Solve Factoring Problems: 2 Examples

[theout]

Factoring problems...

1.
Having problems with factoring almost got it except this 2
(3x^3+21x^2+36x)/ 3x^4+12x^3-27x^2-108x ( Edited)

The Attempt at a Solution

the numerator 3x^3+21x^2+36x I can see the factor right away its 3x so 3x(x^2+7x+12) and for new denominator that I mixed (sorry for that) it has the same factor 3x inside so 3x(x^3+4x^2-9x-36) so we can cancel the 3x on both sides so now we have
(x^2+7x+12)/(x^3+4x^2-9x-36)
back to the numerator we can factor it into (x+4)(x+3) and for the denominator we can us eone of the factors from the top and after trial and error I use (x+4) and (x^2-9) because after u multiply them you get the denominator so now we have
(x+4)(x+3)/(x+4)(x^2-9)
we can cross x+4 now so we have (X+3)/(x^2-9) and after factoring the denaminator we have (x-3)(x+3) so now we cross out (x+3)
and we have our answer 1/(x-3)
another similar problem that I don't get
2.
4x^4-25
---------
6x^3-4x^2+15x-10

The Attempt at a Solution

numerator can be factored into (2x^2-5)(2x^2+5) and since there no further way to factor it I will try to use one of the factors from it for the denaminator so after trial and error I used
(2x^2+5) and 3x-2 so now I can cross out 2x^2+5 and I have my answer because I cannot factor any further 2x^2-5/3x-2

Sorry for my careless mistakes even in the title

 

[HallsofIvy]

It is almost impossible to tell what you are doing. The internet does not "respect" tabes or multiple spaces so trying to write fractions over several lines leads to madness. Either use LaTex or just ( )/( ). Also you never do say what you are trying to do! Reduce the fraction? It does help to make it clearer that you write the numerator and denominator separately and show how you are factoring them. Thank you for that!

I think you have (3x^3+21x^2+36x)/(6x^3-4x^2+15x-108)

And you recognize immediately that "3x" is a factor of that: (3x^3+ 21x^2+ 36x)= 3x(x^2+ 7x+ 12). You also recognize that x^2+ 7x+ 12 factors as (x+ 3)(x+ 4) so the numerator can be written 3x(x+3)(x+4) Good! (Although you way of writing that "3x(x^2+7x+12) ........(x+4)(x+3)" is less than transparent!). Now the 6x^3- 4x^2+ 16x- 108 which may be more trouble! Fortunately, the only reason we want to factor it is in the hope that something in the numerator will cancel something in the denominator so we might as well just look for those factors.
Obviously neither "3" nor "x" is a factor. 3 does not divide -4 and x does not divide -108.

What about x+ 4? Well, if so then the other factor would have to start with 6x^3/x= 6x^2 and the last term would have to be -108/4= -27. Will (x+4)(6x^2+ ?x- 27)= 6x^3- 4x^2+ 16x- 108? If so what would "?" be?

What about x+ 3? Again, it that is a factor, the other factor would have to start 6x^3/x= 6x^2 and the last term would have to be -108/3= -36. Will (x+3)(6x^2+ ?x- 36)= 6x^3- 4x^2+ 16x- 108? If so what would "?" be?

For the next problem (also numbered "1"?)
(4x^4-25)/(6x^3-4x^2+15x-10)

Yes, 4x^4- 25 factors easily into (2x^2- 5)(2x^2+ 5). Neither of those has further factors with integer coefficients. 2x^2- 5 could be factored with irrational numbers (square roots) as coefficients and 2x^2+ 5 has no factors with real coefficients.

The denominator is 6x^3- 4x^2+ 15x- 10 which is, again, hard to factor. But, again, we are only interested in whether 2x^2- 5 and 2x^2+ 5 are factors. Try them! We could do that by "long division": (6x^3- 4x^2+ 15x- 10) divided by 2x^2- 5 or 2x^2= 5, but we can also argue just as before:

If 2x^2- 5 is a factor of 6x^3- 4x^2+ 15x- 10, then the other factor must have leading term 6x^3/2x^2= 3x and last term -10/-5= + 2. There are no terms between 3x and + 2 so- is (2x^2- 5)(3x+ 2) equal to 6x^3- 4x^2+ 15x- 10?

If 2x^2+ 5 is a factor of 6x^3- 4x^2+ 15x- 10, then the other factor must have leading term 6x^3/2x^2= 3x and last term -10/5= -2. There are no terms between 3x and -2 so- is (2x^2+ 5)(3x- 2) equal to 6x^3- 4x^2+ 15x- 10?

 

[theout]

It is almost impossible to tell what you are doing. The internet does not "respect" tabes or multiple spaces so trying to write fractions over several lines leads to madness. Either use LaTex or just ( )/( ). Also you never do say what you are trying to do! Reduce the fraction? It does help to make it clearer that you write the numerator and denominator separately and show how you are factoring them. Thank you for that!

I think you have (3x^3+21x^2+36x)/(6x^3-4x^2+15x-108)

And you recognize immediately that "3x" is a factor of that: (3x^3+ 21x^2+ 36x)= 3x(x^2+ 7x+ 12). You also recognize that x^2+ 7x+ 12 factors as (x+ 3)(x+ 4) so the numerator can be written 3x(x+3)(x+4) Good! (Although you way of writing that "3x(x^2+7x+12) ........(x+4)(x+3)" is less than transparent!). Now the 6x^3- 4x^2+ 16x- 108 which may be more trouble! Fortunately, the only reason we want to factor it is in the hope that something in the numerator will cancel something in the denominator so we might as well just look for those factors.
Obviously neither "3" nor "x" is a factor. 3 does not divide -4 and x does not divide -108.

What about x+ 4? Well, if so then the other factor would have to start with 6x^3/x= 6x^2 and the last term would have to be -108/4= -27. Will (x+4)(6x^2+ ?x- 27)= 6x^3- 4x^2+ 16x- 108? If so what would "?" be?

What about x+ 3? Again, it that is a factor, the other factor would have to start 6x^3/x= 6x^2 and the last term would have to be -108/3= -36. Will (x+3)(6x^2+ ?x- 36)= 6x^3- 4x^2+ 16x- 108? If so what would "?" be?

For the next problem (also numbered "1"?)
(4x^4-25)/(6x^3-4x^2+15x-10)

Yes, 4x^4- 25 factors easily into (2x^2- 5)(2x^2+ 5). Neither of those has further factors with integer coefficients. 2x^2- 5 could be factored with irrational numbers (square roots) as coefficients and 2x^2+ 5 has no factors with real coefficients.

The denominator is 6x^3- 4x^2+ 15x- 10 which is, again, hard to factor. But, again, we are only interested in whether 2x^2- 5 and 2x^2+ 5 are factors. Try them! We could do that by "long division": (6x^3- 4x^2+ 15x- 10) divided by 2x^2- 5 or 2x^2= 5, but we can also argue just as before:

If 2x^2- 5 is a factor of 6x^3- 4x^2+ 15x- 10, then the other factor must have leading term 6x^3/2x^2= 3x and last term -10/-5= + 2. There are no terms between 3x and + 2 so- is (2x^2- 5)(3x+ 2) equal to 6x^3- 4x^2+ 15x- 10?

If 2x^2+ 5 is a factor of 6x^3- 4x^2+ 15x- 10, then the other factor must have leading term 6x^3/2x^2= 3x and last term -10/5= -2. There are no terms between 3x and -2 so- is (2x^2+ 5)(3x- 2) equal to 6x^3- 4x^2+ 15x- 10?

Thanks for your help and tips:) And sorry for my carelessness I was even embarassed when I saw my tread called factoring "peoblems"
I even wrote the wrong denominator in the first one...I think carelessness is the mother of my mistakes

Thanks for the tips they really helped a lot