Master Factorization with Step-by-Step Guide for x^4 - 3x^2 - 4

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In summary: I actually missed that. So the OP already had the trick that we all pointed out. There was just a factorization mistake. I hinted at the part he already had. (sigh) :-)The usual procedure is to guess a zero, e.g. ##x=2##, and then divide ##(x^4-3x^2-4):(x-2)## per long division (see https://www.physicsforums.com/threads/constructing-the-splitting-field-for-a-polynomial-over-z-z3.889140/#post-5595083).Then we get a polynomial of degree
  • #1
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TL;DR Summary: I know the answer is (x -2)(x+2)(x^2+1), but I don't know how that is got. I did this... lol

so it put brackets around (x^4 - 3x^2) and went

x^2(x^2 -3) -4

So i thought you could maybe do this.

(x^2 -4) (x^2 -3)

This gives me the (x -2)(x+2) part at least for the difference of squares, but that gives me the answer of (x -2)(x+2) (x^2 -3)

Eeek.. Any help appreciated.
 
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  • #2
Nathi ORea said:
TL;DR Summary: I know the answer is (x -2)(x+2)(x^2+1), but I don't know how that is got. I did this... lol

so it put brackets around (x^4 - 3x^2) and went

x^2(x^2 -3) -4

So i thought you could maybe do this.

(x^2 -4) (x^2 -3)

This gives me the (x -2)(x+2) part at least for the difference of squares, but that gives me the answer of (x -2)(x+2) (x^2 -3)

Eeek.. Any help appreciated.
The usual procedure is to guess a zero, e.g. ##x=2##, and then divide ##(x^4-3x^2-4):(x-2)## per long division (see https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083).

Then we get a polynomial of degree ##3## and we repeat that procedure.
 
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  • #3
Nathi ORea said:
So i thought you could maybe do this.

(x^2 -4) (x^2 -3)
Where did this come from? It's wrong. Let ##q=x^2## and factor ##q^2-3q-4##. Proceed from there.
 
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  • #4
For this problem I would start with a substitution.

##u=x^2##

To get a quadratic

##u^2-3u-4##

From here you can factor it however you want. Then plug ##x^2## back in and you end up with a product of quadratics, one of which you can factor again.
 
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  • #5
Nathi ORea said:
so it put brackets around (x^4 - 3x^2) and went

x^2(x^2 -3) -4
That doesn't get you anywhere at all.
Nathi ORea said:
So i thought you could maybe do this.

(x^2 -4) (x^2 -3)
But do these factors multiply back to ##x^4 - 3x^2 - 4##?
 
  • #6
Office_Shredder said:
For this problem I would start with a substitution.

##u=x^2##

To get a quadratic

##u^2-3u-4##

From here you can factor it however you want. Then plug ##x^2## back in and you end up with a product of quadratics, one of which you can factor again.
OMG! I totally get that! Thanks so much! I just worked it out :)
 
  • #7
Mark44 said:
That doesn't get you anywhere at all.

But do these factors multiply back to ##x^4 - 3x^2 - 4##?
ha ha.. i know! I was just mucking around.. i didn't get it at all.. lol
 
  • #8
Nathi ORea said:
ha ha.. i know! I was just mucking around.. i didn't get it at all.. lol
Try a different set of factors. This is a fairly easy problem.
 
  • #9
Nathi ORea said:
ha ha.. i know! I was just mucking around.. i didn't get it at all.. lol
Not sure, but you understood and know how? If not, say. This appears to be answered well in post #4.
 
  • #10
Surprised no one has said this yet but would you have had any difficulty with factorising

$$x^2 - 3x - 4$$​

?...
 
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  • #11
epenguin said:
Surprised no one has said this yet but would you have had any difficulty with factorising

$$x^2 - 3x - 4$$​

?...
I didn't think it was necessary. The OP had the basic idea this factorization -
Nathi ORea said:
So i thought you could maybe do this.

(x^2 -4) (x^2 -3)
but got one of the factors wrong. With the correct factorization, the rest of the problem would have been a piece of cake.
 
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  • #12
Mark44 said:
but got one of the factors wrong. With the correct factorization, the rest of the problem would have been a piece of cake.
I actually missed that. So the OP already had the trick that we all pointed out. There was just a factorization mistake. I hinted at the part he already had. (sigh) :-)
 
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  • #14
haider said:
Wouldn't synthetic division be much quicker than long division?
I personally prefer long division, not because it's faster or simpler, but because I am commonly dividing by higher degree polynomials. But yes, it's probably simpler.

-Dan
 
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  • #15
haider said:
Wouldn't synthetic division be much quicker than long division?
I didn't make any complexity considerations and do not really like to guess. It is finally always the Euclidean algorithm in different versions which is ##O(\log^3 n )## (Wikipedia) without tuning. My suspicion is that different versions differ only in the constant.

I find long division more natural as we use this version for numbers anyway. All it takes to know is to realize that it works for polynomials alike. This message is useful whenever polynomials have to be divided.

The special case here is, of course, faster solved by substitution as @Office_Shredder has noted in post #4.
 
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  • #16
Regarding polynomial long division, based on the difficulty level of this problem, it's highly unlikely that the OP has been introduced to this type of division. Only after the student has had practice with "easy" factorizations is poly long division finally introduced. That's my experience in teaching from a number of precalculus textbooks over a span of 20+ years.
 
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  • #17
Mark44 said:
Regarding polynomial long division, based on the difficulty level of this problem, it's highly unlikely that the OP has been introduced to this type of division. Only after the student has had practice with "easy" factorizations is poly long division finally introduced. That's my experience in teaching from a number of precalculus textbooks over a span of 20+ years.
And isn't it a pity? Again a sad example of where students are kept artificially ignorant. The procedure is exactly the same, and if we plug in ##x=10## is also literally the same. That's why I posted it. I really dislike this attitude, and I even heard it myself from a teacher: "This will be taught at college." A shame in my opinion.
 
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  • #18
fresh_42 said:
And isn't it a pity?
Not in my opinion. See below.
fresh_42 said:
Again a sad example of where students are kept artificially ignorant.
They have to learn to crawl before they can walk, and to walk before they can run.

I'm most familiar with college textbooks, although I taught in a high school for two years long ago. At least at the college level, students would be taught polynomial long division and synthetic division after they had worked through a bunch of exercises in factoring monomials, binomials, and trinomials.
 
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  • #19
fresh_42 said:
And isn't it a pity? Again a sad example of where students are kept artificially ignorant. The procedure is exactly the same, and if we plug in ##x=10## is also literally the same. That's why I posted it. I really dislike this attitude, and I even heard it myself from a teacher: "This will be taught at college." A shame in my opinion.
Polynomial Long Division, taught in Algebra, or no later than Algebra 2, regardless of in high school or college; right?
 
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  • #20
symbolipoint said:
Polynomial Long Division, taught in Algebra, or no later than Algebra 2, regardless of in high school or college; right?
I think I had it in calculus one, but it's too long ago to remember exactly. One can put it under ring theory the moment someone says that polynomial rings over fields are Euclidean. This is in my opinion not necessary. Polynomials are taught at school, e.g. partial fraction decomposition for integration. It would take at most an hour to divide them there and to teach what the extended Euclidean algorithm actually is.

I think that the gap between school mathematics and university mathematics is artificially big - for whatever reasons. At the end of school, kids are barely on the level of Pythagoras. Why did I have to learn about an endoplasmatic reticulum in biology, but the correct definition of a continuous function or a partial derivative in mathematics is too complicated? ##^{*})##

However, this would be the subject of a new thread in the education forum. Long polynomial division can be discussed as it is one possible solution to the OP. Whether such a solution is "too complicated or not" is already a meta-subject and we cannot even decide whether it is true or not for the OP.

We should close the thread here since the question is answered in multiple ways.
Discussing ##^*)## in the education forum is still an option.
 
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What is factorization?

Factorization is the process of breaking down a mathematical expression into smaller, simpler expressions. It involves finding the factors (numbers or variables) that can be multiplied together to get the original expression.

Why is factorization important?

Factorization is important in mathematics because it helps simplify complex expressions and make them easier to work with. It is also a fundamental concept in algebra and is used in various areas of science and engineering.

How do I factorize x^4 - 3x^2 - 4?

To factorize x^4 - 3x^2 - 4, we can use the grouping method. First, we can factor out the common factor x^2, leaving us with x^2(x^2 - 3) - 4. Then, we can further factor the expression inside the parentheses as (x + 1)(x - 4). Therefore, the final factorization is x^2(x + 1)(x - 4).

What is the step-by-step guide for factorizing x^4 - 3x^2 - 4?

The step-by-step guide for factorizing x^4 - 3x^2 - 4 is as follows:

  1. Factor out the common factor x^2, leaving x^2(x^2 - 3) - 4.
  2. Factor the expression inside the parentheses as (x + 1)(x - 4).
  3. Combine the factors to get the final factorization of x^2(x + 1)(x - 4).

What are some applications of factorization?

Factorization has various applications in mathematics, science, and engineering. It is used to simplify complex equations, find solutions to equations, and solve real-world problems. It is also used in cryptography, data compression, and in the study of prime numbers.

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