Factoring a four term polynomial

In summary, the Rational Root Theorem can be used to determine whether there are factors of the form (x - p/q). This theorem says that any such roots p/q are such that p divides the constant (-5) and q divides the coefficient...

Homework Statement

I just want to know how get from ##4x^3+3x^2-6x-5=0 ##
to ##(x+1)^2(4x-5)=0##. What's the trick when dealing with these nasty polynomials? I got the answer by taking another approach (solving a root equation) but I noted this is also a way to go, but I can't figure out the method

-[/B]

The Attempt at a Solution

Well I can't seem to get anywhere. The assignement is solved and correct, but I would like to learn how to make the factoring that out! =)Edit: corrected accidental + instead of - in the last factor.
Edit 2: bad reception on smartphone didn't save the first correction. Now it should be fine.

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Homework Statement

I just want to know how get from ##4x^3-3x^2-6x-5=0 ##
to ##(x+1)^2(4x+5)=0##. What's the trick when dealing with these nasty polynomials? I got the answer by taking another approach (solving a root equation) but I noted this is also a way to go, but I can't figure out the method

-[/B]

The Attempt at a Solution

Well I can't seem to get anywhere. The assignement is solved and correct, but I would like to learn how to make the factoring that out! =)
There must be some typos. Where do you get the negative signs from when multiplying ##(x+1)^2(4x+5)\,##?

In general, you have basically two possibilities: you can look up the formulas to solve for ##4x^3 + \ldots = 0## which are a bit nasty, or, which is more promising in such designed cases, guess some roots. That is, if ##4a^3 + \ldots = 0## for some number ##a##, then ##(x-a) \,\vert \, (4x^3 + \ldots )## and you can perform a long division and solve for the remaining polynomial ##4x^2 + \ldots = 0##. To guess zeros, you can simply try low numbers as ##\{-2,-1,0,1,2\}## or you can look at the absolute term: ##4x^3 + \ldots - 5 = 4(x-a)(x-b)(x-c) = 0## shows that ##abc= 5## and you can try some divisors of ##5##.

Homework Statement

I just want to know how get from ##4x^3-3x^2-6x-5=0 ##
to ##(x+1)^2(4x-5)=0##. What's the trick when dealing with these nasty polynomials? I got the answer by taking another approach (solving a root equation) but I noted this is also a way to go, but I can't figure out the method
...
Edit: corrected accidental + instead of - in the last factor.
Perhaps it should be ##\ 4x^3+3x^2-6x-5=0 \,.##

to ##(x+1)^2(4x-5)=0##. What's the trick when dealing with these nasty polynomials?
Well I can't seem to get anywhere. The assignement is solved and correct, but I would like to learn how to make the factoring that out! =)Edit: corrected accidental + instead of - in the last factor.

We get it, it's something like that except the opposite plus an accidental - instead of + in the second term.

SammyS said:
Perhaps it should be ##\ 4x^3+3x^2-6x-5=0 \,.##
Yes, that's correct! (On mobile unit with bad reception). Edited it now! :)

fresh_42 said:
There must be some typos. Where do you get the negative signs from when multiplying ##(x+1)^2(4x+5)\,##?

In general, you have basically two possibilities: you can look up the formulas to solve for ##4x^3 + \ldots = 0## which are a bit nasty, or, which is more promising in such designed cases, guess some roots. That is, if ##4a^3 + \ldots = 0## for some number ##a##, then ##(x-a) \,\vert \, (4x^3 + \ldots )## and you can perform a long division and solve for the remaining polynomial ##4x^2 + \ldots = 0##. To guess zeros, you can simply try low numbers as ##\{-2,-1,0,1,2\}## or you can look at the absolute term: ##4x^3 + \ldots - 5 = 4(x-a)(x-b)(x-c) = 0## shows that ##abc= 5## and you can try some divisors of ##5##.
Yes! Two sign errors I'm afraid! Edited now. Oh okay, so if I guess a root and divide the polynomial with that factor (x-a) I'd be done? :) The factor theorem, right? (Still on mobile, going to try this when I get home!)

Yes! Two sign errors I'm afraid! Edited now. Oh okay, so if I guess a root and divide the polynomial with that factor (x-a) I'd be done? :) The factor theorem, right? (Still on mobile, going to try this when I get home!)
Yes. ##-1## as a root can easily be checked. Then you can calculate ##(4x^3+3x^2−6x−5):(x+1)## and get a quadratic polynomial for which you probably know the formulas to find the remaining zeroes (Vieta's formulas). Otherwise you can also repeat the process and in this case with ##-1## again. You have to divide by ##x+1 = x-(-1)## here, in general by ##x-a## if ##a## is a zero.

For your polynomial ##4x^3 + 3x^2 - 6x - 5##, the Rational Root Theorem can be used to determine whether there are factors of the form (x - p/q). This theorem says that any such roots p/q are such that p divides the constant (-5) and q divides the coefficient of the highest degree term (4). So the possibilities are ##\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4##. In all there are 12 possible rational roots. The easiest way to check each of them is by using Synthetic Division.

Edit: Added my missing LaTeX end tag.

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You can do it in an intuitive way
##4x^3+3x^2-6x-5=3x^3+3x^2+x^3-x-5x-5=3x^2(x+1)+x(x-1)(x+1)-5(x+1)=(x+1)(4x^2-x-5)##

SammyS
##4x^3+3x^2-6x-5=0 ##
to ##(x+1)^2(4x-5)=0##. What's the trick when dealing with these nasty polynomials?

Still simpler than the methods suggested is simpe substitution of a likely looking root. In theexamples gi , -4 +3 -6 ven at your level roots are ften simple small integers. Look at x = 1 for instance the polynomial would be 4 + 3 - 6 - 5 ≠ 0, so not that, now try x = -1, -4 +3 +6 -5 = 0, yes -1 is a root, and (x + 1) is a factor, divide by (x + 1)...

Also if the factors are simple rationals, I mean of form (Ax + B) with A, B integers, then the number of possibles to look at is limited, e.g. here the first term in that case must have come from
(4x +...)(x +...)(x +...) or (2x +...)(2x +..)(x +...)
and the last term from (... + 5)(.. + 1)(... -1) or (...- 5)(.. + 1)(... -1) or (... - 5)(.. - 1)(... -1),.

(You only need tconsider the sign possibilities for one of these tw ,terms OK? )

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Mark44 said:
##\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4
##\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4##

haruspex said:
##\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4##
Thanks! I didn't notice that I had neglected to add the closing tag. I have fixed it in my earlier post.

1. What is factoring a four term polynomial?

Factoring a four term polynomial involves breaking down a polynomial expression with four terms into its simplest form by finding common factors among the terms.

2. Why is factoring a four term polynomial important?

Factoring a four term polynomial is important because it allows us to simplify complicated expressions and solve equations more easily.

3. What are the steps to factor a four term polynomial?

The steps to factor a four term polynomial are:
1. Group the terms into pairs based on common factors.
2. Factor out the common factor from each pair.
3. Look for any remaining common factors and factor them out.
4. Use the distributive property to combine the factors and simplify the expression.

4. Can all four term polynomials be factored?

No, not all four term polynomials can be factored. Some may have complex roots or may not have any common factors among the terms.

5. How do I check if my factoring is correct?

You can check if your factoring is correct by multiplying the factors back together using the distributive property. The result should be the original polynomial expression.

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