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[The Abominable Snowman]

We're doing a section where we have to find real and imaginary roots of functions. I'm completely sure I've never done synthetic division before.

This question in particular asks me to use synthetic substitutions to find F(-3) where the F(x)=x^3-2x^2+7x+1
Edit:I got 25. Is this correct?

the second part of this problem asks me to use the same F(-3) on this problem;
F(x)=x^4-3x^3+4x^2+2x-10
Edit: I got 182. does this look correct?

Does anyone know how to do this?

Furthermore, It asks me to find the positive, negative, and imaginary zeroes for the following equations;
Find positives;
f(x)=2x^4-2x^3+2x^2-x-1
F(x)=4x^3-2x^2+x+3

find negatives;
f(x)=3x^4+x^3-3x^2+7x+5
f(x)=7x^4+3x^3-2x^2-x+1

Find imaginaries;
f(x)=5x^6+7x^4+8x^2+3
f(x)=x^5-x^4+x^3+x-7

and finally, the directions state for me to find the simplest polynomial function with integral coefficients that have the given zeroes;
-2, 2i
3+2i, 1

Does anyone know how to do ANY of this. You don't even have to do all of this, just sorta guide me around on doing this. If you have a link to a site, or a program that'd help, I'd love that too. If you know how to find the answers on a TI-83, I'd appreciate that as well. Thanks in advance

 

[Lathentar]

To find the zeros of a function graph it and see where f(x) = 0.

 

[Dilbert]

I can't type out synthetic substitution online...sorry. Hopefully that will make sense.

A "zero" is the same thing as a "root." For some function f(x), the zeroes are the values for which f(x) = 0. According to the fundamental theorem of algebra, a polynomial equation of degree n will have n roots. (In case you don't remember, the degree of a polynomial is the highest exponent on any of its terms.) The n roots can be positive, negative, zero, or imaginary, and the imaginary roots always appear as conjugate pairs. (An example of conjugate pairs: if 2+3i is a root, then 2-3i is a root as well.)

So, to solve some of your problems as examples...

Find all zeroes of f(x)=2x^4-2x^3+2x^2-x-1

In order to find the zeroes, we want to factor the polynomial completely. According to the root-factor theorem (that might not be the proper name, it's been a while), if the roots of an nth degree polynomial are a1, a2, a3, ..., an, you can express the polynomial as the product of factors in the form (x - a1)(x - a2)(x - a3)...(x - an) = 0. You can also solve it graphically by observing where the graph crosses the x-axis...but realize that imaginary zeroes are not visible when observing the graph, except to note that the number of x-axis crossings are fewer than n by some multiple of 2.

In this case, you want to group some terms together and factor out common monomials:

f(x) = 2x^4 - 2x³ + 2x² - x - 1
f(x) = 2x³(x - 1) - (2x + 1)(x - 1)
f(x) = (2x³ - 2x + 1)(x - 1)

Since the zeroes happen when f(x) = 0, and since we know that the product of any number of factors is zero when any of the factors is zero, we get the following:

x - 1 = 0 or 2x³ - 2x + 1 = 0

The first equation is trivial: x = 1 is a zero of f(x). The second equation can be solved exactly -- there ought to be a formula which solves for all zeroes of a cubic polynomial, though I don't remember it off the top of my head.

Find the simplest polynomial function with zeroes 3+2i, 1.

As mentioned, imaginary roots always appear in conjugate pairs. If 3+2i is a root, 3-2i is also a root. Using the root-factor theorem:

f(x) = [x - 1][x - (3+2i)][x - (3-2i)]
f(x) = (x - 1)(x² -6x + 13)
f(x) = x³ - 6x² +13x - x² + 6x - 13
f(x) = x³ - 7x² + 19x - 13

Hope this gives you some examples...

 

[ghostface]

Jinx, if you don't mind my asking, what do you do in life?

 

[Dilbert]

Jinx, if you don't mind my asking, what do you do in life?

I'm a poet.

 

[Desperado]

hooray for high school math?

uhh, yeah what jinx said.

synthetic division is really simple once you learn it. but

can't type out synthetic substitution online

 

[The Abominable Snowman]

This question in particular asks me to use synthetic substitutions to find F(-3) where the F(x)=x^3-2x^2+7x+1
Edit:I got 25. Is this correct?

the second part of this problem asks me to use the same F(-3) on this problem;
F(x)=x^4-3x^3+4x^2+2x-10
Edit: I got 182. does this look correct?

Edited. Does that look correct?

 

[MechaSnake]

Edited. Does that look correct?

F(-3)=x^3 - 2x^2 + 7x + 1
=(-3)^3 - 2*(-3)^2 + 7*(-3) + 1
=-27 - 18 + (-21) + 1
=-65

F(-3)=x^4 - 3x^3 + 4x^2 + 2x - 10
=(-3)^4 - 3*(-3)^3 + 4*(-3)^2 + 2*(-3) - 10
=81 - (-81) + 36 + (-6) - 10
=182

 

[nitewulf]

Edited. Does that look correct?

http://webpages.charter.net/thejacowskis/chapter6/synthsub.html
that should help you with the synthetic substitutions, if you dont recall doing it in class.
and a note about factoring higher order polynomials. sometimes you cant easily see a real root by looking at the equation. but you can always use some numbers to check it it's a root or not. graph it to spot the roots easily. math text problems usually always had 1, -1, 2, -2 as zeroes for these problems. it's sorta assumed that you'll substitute those numbers and check if they are roots or not. and if so, then you can follow the latter steps from jinx's solution.

 

[MechaSnake]

Edited. Does that look correct?

Edit: Don't remember what synthetic substitutions is, but i'm just using substitutions here.

F(-3)=x^3 - 2x^2 + 7x + 1
=(-3)^3 - 2*(-3)^2 + 7*(-3) + 1
=-27 - 18 + (-21) + 1
=-65

F(-3)=x^4 - 3x^3 + 4x^2 + 2x - 10
=(-3)^4 - 3*(-3)^3 + 4*(-3)^2 + 2*(-3) - 10
=81 - (-81) + 36 + (-6) - 10
=182

 

[Dilbert]

and a note about factoring higher order polynomials. sometimes you cant easily see a real root by looking at the equation. but you can always use some numbers to check it it's a root or not. graph it to spot the roots easily. math text problems usually always had 1, -1, 2, -2 as zeroes for these problems. it's sorta assumed that you'll substitute those numbers and check if they are roots or not. and if so, then you can follow the latter steps from jinx's solution.

Good point which I hadn't remembered. Once you have a root obtained from WHATEVER method (graphing, factoring, sheer dumb luck), you can use synthetic substitution to reduce the degree of the polynomial by one and hopefully see a way to factor what remains.

+1 nitewulf

 

[ghostface]

I'm a poet.

Ah cool, I always figured from your posts that you had a background in science/mathematics.

 

[Dilbert]

Ah cool, I always figured from your posts that you had a background in science/mathematics.

I was mostly kidding. I majored in physics, and work as an engineer...though, strangely, I don't use math very often at my job.

 

[The Abominable Snowman]

Jinx- One more question: When finding the negative zeroes, do you just plug in the negative x? Also, when you do that, do you switch the signs (From positive to negative, and vice versa) or only switch signs when the exponent changes signs (Like, in positive and negative, -3x^2 stays the same, yet in positive real zeroes it's 5x^3 , but in negative zeroes, it would be -5x^3

 

[Dilbert]

Jinx- One more question: When finding the negative zeroes, do you just plug in the negative x? Also, when you do that, do you switch the signs (From positive to negative, and vice versa) or only switch signs when the exponent changes signs (Like, in positive and negative, -3x^2 stays the same, yet in positive real zeroes it's 5x^3 , but in negative zeroes, it would be -5x^3

A zero is always a number, not a variable. Negative zeroes are simply solutions of f(x) = 0 for which x < 0.

As for how to write the associated factor: If -2 is a zero, the factor would be (x - (-2)) or (x + 2).

 

[The Abominable Snowman]

A zero is always a number, not a variable. Negative zeroes are simply solutions of f(x) = 0 for which x < 0.

As for how to write the associated factor: If -2 is a zero, the factor would be (x - (-2)) or (x + 2).

I'm talking about finding the negative real zeros...

 

[Dilbert]

I'm talking about finding the negative real zeros...

That's exactly what I'm talking about as well. When you find the set of all zeroes, you will have found the negative ones as well. I don't know of any technique to find ONLY the negative zeroes, except for examining a graph.

 

[slayn]

I didn't read everyone's post to see if this had been mentioned or not, but theres a theorum for listing possible roots thats useful for higher order functions.

if a polynomial has the form:

An*x^n+ ... + A1*x + A0

its something like (man its been a while) all the possible real roots will follow the patter of being + or - some factor of A0 divided by + or - some factor of An.

so if the polynomial is

F(x)=4x^3-2x^2+x+3

possible roots to check are:

+ or - 1,1/2,1/4,3,3/2,3/4

If I got that wrong... which I probably did, there is a theorum *like* that if not exactly that ;P