Can this polynomial be factored into two integer products

[timetraveller123]

Homework Statement

Homework Equations

none

The Attempt at a Solution

i assumed it can be factored into the form
$(x^2 + m_1 x + m_0)(x + n_0)$
by comparison of coefficients
$m_0 n_0 = -abc -1\\ m_1 + n_0 = -a -b -c\\ m_0 + m_1 n_0 = ab +ac + bc\\$
the only other information i have is all the six variables are integers
is there a way from here

 

[fresh_42]

Can you imagine the graph of $y=(x-a)(x-b)(x-c)$ and what may happen to it, if you shift this curve by $-1\,$?

Edit: My bad, I've forgotten the integer restriction. Seems to be less easy that I first thought.

 

[timetraveller123]

@fresh_42 42
i think i have made a little progress using a different method help me verify
once again assuming it can be factored
into the form with integer coefficients
$(x^2 + m_1x + m_0)(x + n_0)$
then n_0 becomes a root of
$(x-a)(x-b)(x-c) -1 = 0\\ (-n_0 - a)(-n_0 - b)(-n_0 -c) =1\\$
but each term is a distinct integer not possible
hence not factorable
is this valid

 

[ehild]

@fresh_42 42
i think i have made a little progress using a different method help me verify
once again assuming it can be factored
into the form with integer coefficients
$(x^2 + m_1x + m_0)(x + n_0)$
then n_0 becomes a root of
$(x-a)(x-b)(x-c) -1 = 0\\ (-n_0 - a)(-n_0 - b)(-n_0 -c) =1\\$
but each term is a distinct integer not possible
hence not factorable
is this valid

It is very clever! But you should elaborate to make a real solution. Why can not be the factors (-n_0 - a), (-n_0 - b), (-n_0 -c), all distinct?

 

[timetraveller123]

@ehild
ok
for 1 to have 3 different integer factors is if it is
1,1,1
1,-1,-1
which is all the possibles
each of which has at least one repeated factor which is not allowed

 

[ehild]

@ehild
ok
for 1 to have 3 different integer factors is if it is
1,1,1
1,-1,-1
which is all the possibles
each of which has at least one repeated factor which is not allowed

OK. So you can choose 2 different number from a, b, c (-1-no and 1-no), but the third should be also 1-no.

 

[timetraveller123]

thanks

 

[Jamison Lahman]

Does a*b*c-1 always equal a prime number?

1*2*5-1 = 9
:/

 

[Jamison Lahman]

Let me know what you think about this:
$(x-a)(x-b)(x-c)-1 = (1)x^3-(a+b+c)x^2+(ab+ac+bc)x-(abc-1)$
Now, let's introduce coefficients for each term, i.e. $y_1 = 1$, $y_2=(a+b+c)$, $y_3 = (ab+ac+bc)$, and $y_4=(abc-1)$. The only way we will be able to factor this into the product of two polynomials is if the total parity of the coefficients is an even number.
Creating a table:
$\begin{array}{c|cccc} \text{of even numbers} (abc) & y_1 \text{ parity} & y_2 \text{ parity} & y_3 \text{ parity} & y_4 \text{ parity} \\ \hline 0 & odd & odd & odd & even \\ \hline 1 & odd & even & odd & odd \\ \hline 2 & odd & odd & even & odd \\ \hline 3 & odd & even & even & even \\ \hline \end{array}$
As you can see, the parities never match and therefore it will never factor into the product of two polynomials.

 

[Jamison Lahman]

Some not-so-relevant fun with algebra: $$x^3-(a+b+c)x^2+(ab+bc+ac)x-(abc-1)=0$$ if we factor that...
$$(x^2+(ab+bc+ac))\left(x-\frac{abc-1}{ab+bc+ac}\right)=0$$ which implies, $$(a+b+c)=\frac{abc-1}{ab+bc+ac}$$
(Using WolframAlpha) we find the solutions for which are: a=-3/2, b=-5/2, and c=2 or a=-3/2, b=1/2, and c=1/2.
Using the first set, we are finally able to factor into a product of two polynomials:
$$\left(x^2-\frac{17}{2}\right)(x+2)=0$$Looking at the graph it obviously has only two roots.

This isn't a solution to the original problem since the values for the coefficients are not integers, but I've thoroughly enjoyed tinkering with this problem and thought it would be educational to share.

 

[timetraveller123]

Let me know what you think about this:
$(x-a)(x-b)(x-c)-1 = (1)x^3-(a+b+c)x^2+(ab+ac+bc)x-(abc-1)$
Now, let's introduce coefficients for each term, i.e. $y_1 = 1$, $y_2=(a+b+c)$, $y_3 = (ab+ac+bc)$, and $y_4=(abc-1)$. The only way we will be able to factor this into the product of two polynomials is if the total parity of the coefficients is an even number.
Creating a table:
$\begin{array}{c|cccc} \text{of even numbers} (abc) & y_1 \text{ parity} & y_2 \text{ parity} & y_3 \text{ parity} & y_4 \text{ parity} \\ \hline 0 & odd & odd & odd & even \\ \hline 1 & odd & even & odd & odd \\ \hline 2 & odd & odd & even & odd \\ \hline 3 & odd & even & even & even \\ \hline \end{array}$
As you can see, the parities never match and therefore it will never factor into the product of two polynomials.

i don't understand what you mean the total parity of coefficients is an even number can please help me explain

 

[Jamison Lahman]

i don't understand what you mean the total parity of coefficients is an even number can please help me explain

Sorry, I am still working on explaining things better. Using the term "even number" when dealing with parity is reasonably confusing and I apologize. I think it is best to describe this with an example. Say you are given the following equation and are asked to factor it,
$$ x^3+2x^2+3x+5$$
You may want to try to factor out an x² from the first two terms, but then you would be left with$x^2(x+2)+(3x+5)$. As you can see, the term on the left has one even number and the term on the right has no even numbers. It may be possible to factor this given the initial restrictions using the technique known as Completing the Square, but I am unsure how to disprove that. I don't think completing the square will give you the product of two terms in this scenario, so perhaps that is why you can ignore it. At any rate, factoring by grouping is not possible unless there are the same ratio of even-to-odd coefficients in the two grouped terms. In the example, this would be the (x+2) and (3x+5) terms.

 

[Jamison Lahman]

According to https://www.mathsisfun.com/algebra/completing-square.html, the remainder outside of the product, e, is given by the equation,
$$e = c - \frac{b^2}{4a}$$
In order to factor it into the product of two terms, we want e = 0. Since we can make c either even or odd, the parity of $\frac{b^2}{4a}$ seems irrelevant. With that being said, it looks like completing the square can be an alternative to factoring by grouping to make the product of two terms and therefore the proof is invalid. :/

Edit: I have mistaken myself. This is solely for quadratic functions, not cubic functions. I am unsure if completing the square works for cubic functions. Perhaps someone else may be able to help past here.

 

[timetraveller123]

i am sorry i still don't get it i think your method beyond my scope of understanding but anyways seems like a interesting method to me thanks for sharing
once again sorry i hope you are not offended

 

[Jamison Lahman]

i am sorry i still don't get it i think your method beyond my scope of understanding but anyways seems like a interesting method to me thanks for sharing
once again sorry i hope you are not offended

Don't worry about it then. I am wrong anyway.
$$(x^3+4x^2)+(2x+8) = x^2(x+4)+2(x+4) = (x^2+2)(x+4)$$
That is an example of 3 even coefficients being factored into the product of two terms with integer coefficients :/

 

[Jamison Lahman]

Don't worry about it then. I am wrong anyway.
$$(x^3+4x^2)+(2x+8) = x^2(x+4)+2(x+4) = (x^2+2)(x+4)$$

That is example of 3 even coefficients being factored into the product of two terms with integer coefficients :/