Factoring a four term polynomial

[Quadrat]

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

Homework Equations

-[/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.

 

[fresh_42]

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

Homework Equations

-[/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 $5$.

 

[SammyS]

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 $\ 4x^3+3x^2-6x-5=0 \,.$

 

[epenguin]

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.

 

[Quadrat]

Perhaps it should be $\ 4x^3+3x^2-6x-5=0 \,.$

Yes, that's correct! (On mobile unit with bad reception). Edited it now! :)

 

[Quadrat]

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!)

 

[fresh_42]

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.

 

[Mark44]

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.

 

[ehild]

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

 

[epenguin]

$4x^3+3x^2-6x-5=0$
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? )

 

[haruspex]

##\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4

$\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4$

 

[Mark44]

$\pm 1, \pm, 1/2, \pm 1/4, \pm 5, \pm 5/2, \pm 5/4$

Thanks! I didn't notice that I had neglected to add the closing tag. I have fixed it in my earlier post.