Unknown Constants

RickyOswaldIOW · 2006-03-05 07:01:53

In this question I am given the polynomial f(x) = 3x^4 + ax^3 + bx^2 - 7x - 4 and the factor (x - 1). I need to find the values of a and b. If I put the factor into the polynomial I get

a + b = 8

but there are many numbers that can be added to make 8, I do not know how to get another factor since there are two unknown constants in the polynomial!

fgarb · 2006-03-05 08:40:48

rickyoswaldiow,

It looks to me as though you're saying the question is:

f(x) = 3x^4 + ax^3 + bx^2 - 7x - 4, find a and b such that f(x) has (x-1) as a factor. Are you sure that is what the question is asking, because I don't think that's enough information to solve it.

In general, I'd say you can't pin down two unknowns from one factor. When you work through the polynomial division, you need to end up with a remainder of zero at the end, but your equation for the remainder will be in terms of both a and b. And one equation just isn't enough to solve for two unknowns.

RickyOswaldIOW · 2006-03-05 09:45:41

That is the question, aye. The other questions give you two factors to work from. Let my type the question exactly as it is written (I've just noticed somthing that I did not see before as it was smudged over with ink!);

"Given that (x² - 1) is a factor of the polynomial f(x), where f(x) = 3x^4 + ax^3 + bx^2 - 7x - 4, find the values of a and b and hence factorise f(x) completely."

I'm assuming that the x² instead of simply x in the factor plays a vital role here, I have not seen this before.

fgarb · 2006-03-05 11:39:49

Ah, that would explain it then. Saying something has (x^2 - 1) as a factor is actually saying it has two different factors - that is, anything that is a factor of (x^2-1) must also be a factor of the equation you're trying to solve for. Try breaking (x^2-1) up into its constituent factors and applying them part by part and see if that gets you anywhere. Good luck, and feel free to ask for more help if you get stuck with that.

Last edited by fgarb (2006-03-05 11:40:38)

RickyOswaldIOW · 2006-03-05 14:55:26

Try breaking (x^2-1) up into its constituent factors and applying them part by part

How do I do this? I've never seen such a factor before.

fgarb · 2006-03-05 15:21:24

One of the factors of x^2-1 is what you were using before: x-1. You should be able to figure out the other one either by division or by trial and error, the answer isn't complicated.

Then, you know that f(x) has to be divisible by both x-1 and the other factor that you found. See if you can use both of those conditions to find a and b. Good luck!

RickyOswaldIOW · 2006-03-06 05:57:51

I do indeed know how to find a and b if I have two factors but I really have no idea how to get the second factor. What must I do to x - 1 to make x^2 - 1?
Maybe I must simply guess the other factors and put them into f(x) and test different values of a and b from f(1) till I find another that makes f(x) = 0?

(x^2 - 1) / (x - 1) = ???

Last edited by rickyoswaldiow (2006-03-06 06:07:19)

RickyOswaldIOW · 2006-03-06 06:08:14

(x^2 - 1) / (x - 1) = x + 1!

krassi_holmz · 2006-03-06 06:28:34

Good!
Well done!!
"Given that (x² - 1) is a factor of the polynomial f(x), where f(x) = 3x^4 + ax^3 + bx^2 - 7x - 4, find the values of a and b and hence factorise f(x) completely."

Do you know what is the polynomial division?
It may help you to reduce the power of f(x).
Just divide f(x) by (x^2-1) and then the remainder which you will get must be divisible to (x^2-1).

RickyOswaldIOW · 2006-03-06 06:40:54

Do you know what is the polynomial division?

Nope, no idea

I just looked at what I could do to (x - 1) to make (x^2 - 1). I only got (x - 1) in the first place beause I misread the book! So that was luck, then fgarb pointed out that it was one of the factors anyway.

f(1) = 3(1)^4 + a(1)^3 + b(1)^2 - 7(1) - 4 = 0
a + b = 8

f(-1) = 3(-1)^4 + a(-1)^3 + b(-1)^2 - 7(-1) - 4 = 0
-a + b = -6

Thus: a = 7 and b = 1.
How do I go about factorising f(x) now? Do I use the same long division method I have been using on cubic polynomials?

RickyOswaldIOW · 2006-03-06 06:43:48

long division of x^2 - 1/f(x) just like you said krassi?

I cannot seem to make this work

x^2 - 1 / 3x^4 + 7x^3 (just to start)

so I divide 3x^4 by x^2 to give me 3x^2.

I then multiply 3x^2 by x^2 and then by -1 to give me 3x^4 - 3x^2.

I place this underneath 3x^4 + 7x^3 and subtract it, 3x^4 - 3x^4 = 0 as I would expect but I cannot subtract -3x^2 from 7x^3...

Last edited by rickyoswaldiow (2006-03-06 06:47:59)

krassi_holmz · 2006-03-06 06:53:19

Good. but if a=7 and b=1 we get:
f(x)= (x^2-1)(4+3x)(1+x)= (x-1)(4+3x)(1+x)^2,
which is divisible by x^2-1
Well done ricky and fgarb!

Last edited by krassi_holmz (2006-03-06 06:55:33)

RickyOswaldIOW · 2006-03-06 10:42:23

how do I divide f(x) by x^2-1?

mathsyperson · 2006-03-06 10:44:23

Using the difference of 2 squares rule, (x² - 1) = (x+1)(x-1).

Those two terms are both included in f(x), so dividing cancels them out.

f(x)/(x² - 1) = (4+3x)(x+1)

RickyOswaldIOW · 2006-03-07 09:13:22

I cannot divide 3x^4 + 7x^3 + x^2 - 7x - 4 by x^2 - 1 directly (not using my method of long division at least). If I divide by (x + 1) I am left with a cubic expression 3x^3 + 4x^2 - 3x - 4. Do I need to divide this by the other factor (x - 1)?

RickyOswaldIOW · 2006-03-07 11:27:33

I went back to this question and tried it out myself.
firstly I divide f(x) by one of the factors (x + 1). Then I take the quotient and divide that by the other factor (x - 1). I am now left with the quadratic 3x^2 + 7x + 4 which I factorise into (3x + 4) and (x + 1).
Thus the factors are:
(3x + 4)(x - 1)(x + 1)(x + 1) == (3x + 4)(x - 1)(x + 1)^2

As always, thanks for the help

fgarb · 2006-03-07 15:09:18

Looks right to me! Let us know if anything about the technique you used doesn't make sense.

krassi_holmz · 2006-03-07 18:43:49

I told you.

Math Is Fun Forum

#1 2006-03-05 07:01:53

Unknown Constants

#2 2006-03-05 08:40:48

Re: Unknown Constants

#3 2006-03-05 09:45:41

Re: Unknown Constants

#4 2006-03-05 11:39:49

Re: Unknown Constants

#5 2006-03-05 14:55:26

Re: Unknown Constants

#6 2006-03-05 15:21:24

Re: Unknown Constants

#7 2006-03-06 05:57:51

Re: Unknown Constants

#8 2006-03-06 06:08:14

Re: Unknown Constants

#9 2006-03-06 06:28:34

Re: Unknown Constants

#10 2006-03-06 06:40:54

Re: Unknown Constants

#11 2006-03-06 06:43:48

Re: Unknown Constants

#12 2006-03-06 06:53:19

Re: Unknown Constants

#13 2006-03-06 10:42:23

Re: Unknown Constants

#14 2006-03-06 10:44:23

Re: Unknown Constants

#15 2006-03-07 09:13:22

Re: Unknown Constants

#16 2006-03-07 11:27:33

Re: Unknown Constants

#17 2006-03-07 15:09:18

Re: Unknown Constants

#18 2006-03-07 18:43:49

Re: Unknown Constants

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