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## Topic review (newest first)

krassi_holmz
2006-03-08 17:43:49

I told you.

fgarb
2006-03-08 14:09:18

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

RickyOswaldIOW
2006-03-08 10:27:33

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

RickyOswaldIOW
2006-03-08 08:13:22

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)?

mathsyperson
2006-03-07 09:44:23

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 09:42:23

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

krassi_holmz
2006-03-07 05:53:19

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!

RickyOswaldIOW
2006-03-07 05:43:48

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...

RickyOswaldIOW
2006-03-07 05:40:54

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?

krassi_holmz
2006-03-07 05:28:34

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
2006-03-07 05:08:14

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

RickyOswaldIOW
2006-03-07 04:57:51

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) = ???

fgarb
2006-03-06 14:21:24

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 13:55:26

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
2006-03-06 10:39:49

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.