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

• eureka_beyond
We are supposed to factorize it ourselves. The only information given is the first expression ((x-1)(x-2)(x-3)(x-4) -48) and the fact that the final answer should be (x2-5x+12)(x2-5x-2).In summary, the problem is asking to factorize the expression (x-1)(x-2)(x-3)(x-4) -48 into the form (x2-5x+12)(x2-5x-2). This cannot be done without knowing the actual quadratic factored form, as the given expression does not have a factorization into first degree binomials.

## 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?

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.

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

To factorize this expression, you can start by grouping the terms in pairs:
[(x-1)(x-4)][(x-2)(x-3)] -48
Next, you can use the difference of squares formula to expand each pair:
[(x^2-5x+4)][(x^2-5x+6)] -48
Then, you can factor out the common factors of (x^2-5x) from each pair:
(x^2-5x)(x^2-5x+2) -48
Finally, you can use the difference of squares formula again to factor out the remaining terms:
(x^2-5x)(x^2-5x+2) - (4)(12)
And the fully factorized expression is (x^2-5x+2)(x^2-5x-12).

## 2. Can I use the quadratic formula to factorize (x-1)(x-2)(x-3)(x-4) -48?

No, the quadratic formula is used to find the roots of a quadratic equation (in the form of ax^2 + bx + c = 0). The expression (x-1)(x-2)(x-3)(x-4) -48 is a quartic polynomial (in the form of ax^4 + bx^3 +cx^2 + dx + e).

## 3. Can I use the distributive property to factorize (x-1)(x-2)(x-3)(x-4) -48?

Yes, you can use the distributive property to expand the expression, but it will not give you the fully factorized form. It will only expand the expression into a longer form.

## 4. How can I check if my factorization is correct for (x-1)(x-2)(x-3)(x-4) -48?

You can use the distributive property to expand your factorization and see if it simplifies back to the original expression [(x-1)(x-2)(x-3)(x-4) -48]. If it does, then your factorization is correct.

## 5. Can I factorize (x-1)(x-2)(x-3)(x-4) -48 into complex factors?

No, the expression (x-1)(x-2)(x-3)(x-4) -48 only has real number solutions. Therefore, the factors will also be real numbers.