Techniques for Factoring Large Polynomial

In summary, the given complicated polynomial can be simplified by noticing that all coefficients are divisible by 8, resulting in the factor of 8(4t^2+1)(4t^2+4t+5). If the root is real, it can be found using the Rational Root theorem. If it is a complex root, it will have a complex conjugate and can be factored into real quadratic factors, with the possibility of one additional real linear factor if the polynomial has an odd order.
  • #1
Seydlitz
263
4
Hello guys,

I'd like to ask you how to efficiently factorize complicated polynomial like this one for example:

$$\frac{128t^4+128t^3+192t^2+32t+40}{(4t^2+1)^4}$$

I've spend more than a hour trying to decrypt and decompose the polynomial, but to no avail. For simple cubic polynomial I know it might be possible to try ##-1## or ##1## or the factor of the lowest term to get the correct linear factor, which can then be used to do long division to get the other factor if there's any.

Eventually I thought it's the end of the line, but then when I checked to WA. the nominator decomposed to this.

$$8(4 t^2+1)(4 t^2+4 t+5)$$

How can we know the factor if it's not linear i.e ##(x-a)##? For the note I'm currently reading the Complex Numbers section in Boas book. So far the author has not discussed yet techniques to factor polynomial like this.

Thank You
 
Last edited by a moderator:
Mathematics news on Phys.org
  • #2
Hello Seydlitz! :smile:

(btw, it's "numerator" and "denominator" :wink:)

I don't understand why you don't simply use long division, it's reasonably quick.

Alternatively, if you know that (4t2+1) is a factor, then you can find the other quadratic more or less immediately, by inspection.
 
  • #3
As an alternative, the first thing would be to notice that all of the terms in the numerator are even, so bring out a factor of 2. Better yet, notice that the coefficients are all divisible by 8. This means that you can write the numerator as 8(16t4 - 16t3 + 24t2 + 4t + 5).

The Rational Root theorem can be used to find roots of the form p/q, where p = ±1 or ±5, and q = ±1, ±2, ±4, ±8 or ±16.
 
  • #4
tiny-tim said:
Hello Seydlitz! :smile:

(btw, it's "numerator" and "denominator" :wink:)

I don't understand why you don't simply use long division, it's reasonably quick.

Alternatively, if you know that (4t2+1) is a factor, then you can find the other quadratic more or less immediately, by inspection.

Ouhh my bad. I typed it at night.

I know it's possible to use long division because the polynomial is rational, but what if I've only been given the numerator. How can I test quickly that it can be decomposed?

Mark44 said:
As an alternative, the first thing would be to notice that all of the terms in the numerator are even, so bring out a factor of 2. Better yet, notice that the coefficients are all divisible by 8. This means that you can write the numerator as 8(16t4 - 16t3 + 24t2 + 4t + 5).

The Rational Root theorem can be used to find roots of the form p/q, where p = ±1 or ±5, and q = ±1, ±2, ±4, ±8 or ±16.

If the root is real then I think I can guess it fine, but then in this case the polynomial has complex roots. ##4t^2+4t+5## and the ##4t^2+1##.
 
  • #5
I guess the minus sign in the first fraction is a typo? For the factorization, you need a plus there.

Seydlitz said:
I know it's possible to use long division because the polynomial is rational, but what if I've only been given the numerator. How can I test quickly that it can be decomposed?
Well, it has order>2, it has to be composite.

To make numbers smaller, I would substitute 2t=s:
8(s^4 + 2s^3 + 6s2 + 2s + 5).

You can test imaginary numbers together with real numbers, it will need more time but it is possible.
You can even see ±i as a solution here.
 
Last edited:
  • #6
(just got up :zzz:)
Seydlitz said:
I know it's possible to use long division because the polynomial is rational, but what if I've only been given the numerator. How can I test quickly that it can be decomposed?

no way that i know of :redface:

(but don't worry … they won't give you anything that horrible in the exam! o:))
 
  • #7
mfb said:
I guess the minus sign in the first fraction is a typo? For the factorization, you need a plus there.

Yes it should be positive for the second term, sorry for that. I can't no longer edit the first post.

mfb said:
Well, it has order>2, it has to be composite.

To make numbers smaller, I would substitute 2t=s:
8(s^4 + 2s^3 + 6s2 + 2s + 5).

You can test imaginary numbers together with real numbers, it will need more time but it is possible.
You can even see ±i as a solution here.

Ah so it's possible to check it with imaginary number as well. I'll try to keep that in mind.

tiny-tim said:
(just got up :zzz:)

no way that i know of :redface:

(but don't worry … they won't give you anything that horrible in the exam! o:))

Thanks for your assurance tiny-tim. It's not everyday I see polynomial like it.

I just need a bit of clarification there finally, if a polynomial has order>2 then it has to be composite but it's not necessarily possible to factorize them into simpler factor. Is that correct?
 
  • #8
Seydlitz said:
… finally, if a polynomial has order>2 then it has to be composite but it's not necessarily possible to factorize them into simpler factor. Is that correct?

no

every real polynomial can be factored into linear factors: (x -a1)(x -a2)…(x -an)

but the roots may be complex, except:

if a is a complex root, then so is a* (its complex conjugate), and so a + a* and aa* are both real, and so (x -a)(x - a*) = x2 - (a + a*)x + aa* is real

so any real polynomial can be factored into real quadratic factors, plus (if it has odd order) one real linear factor :wink:
 
  • #9
tiny-tim said:
no

every real polynomial can be factored into linear factors: (x -a1)(x -a2)…(x -an)

but the roots may be complex, except:

if a is a complex root, then so is a* (its complex conjugate), and so a + a* and aa* are both real, and so (x -a)(x - a*) = x2 - (a + a*)x + aa* is real

so any real polynomial can be factored into real quadratic factors, plus (if it has odd order) one real linear factor :wink:

Ah ok, so every real polynomial can be factored into real quadratic factors or with the addition of one linear factor, the quadratic factors can then be further factorized to complex linear root if it has no real linear root. Is this correct?
 
  • #10
correct :smile:
 
  • #11
If you're only interested in real roots, it's worth noting that we can see instantly that it has no positive real roots, since none of the terms can be negative.

If we look at the dominance of the terms for the even coefficients, we should be suspicious that it has no negative real roots either. If you mirror your equation about t=0, by replacing t with -t, then you may be able to convince yourself of this.

Your lowest order term is a constant. The next term linear. As the order increases their dominance as we move outside of the range of -1<t<+1 increases. Even terms are always offer a positive contribution and odd terms offer a positive contribution in for t>0 and a negative contribution for t<0.

At a guess, for even order polynomials, I would expect there to be a heuristic that suggests there are no real roots where all coefficients are positive; odd coefficients are less than the previous even coefficient and even coefficients are greater than or equal to, the previous odd coefficient. Which matches your equation.

It's far from a proof but getting a bit of an intuitive understanding of the problem that you're trying to solve can save you a lot of time.

Analytical solutions for the roots of higher order polynomials get very complicated fast. If you need a solution then practically, you would turn to numerical methods.
 
Last edited:
  • #12
@craigi: To see that there are no negative real roots, consider the cases -1 < t < 0 and t <= -1:
t>-1: 192t^2 + 128t^3 > 0 and 40 + 32t > 0 and 192t^4>0
t<=-1: 192t^4 + 192t^3 > 0 and 192t^2 + 32t > 0 and 40>0
 

1. What are the most common techniques for factoring large polynomials?

The most common techniques for factoring large polynomials include the following:

  • Factoring by grouping: This involves grouping terms in the polynomial and finding common factors to simplify the expression.
  • Factoring by using the GCF (Greatest Common Factor): This method involves finding the largest factor that divides all terms in the polynomial.
  • Factoring by using the difference of squares: This technique is used when the polynomial is in the form of a2 - b2, where a and b are integers.
  • Factoring by using the sum or difference of cubes: This method is used when the polynomial is in the form of a3 ± b3, where a and b are integers.
  • Factoring by using the quadratic formula: This technique is used when the polynomial is in the form of ax2 + bx + c, where a, b, and c are integers.

2. How do I know which technique to use for factoring a large polynomial?

The best way to determine which technique to use for factoring a large polynomial is to first look for any common factors that can be factored out using the GCF method. If the polynomial cannot be factored using the GCF method, then try factoring by grouping. If neither of these methods work, then try factoring by using the difference of squares or the sum/difference of cubes. If the polynomial is a quadratic expression, then use the quadratic formula to factor it.

3. Can I use a calculator to factor large polynomials?

Yes, there are many online calculators and software programs available that can help you factor large polynomials. However, it is important to understand the steps and techniques involved in factoring, as blindly relying on a calculator may not always provide the correct answer.

4. Is there a specific order in which I should factor a large polynomial?

There is no specific order in which you should factor a large polynomial. However, it is recommended to first look for any common factors using the GCF method, then try factoring by grouping, and finally move on to the other techniques mentioned above.

5. Are there any tricks or shortcuts for factoring large polynomials?

There are some common patterns and tricks that can help with factoring large polynomials, such as the difference of squares and the sum/difference of cubes. However, these are not applicable to all polynomials, so it is important to understand the underlying concepts and techniques for factoring. Practice and familiarity with different types of polynomials can also help in identifying patterns and making the factoring process more efficient.

Similar threads

Replies
0
Views
495
Replies
1
Views
1K
  • Programming and Computer Science
Replies
3
Views
748
  • General Math
Replies
4
Views
2K
  • General Math
Replies
13
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
1
Views
2K
  • General Math
Replies
8
Views
3K
Replies
6
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
913
Back
Top