More College Algebra Help

[BlancoNino]

I have a polynomial

2x^5 - 14x^4 + 28x^3 - 24x^2 + 48x - 64

I used synthetic division to find the two zeros which are :

2, 4

Using the quotient polynomial means to use the numbers obtained when doing synthetic division (the quotient of the zero tested times the number in the original polynomial). This also means removing one number from each of the powers across the equation. Since I can only solve for two zeros, this only brings my leading coefficient down to 2x^3.

HOWEVER, I'm supposed to make the polynomial so that the leading co-efficient is x^2 and use the quadratic formula or regular factoring to find the rest of the zeros.

SO, how do I bring the polynomial down one more level if I've already found both zeros? Do I plug in another number and use the remainder somehow?

THanks in advance for all help!

 

[Acanthus]

Im kind of lost as to what youre asking for...

You are supposed to change the polynomial so that x^2 is the highest power?

 

[BlancoNino]

yeah supposedly by finding the zeros it will help simplify the equation somehow and I'm supposed to find the rest of the zeros after I "get to x^2"

I'm kinda lost too.

 

[BlancoNino]

bump

 

[BlancoNino]

Okay, using the quotient polynomial from using synthetic division, I rewrite the equation except with one less power across the board (becomes 2x^4, etc) and using the numbers from the quotient. I do that twice because there are two real zeros. However this only brings it down to 2x^3 + the rest. Do I use the remainder somehow? The instructions clearly says to work down to x^2.

 

[DrawninwarD]

Enter your equation here: http://www.e-tutor.com/et3/graphing

You have two roots at x = 2 and another at x = 4.

The polynomial is 5th degree. It can have a maximum of 5 roots. This one only has 3 (of which 2 are distinct).

I don't know exactly what you're asking for either.

 

[dullard]

1) First of all, you don't have an equation, so you don't have any zeros. I assume you meant to write this:

2x^5 - 14x^4 + 28x^3 - 24x^2 + 48x - 64 = 0

2) Always start by simplifiying any equation. It takes almost no time at all, and it just makes things easier. I'll divide by two:

(simplified polynomial) = x^5 - 7x^4 + 14x^3 - 12x^2 + 25x - 32 = 0

3) You said that you already found two roots, x=2 and x=4. Those are both correct.

4) Thus, if you have a root, you can divide it out. You want to write this:
(x-2) * (something) = (simplified polynomial) = 0

Thus, (something) = (simplified polynomial) / (x-2). Do the division and you get:

(x-2) * (x^4 - 5x^3 + 4x^2 - 4x +16) = (simplified polynomial) = 0

Repeat with your x=4 root.

(x-2) * (x-4) * (something #2) = (x-2) * (x^4 - 5x^3 + 4x^2 - 4x +16) = 0

Thus, (something #2) = (x^4 - 5x^3 + 4x^2 - 4x +16) / (x-4). Do the division and you get:

(x-2) * (x-4) * (x^3 - x^2 - 4) = (simplified polynomial) = 0

5) Now you have to work with (x^3 - x^2 - 4) = 0. This is where I assume you are stuck. There are many ways to proceed. You can randomly guess until you get the answer. You can just graph it and see the answer. You can use the fact that repeating answers are common and try those. Your teacher probably gave you several other equally correct ways to continue. I'll let you do your own homework and figure out which method you want.

6) Lets try the repeating answers are common method (ie this is a good start for the random guessing method). So, lets try x=2 again. Plug x=2 into (x^3 - x^2 - 4) and yep, that works, the answer is zero.

So now we repeat the long division to get:

(x-2) * (x-4) * (x-2) * (something #3) = (x-2) * (x-4) * (x^3 - x^2 - 4) = 0

or,

(x-2) * (x-4) * (x-2) * (x^2 + x +2) = 0.

7) Now use the quadratic method to solve the last bit (x^2 + x + 2) = 0

 

[sao123]

hmm... i remember ther ebeing a really easy way to do this with derivitives...

basic algebra sux.

 

[hypn0tik]

Originally posted by: dullard
1) First of all, you don't have an equation, so you don't have any zeros. I assume you meant to write this:

2x^5 - 14x^4 + 28x^3 - 24x^2 + 48x - 64 = 0

2) Always start by simplifiying any equation. It takes almost no time at all, and it just makes things easier. I'll divide by two:

(simplified polynomial) = x^5 - 7x^4 + 14x^3 - 12x^2 + 25x - 32 = 0

3) You said that you already found two roots, x=2 and x=4. Those are both correct.

4) Thus, if you have a root, you can divide it out. You want to write this:
(x-2) * (something) = (simplified polynomial) = 0

Thus, (something) = (simplified polynomial) / (x-2). Do the division and you get:

(x-2) * (x^4 - 5x^3 + 4x^2 - 4x +16) = (simplified polynomial) = 0

Repeat with your x=4 root.

(x-2) * (x-4) * (something) = (x-2) * (x^4 - 5x^3 + 4x^2 - 4x +16) = 0

Thus, (something) = (x^4 - 5x^3 + 4x^2 - 4x +16) / (x-4). Do the division and you get:

(x-2) * (x-4) * (x^3 - x^2 - 4) = (simplified polynomial) = 0

5) Now you have to work with (x^3 - x^2 - 4) = 0. This is where I assume you are stuck. There are many ways to proceed. You can randomly guess until you get the answer. You can just graph it and see the answer. You can use the fact that repeating answers are common and try those. Your teacher probably gave you several other equally correct ways to continue. I'll let you do your own homework and figure out which method you want.

6) Lets try the repeating answers are common method (ie this is a good start for the random guessing method). So, lets try x=2 again. Plug x=2 into (x^3 - x^2 - 4) and yep, that works, the answer is zero.

So now we repeat the long division to get:

(x-2) * (x-4) * (x-2) * (something) = (x-2) * (x-4) * (x^3 - x^2 - 4) = 0

or,

(x-2) * (x-4) * (x-2) * (x^2 + x +2) = 0.

7) Now use the quadratic method to solve the last bit (x^2 + x + 2) = 0

If you're looking to find the roots of something like ax^3 + bx^2 + cx + d = 0, the first thing you should do is find all the factors of a and d. Then, the roots you want to try are the factors of d divided by the factors of a.

For example, in our case we have x^3 - x^2 - 4 = 0. a = 1, so it's factors are 1
d = 4, so the factors are 1, 4, 2. Then you try x = 1/1, 1/-1, 4/1, 4/-1, 2/1, 2/-1, and so on. You should find that x = 2/1 will work.

If you find that none of those combinations work, then you probably have something that doesn't factor nicely and you'll need alternate methods for computing the root.

In fact, you could have done this with your 5th order polynomial and you'll see that the factors of 32 are 1, 2, 4, 8, 16, 32. You start trying numbers and you'll see that 2 and 4 work and so on...

You can continue this method when you get down to a quadratic (2nd order polynomial), but in this case you also have the option of using the quadratic formula.

Hope this helps!

 

[DrPizza]

Re: Dullard's #5. If you're allowed to graph it in the first place to see where the real roots are, then if the function "touches" the x-axis and turns around, i.e. like y=x², then that particular root repeats an even number of times. If you get an s-shape (i.e. y = x^3) where it passes through the x-axis, then you have a root repeated an odd number of times.

And, of course, you know about the possible rational roots - factors of the constant divided by factors of the leading coefficient. (in case you're just using the trial and error method.)