Question

Dividing polynomials

i'll post the post :)

Answer 1

@e.mccormick how do i do 57 and 61 ?? >.<

Answer 2

With 57, if you do synthetic div. and get a remainder of 0, it is a root (zero). Then you have a simpler poly to factor for the others.

Answer 3

can you show me how?

Answer 4

Can you do the synthetic division?

Answer 5

is this where c=3 turns into x-3?

Answer 6

With synthetic, just use it as 3. Less writing.

Answer 7

how?

Answer 8

Trying to find a latex command to show this better.

Answer 9

Sorry bout the delay. Seen things like this:\[\begin{matrix}3\\ \\ \end{matrix}
\left|\begin{matrix}
1 & -1 &-11&15 \\
& & \\
\hline
\end{matrix}\right.\]

Answer 10

1 -2 -17 -36 ?

Answer 11

hmmm... Lets see where that got off the tracks.... ah. How did you get -2?

Answer 12

its 1 2 5 0

Answer 13

1 2 -5 0, you lost the - in what you typed, but I can tell you did it right because you got the 0.

Answer 14

That 0 means there is no remainder. Therefore 3 is a root! (Or zero... they mean the same thing)

Answer 15

Now, we started with \(x^3-x^2-11x+15\) and what we got was:
\((x-3)(x^2+2x-5)\)
Do you see the relationship between what I just wrote and the synthetic division?

Answer 16

in the back it says that the answer is -1 plus/minus square root of 6

Answer 17

Well, if you put \(x^2+2x-5\) into the quadratic formula, I bet that is what it becomes.

See, the goal of the synthetic was to get it from something with a 3rd power to something smaller.

Answer 18

Here is another way to look at it:
https://www.desmos.com/calculator/vrrjug5ned

The top graph is what we got after we divided out 3, AKA: \((x-3)\). The bottom one is the other equation. See how they share two points on the x axis?

Answer 19

ohh okay. thank you very much :D

Answer 20

Yah, I find sometimes seeing it helos show what they mean by a root or a 0. It is when the y would become 0! I mean, that is simple enough, but when you see it and how one equation becomes the other as far as where they cross the 0 line, it makes more sense.

Answer 21

Now, you know how to finish that by putting the new equation in the quadratic... I hope. So the other question. It is simple... but will take a little work.

Answer 22

4th degree. You know what that part means?

Answer 23

i get how to do the rest of my hw but i really didn't understand how the degree works v.v

Answer 24

Degree is the highest power.

Answer 25

so like the max in a quadratic?

Answer 26

Hmmm... max would be the highest point on the graph.

Answer 27

x^10?

Answer 28

\(x^2\) is second degree.

Answer 29

\(x^{10}\) is 10th degree. So is \(92x^{10}+7x^5-x^3+7\). The degre is just the number of the highest exponent.

Answer 30

okay. i get it now

Answer 31

The 4th degree is in the form of \(ax^4+bx^3+cx^2+dx^1+ex^0\) And since \(x^1=x\) and \(x^0=1\) this is commonly shown as: \(ax^4+bx^3+cx^2+dx+e\) For it to be the 4th degree, a must be anything other than 0. If b, c, d, or e are 0, that term goes by-by, but it is still 4th degree.

Answer 32

So, now that we have flogged the degree issue to death! How to turn a bunch of roots into an equation. Do you have a guess? It is not hard. Really simple way to do it.

Answer 33

i understand the way you are explaining it to me. (:

Answer 34

Good!

Answer 35

you're a great teacher!!!!

Answer 36

Now, they gave you a list of zeros, right? -1 1 3 5. But what do you get your zeros from? When you find a zero and you say x=-1, how do you find that?

Answer 37

-1 l -1 1 3 5
-1 1 -2 -1
2 1 4


like that

Answer 38

No... this is the result of doing factoring. So like when they tell you to factor some equation and then say \(x=\{-1, 1, 3, 5\}\) at the end.

Answer 39

you plug it in

Answer 40

Hmm... Let me try a different direction. Lets start with this:

If I said \(x-3=0\) solve for x, what would you do?

Answer 41

x=3

Answer 42

Right... so, if I said, x=3, solve for 0, you could see how this was taking that and going backwards so x=3, solve for 0, is x-3=0.

Answer 43

ohh i see

Answer 44

So the solution for 61 is:
\(x=\{-1, 1, 3, 5\}\) solve for zeros, there are four of them! That gives you the four factors you need to multiply.

Answer 45

(x+1)(x-)(x-3)(x-5)
x^(4)-8x^(3)+14x^(2)+8x-15

Answer 46

I didn't multiply it all out myself, but if you did that stp right, it should be the answer.

Answer 47

okay :D

Answer 48

Tossed it in an algebra solver. Looks good.

Now, the first part of this exercise was all about finding zeros from a polynomial. This second part was about finding a polynomial from the zeros. This is a big deal in math: For any process, there is a reverse!

So a+b is addition, and a-b is subtraction. Addition is the reverse of subtraction or subtraction is the reverse of addition. All this is doing is showing you the same sort of forward and backwards relationship between polynomials and zeros.

Answer 49

ohh okay

Answer 50

I bring up the + and - to show you how far back they have been doing this. \(\times\) and \(\div\) are the same sort of thing. Functions and inverse functions. You might have seen trig functions like sine. That has an inverse called arcsine. If not, you will get it taught to you later. In every part of math, they do this.

Answer 51

sec, cot, cos, tan, sin, i'm missing one

Answer 52

csc

Answer 53

ohh yeah!!

Answer 54

Kk. My point is, remember that when you learn things in math. If they show you doing something forwards, you can count on it that backwards is soon to follow. It can help the learning if you know what is coming next.

Answer 55

okay. i'll remember. you taught me a lot!!!

Answer 56

Well, the goal is to have you connect with the math a bit better. Then you can do the work yourself.

Have fun!

Answer 57

haha i will and thank you so much!!

Answer 58

nice writeup @e.mccormick