Question

If you are trying to find all the zeros of an equation and your last "integer" doesn't have an "X" do you just leave it has is and not simple before putting it into the quadratic equation? @mathmale

Answer 1

Here's my interpretation of what you're asking: "If you're given a polynomial equation in which the last COEFFICIENT is a constant (that is, it doesn't stand before any power of x), do you leave that CONSTANT COEFFICIENT as is?"
Actually, Nicole, if I'm right on this interpretation, this is the situation you'll almost always encounter.

Answer 2

For example: Find all the zeros of the polynomial equation x^2 - x - 6 = 0. You'd definitely want to leave the equation as is. What methods of factoring quadatric equations are you familiar with ?

Answer 3

This being an example:
After the first arrow it is simplified cause the last "integer" has an x.

Answer 4

So, would it be the same or do I have to do it differently since I don't have an "X" on the last bit?

Answer 5

Oh, it helps so much to see what you've already done.
Did you mean y - x^3 - 3x^2 - 9x or y = x^3 - 3x^2 -ax?

Answer 6

I didn't want to put up the equation I'm working with because I wanted to do it on my own but if you need to see it I can send it to you. The picture is of an equation I have already done - the one I am working with now doesn't have an ending "X" so I can't factor it out into say.. x(x^3 + 7x^2 and continued because if I did and then you times it, it would then have an "X" that wasn't there before..

Answer 7

May I refer to the most recent photo that you've shared with me? is that the problem you're working with? Just one look at that photo and I pretty much knew how to respond.

Answer 8

In the long run I'd prefer that you just post on OpenStudy or send me privately photos of any problem(s) you'd like to discuss. You are to be commended for working on these problems on your own and having something to share because of your having done so.

Answer 9

The equation I'm working with now is: -3x^4 + 27x^2 + 1200 = 0
I am trying to find all the zeros for the equation but because there is not an "X" on the end of 1200. I also would like to know if I have to put in place holders such as x^3?

Answer 10

Thanks for typing this out. I see what you're asking. Yes, you do have to put in place holders for any missiong powers of x. Represent that missing x^3 with the term 0*x^3 and the missing x with 0*x. This fourth order polynomial should theoretically have FIVE terms on the left side.

Answer 11

Are you familiar with synthetic division? If not, what method were you planning to try to find the zeros of this equation?

Answer 12

Also, leave that 1200 on the end exactly as it is. If you like, you could think of that 1200 as the coefficient of x^0.
Note that x^0 = 1.

Answer 13

Okay, and yes I know synthetic division. I had used the equation x = -b +- the square root of b^2 - 4ac / 2a to solve the last one I did but I was factored out

Answer 14

Hold just a moment...I'm thinking.

Answer 15

First, please share with me the approach you were thinking of using. I don't want to push my own approaches on you if you already have one of your own that would work.

Answer 16

I'm not entirely sure what to use:(
Would this be right for the next step?

Answer 17

Yes, if you'd please write the coefficient ZERO in front of both the x^3 and the x^1.

Answer 18

Those are place holders, remember?

Answer 19

Okay, I wrote them.

Answer 20

May I ask where this problem came from? I see a way to solve it, but it's a challenging problem.

Answer 21

I have to turn in work for school. I hadn't done one like this before and I wanted to make sure I know how and that I did it right before sending it i for scoring. If you would like - I wouldn't mind - changing the number so I would still do the work that has to be turned in on my own.

Answer 22

You're surely very honest!

Answer 23

So if we did the same type of problem so I learned how, I could then do the scoring one on my own. And thank you ha

Answer 24

please excuse me, just a moment.

Answer 25

Okay. Again, thank you for all your help!

Answer 26

I believe I could help you with a similar problem, but must warn you that it'd take a while.
I'd be OK, personally, discussing what to do with this particular problem, with you doing the work and I providing suggestions and guidance.

Answer 27

I'll leave it up to you what we do next, if anything.
My first suggestion would be that you take the original equation and divide it through by -3.
Can you see any value in doing tht?

Answer 28

Okay, that would be great because I also have other work to do and would like to go to bed not to late. Dividing by -3 would make the 27 a -9 and the 1200 a -400 - making the equation smaller.

Answer 29

By the way, my approach worked: one solution / zero is 5.
OK, Nicole, perhaps we'll have the opp to work together again later; I hope so. Admire your perseverance and your sharing your previously completed work with me.
Yes, if you divide that equation through

Answer 30

by -3, you'd get x^4 -27x^2 - 400 = 0, which is significantly easier to solve.

Answer 31

If you have to go, could you please quick tell me what to do next and I'll hope I don't mess up on making my way to the end.

Answer 32

More power to you for being brave enought o do that . I'm thinking, just for a moment, about what to tell you next.

Answer 33

Nicole: try again dividing the original equation by -3. That -27 is incorrect.

Answer 34

27 / -3 = -9 though - I did it on my calculator because I'm not good with negative ha

Answer 35

OK. So, are we in agreement that we have x^4 - 9x^2 - 400 = 0?

Answer 36

Yes, that is what I got after / by -3

Answer 37

Cool. Now you're exactly on the right track. This latest equation is easier to understand and to solve than the original one.

Answer 38

OK: let me ask how comfortable you would be in letting p = x^2, so that
p^2 - 9p - 400 = 0. Remember having done that before?

Answer 39

I didn't do that before, I hadn't see why to since you would just put it back later. I may be mistaken though.

Answer 40

I haven't ever used that

Answer 41

It's a commonly used trick. Which looks easier to solve, x^4 - 9 x^2 - 400 = 0, or
p^2 - 9p - 400 = 0?

Answer 42

To me; x^4 but only because I haven't used P in this certain type of equation but if it makes it easier, then I'll try.

Answer 43

In other words, making the substitution p = x^2 reduces the fourth order polynomial to a 2nd order one (a quadratic), which I'm sure you'd be able to solve. Suppose you were to get p = 25 (which actually is a solution); then p = 25 = x^2 and x would then be 5 or -5. Are you comfortable with this reasoning?

Answer 44

So, p = 25 is just an intermediate result; We'd have to remember that p = x^2 and that 25 = x^2, so that x = plus or minus 5.

Answer 45

I think I get it.

Answer 46

I'm not sure what to do have you put p = x^2 in

Answer 47

I believe there are two more soltions, but could be wrong on that.

Again, the reason I make that substitution is to reduce the order of the original polynomial from 4 to 2.

You do NOT have to do it this way. You could use synthetic division;
Try it:

Let 5 b e

Answer 48

Do you add 400 to both sides?

Answer 49

No. We want to keep that equation in standard form. Adding 400 to both sides would mess up that standard form.

Answer 50

Oh, I see.

Answer 51

If you're comfortable with synthetic division, please go ahead and try verifying that 5 is a root of the original equation.

5 | 1 0 -9 -400

Answer 52

Want to give it a try?

Answer 53

Yes

Answer 54

I've just done that, and get a zero remainder, which verifies that 5 is indeed a root of the original equation.

Answer 55

Do you get the same result thru synthetic division?

Answer 56

I think I did it wrong ha :(
I got -320, NOT good with negatives..

Answer 57

Want to give it another try? Practice makes perfect.

Answer 58

Yes ha

Answer 59

5 | 1 0 -9 0 -400 results in what remainder?

Answer 60

One Second, I think I missed a zero

Answer 61

5 | 1 0 -9 0 -400 (for better visibility)

Answer 62

Do you add or subtract? Just to be clear

Answer 63

Depends on the sign of each particular coefficient {1, 0, -9, 0, -400}.

Answer 64

I think thats where I am going wrong..

Answer 65

Bring that 1 down. Multiply that 1 by 5. Result is positive, right? write that 5 under the first 0. Add 0 and 5: you get 5. Now mult that 5 by 5 and write the result under the -9. In that case you'd have 25-9 = 16.
Try finishing the problem.

Answer 66

OKay

Answer 67

5 times 16 is 80; write the 80 under the 2nd zero coefficient and add the two numbers together. Result: 80. mult that 80 by 5 and write the result under the -400. OK?

Answer 68

:D I got it! I had done it like that but was missing the 2nd zero

Answer 69

So cool! Bully for you, Nicole!!
Now you should see a new string of coefficients : 1 5 16 80. Do you?

Answer 70

Yes

Answer 71

Great. Perfect.

Answer 72

Now test my thesis that -5 is also a root of the original equation

Answer 73

Pu tin -5 as "e"?

Answer 74

*put

Answer 75

by performing the following synthetic division:

-5 | 1 5 16 80.

Go ahead, please.

Answer 76

Okay

Answer 77

I end up with a 0 under the -80

Answer 78

Fantastic. What does that mean?

Answer 79

That both 5 and -5 are zeros for the equation

Answer 80

Absolutely right! Congratulations!!!

Answer 81

Now you have one final row of coeffciients: they are 1 0 16, right?

Answer 82

Yes

Answer 83

So, what does that mean for us? 1 0 16 translates into the quadratic 1x^2 + 16 = 0, which has only imaginary roots. If you were supposed to find ALL of the roots, then you'd need to write the results as {5, -5, 4i, -4i} and you'd be done!

Answer 84

You could, if you wished, verify through synth div that all four of these numerals are indeed roots of the original equation!

Answer 85

Great! Thank you soo much for all your help! I'll will put it all together and hope I do as well on the rest as I did for this one(:

Answer 86

My great pleasure. Really! You're so cool to work with....quick, willing to work, able to express yourself very well and to find your own mistakes.

Answer 87

Post any further questions you have right here in OpenStudy; perhaps I'll have the good fortunate to see them and to work with you further.
Happy New Year to you, Nicole, and good night!

Answer 88

Happy New Years to you as well! Have a goodnight.

Answer 89

Bye!