is no complex number in the equation?

Answer 7

bro your speaking a different language right now.... couldn't you just multiply it by i?

Answer 8

no.. (and sorry for the language)
i was thinking of "\(x^2 = -1\)". polynomial of second order with a negative \(\Delta\) will do the trick.

The theory says that a if \(1+3i\) is a zero of the polynomial \(Q(x)\), then \(1-3i\) is also a zero.
Therefore \(Q(x)=(x-(1+3i))(x-(1-3i))\) is a second order polynomial and if you compute it in standard form, you'll see no trace of "\(i\)".

Answer 9

i ment language as in your speaking math to a chic who speaks highschool XD okay so i basically do foil

Answer 10

did you get that thing about complex zeros?

P(x) is just (x-2)(x+4)Q(x) = (x-2)(x+4)(x-(1+3i))(x-(1)3i)).
You have to obtain P(x) in the form \(a_4x^4+a_3x^3 + a_2x^2 + a_1x + a_0\).
that 's a litle bit tedious.

Answer 11

typo:
the last parenthesis is (x-(1-3i)), not (x-(1)3i)) !

Answer 12

so i foil the first two (x-2)(x+4)=x^2-2x+4x-8=x^2+2x-8
then i foil the last two x=(x-(1+3i))(x-(1-3i))
x(x) + x(-1) + x(3i) - 1(x) - 1(-1) - 1(3i) - 3i(x) - 3i(-1) - 3i(3i)
x^2 - x + 3ix - x + 1 - 3i - 3ix + 3i - 9i^2
x^2 - 2x + 1 - 9(-1)
x^2 - 2x + 1 + 10
x^2 -2x + 11
and then i multiply the two answers together?
so (x^2+2x-8)(x^2-2x+11)

Answer 13

yes.

Answer 14

except that 1 - 9(-1) = 10, not 11.

Answer 15

which will be
x^2+2x-8
x^2-2x+11
------------------------
11x^2+22x-88
-2x^3-4x^2+16x+0
x^4+2x^3-8x^2+0 +0
------------------------
x^4-x^2+38x-88

Answer 16

aww .-. okay ill redo it :P

Answer 17

almost

Answer 18

x^2+2x-8
x^2-2x+10
------------------------
10x^2+20x-80
-2x^3-4x^2+16x+0
x^4+2x^3-8x^2+0 +0
------------------------
x^4+0 -2x^2+36x-80

Answer 19

alright :)

Answer 20

yay can you help me do it backwards too c:

Answer 21

ok

Answer 22

Using the given zero, find all other zeros of f(x).

-2i is a zero of f(x) = x4 - 45x2 - 196

since -2i is a zero 2i is automatically a zero but how do i figure out the other ones? like i get the fact that its wherever the graph stops and isnt hit by a line but how do i find it for imaginary numbers .-.

Answer 23

you're right, 2i is also a zero. therefore the polynomial can be factorized as (x-2i)(x+2i)Q(x) = (x^2+4)Q(x).

you can find Q(x) by euclidean division

Answer 24

horner's rule maybe?

Answer 25

I'd like to say something about the first problem. If 2, -4, and 1 + 3i are the zeroes of a polynomial, then 1 - 3i is also a zero and

x = 2
x = -4
x = 1 + 3i
x = 1 - 3i

Upon moving everything to one side we get
x - 2 = 0
x + 4 = 0
x - 1 - 3i = 0
x - 1 + 3i = 0

From there we can apply zero product property to get
(x - 2)(x + 4)(x - 1 - 3i)(x - 1 + 3i) = 0

Then we can simply multiply and expand to get our 4th degree polynomial.

Answer 26

That's about the simplest anyone can explain it without going into technical jargon.

Answer 27

okayy well i got the answer none the less c:
okay so the euclidean formula is ax+by = gcd(a, b).

Answer 28

I meant


something like that.