How do I factorize (x-1)(x-2)(x-3)(x-4) -48

  • Thread starter Thread starter eureka_beyond
  • Start date Start date
AI Thread Summary
To factorize (x-1)(x-2)(x-3)(x-4) - 48 into (x²-5x+12)(x²-5x-2), one must first expand the left-hand side and simplify it into standard polynomial form. The goal is to verify if the proposed factorization is correct by performing polynomial long division with one of the quadratic factors. If the division yields no remainder, the other factor is confirmed. However, if the factorization is not provided, finding it may not be possible, as the quadratic factors do not appear to be factorable into first-degree binomials. The clarity of the problem statement is crucial for determining the correct approach.
eureka_beyond
Messages
20
Reaction score
0

Homework Statement


how do I factorize (x-1)(x-2)(x-3)(x-4) -48 into (x2-5x+12)(x2-5x-2)

Homework Equations



no relevant equations

The Attempt at a Solution



I've found the value of (x-1)(x-4) and the value of (x-2)(x-3) first. which is (x2-5x+4) abd (x2-5x+6), I put them together and -48 at the end. The problem is, how do I put the -48 inside the brackets?
 
Physics news on Phys.org


eureka_beyond said:

Homework Statement


how do I factorize (x-1)(x-2)(x-3)(x-4) -48 into (x2-5x+12)(x2-5x-2)

Why would you need to put the -48 inside the brackets?

You are not asked to do so, are you?
 


if I don't put the -48 inside, then the answer won't be (x2-5x+12)(x2-5x-2), which is the answer to the question. So yes, I do need to put the -48 inside.
 


Completely multiply the binomials in your first member and simplify, putting this into general form. Look then at one of your degree-two factors from your second member. You should be able to divide your second member by either of these factors. If you find no remainder, then you know the quotient found is the other factor. Read and reread that process very carefully and then do it part by part.

Your entire first member is (x-1)(x-2)(x-3)(x-4) -48 and your entire second member is
(x2-5x+12)(x2-5x-2)
 


symbolipoint said:
Look then at one of your degree-two factors from your second member. You should be able to divide your second member by either of these factors.

I don't get it =[, sorry but can you explain a bit more, I'm really bad at this topic. Thankyou so much !
 


How was the question worded exactly? You said:
"how do I factorize (x-1)(x-2)(x-3)(x-4) -48 into (x2-5x+12)(x2-5x-2)"

But was the factorization (x2-5x+12)(x2-5x-2) given in the problem, or are we looking for this factorization?

symbolipoint is assuming that the factorization was given. So given this:
(x-1)(x-2)(x-3)(x-4) - 48 = (x2-5x+12)(x2-5x-2)
You completely expand the left hand side and write the resulting polynomial in standard form (decreasing powers of x). Then divide it by one of the factors on the right hand side using long division. If this is a true statement, then the quotient you get should be the other factor.

However, if we are NOT given this factorization, and that we are supposed to find it ourselves, then symbolipoint's suggestion won't work.
 


If the actual quadratic factored form is not given, or at least if one of the factors is not given, then we may be unable to find a factorization for the first expression, (x-1)(x-2)(x-3)(x-4) -48 , since both the quadratic factors appear to not be factorable themselves into first degree binomials.

We would need to actually be given the factorization and then show that this factorization is correct.
 


I should have put it clearer in the first place. The actual form (x2-5x+12)(x2-5x-2)
is not given.
 
Back
Top