Factorising Polynomials

[ConnorK]

I am working through a chapter on factorising polynomials and I don't understand one part of an example. The sentence i don't understand is this:
"By considering the constant term, it can be seen that -@c = 6. Therefore @ divides 6 ( since @ is an integer it follows that b and c are too.)."

Below is the part of the textbook I got the sentence from. I don't understand what the books means that "@ Divides 6" and that it follows that b and c are integers because @ is an integer.
All help appreciated

 

[lev888]

I am working through a chapter on factorising polynomials and I don't understand one part of an example. The sentence i don't understand is this:
"By considering the constant term, it can be seen that -@c = 6. Therefore @ divides 6 ( since @ is an integer it follows that b and c are too.)."

Below is the part of the textbook I got the sentence from. I don't understand what the books means that "@ Divides 6" and that it follows that b and c are integers because @ is an integer.
All help appreciated

If 2 polynomials are equal their corresponding coefficients are equal:
-2 = -(a-b)
-5 = -(ab-c)
6 = -ac

We know that a is an integer. Therefore b is an integer because -2 = -(a-b). And then c is an integer because -5 = -(ab-c).
So, a and c are integer factors of 6 because 6 = -ac. This means a divides 6.

 

[fresh_42]

We have [math] x^3-2x^2-5x+6=x^3-(\alpha -b)x^2-(\alpha b-c)x-\alpha c [/math] and by comparison of the coefficients at the [imath] x [/imath] terms [math] -2=-(\alpha -b)\, , \,-5=- (\alpha b-c)\, , \, +6=-\alpha c .[/math] We have assumed that [imath] \alpha [/imath] is an integer. That means [imath] +6=-\alpha c [/imath] or [imath] \alpha =\dfrac{+6}{-c} [/imath] is an integer. So when is the RHS an integer? Only if [imath] c [/imath] divides six. And [imath] \{\pm 1\, , \,\pm 2\, , \,\pm 3\, , \,\pm 6\} [/imath] are all integer divisors of six. These are the possibilities for [imath] c [/imath] and six divided by one of these numbers is again one of them. Hence [imath] \alpha \in \{\pm 1\, , \,\pm 2\, , \,\pm 3\, , \,\pm 6\}[/imath], too. Finally, [imath] -2=-(\alpha -b) [/imath] or [imath] b=\alpha -2. [/imath] With [imath] \alpha [/imath] being an integer, [imath] b [/imath] is as well an integer.

Now we have that [imath] \alpha [/imath] is an integer (per assumption), and [imath] \alpha=-\dfrac{6}{c}\in \mathbb{Z} [/imath] forces [imath] c\in \{\pm 1\, , \,\pm 2\, , \,\pm 3\, , \,\pm 6\},[/imath] and [imath] b=\alpha -2 \in \mathbb{Z}. [/imath]

 

[fresh_42]

You can also try the numbers in [imath] \{\pm 1\, , \,\pm 2\, , \,\pm 3\, , \,\pm 6\} [/imath] and check whether one of them results in [imath] x^3-2x^2-5x+6=0. [/imath] If you found such a number [imath] \alpha\in \{\pm 1\, , \,\pm 2\, , \,\pm 3\, , \,\pm 6\}[/imath] with that property, them you can perform a long division, say [imath] (x^3-2x^2-5x+6) : (x-3) .[/imath]

 

[Dr.Peterson]

I am working through a chapter on factorising polynomials and I don't understand one part of an example. The sentence i don't understand is this:
"By considering the constant term, it can be seen that -αc = 6. Therefore α divides 6 ( since α is an integer it follows that b and c are too.)."

Below is the part of the textbook I got the sentence from. I don't understand what the books means that "α Divides 6" and that it follows that b and c are integers because α is an integer.
All help appreciated

I'll take your question literally:

I don't understand what the books means that "α Divides 6" and that it follows that b and c are integers because α is an integer.​

So, what does this statement mean?

That statement, "α divides 6", means that 6 can be divided evenly by alpha; that is, 6 is an integer multiple of alpha, and alpha is a divisor of 6.

They say that because since αc = -6, then -αc = 6, so 6 is in fact an integer multiple of alpha, namely α(-c).

They mention that b and c are integers because that is necessary for what I just said to be true: if c is an integer, then 6 is an integer multiple of alpha.

And why must b and c be integers? That needs some explanation; it is not clear what sort of thinking they are assuming should be obvious to you. Possibly this comes from something previously discussed, such as that the quadratic factor [imath]x^2+bx+c[/imath] is obtained by polynomial division, and that process can only produce integers. (This can also be expressed as in answer #2.)

Does that help you understand what they mean? The logic is stated backwards, so I can see how it can be hard to follow.

 

[ConnorK]

I'll take your question literally:

I don't understand what the books means that "α Divides 6" and that it follows that b and c are integers because α is an integer.​

So, what does this statement mean?

That statement, "α divides 6", means that 6 can be divided evenly by alpha; that is, 6 is an integer multiple of alpha, and alpha is a divisor of 6.

They say that because since αc = -6, then -αc = 6, so 6 is in fact an integer multiple of alpha, namely α(-c).

They mention that b and c are integers because that is necessary for what I just said to be true: if c is an integer, then 6 is an integer multiple of alpha.

And why must b and c be integers? That needs some explanation; it is not clear what sort of thinking they are assuming should be obvious to you. Possibly this comes from something previously discussed, such as that the quadratic factor [imath]x^2+bx+c[/imath] is obtained by polynomial division, and that process can only produce integers. (This can also be expressed as in answer #2.)

Does that help you understand what they mean? The logic is stated backwards, so I can see how it can be hard to follow.

Sorry about the very late reply.

"That statement, "α divides 6", means that 6 can be divided evenly by alpha; that is, 6 is an integer multiple of alpha, and alpha is a divisor of 6"

By this do you mean that when 6 is divided by alpha an integer is produced? I am unfamiliar with the term "integer multiplier", i looked it up and applying it to this situation would it be correct to say that alpha multiplied by some integer equals 6?


I think i may be missing something here. Correct me if i am wrong, is it true that if the product of 2 quantities produce an integer then one of the quantities has to also be an integer. In other words you cannot multiply two rational numbers that are not integers and get an integer? but i can multiply an integer and a rational number that is not an integer and get a integer?

 

[Dr.Peterson]

"That statement, "α divides 6", means that 6 can be divided evenly by alpha; that is, 6 is an integer multiple of alpha, and alpha is a divisor of 6"

By this do you mean that when 6 is divided by alpha an integer is produced? I am unfamiliar with the term "integer multiplier", i looked it up and applying it to this situation would it be correct to say that alpha multiplied by some integer equals 6?

Yes. We say that one integer, m, divides another, n, when their quotient n/m is an integer. An integer multiple (or, simply, a multiple) of a number means the product of that number and an integer. This is not the same as "multiplier", which would be the integer you multiplied by.

I think i may be missing something here. Correct me if i am wrong, is it true that if the product of 2 quantities produce an integer then one of the quantities has to also be an integer. In other words you cannot multiply two rational numbers that are not integers and get an integer? but i can multiply an integer and a rational number that is not an integer and get a integer?

No. The product of, say, 2/3 and 3/2, neither of which is an integer, is 1, an integer.

But the product of an integer and a non-integer can't be an integer.

Many things like this can be seen just by looking for a counterexample -- an example that shows the statement is not always true.

 

[mario99]

But the product of an integer and a non-integer can't be an integer.

[imath]\displaystyle 2 \times \frac{1}{2} = 1 \ \ \ [/imath] (The result is an integer.)

 

[Dr.Peterson]

[imath]\displaystyle 2 \times \frac{1}{2} = 1 \ \ \ [/imath] (The result is an integer.)

Oops. I was rushing. Can't do that in math.

I suspect what I had in mind was "integer and irrational".

 

[ConnorK]

I'll take your question literally:

I don't understand what the books means that "α Divides 6" and that it follows that b and c are integers because α is an integer.​

So, what does this statement mean?

That statement, "α divides 6", means that 6 can be divided evenly by alpha; that is, 6 is an integer multiple of alpha, and alpha is a divisor of 6.

They say that because since αc = -6, then -αc = 6, so 6 is in fact an integer multiple of alpha, namely α(-c).

They mention that b and c are integers because that is necessary for what I just said to be true: if c is an integer, then 6 is an integer multiple of alpha.

And why must b and c be integers? That needs some explanation; it is not clear what sort of thinking they are assuming should be obvious to you. Possibly this comes from something previously discussed, such as that the quadratic factor [imath]x^2+bx+c[/imath] is obtained by polynomial division, and that process can only produce integers. (This can also be expressed as in answer #2.)

Does that help you understand what they mean? The logic is stated backwards, so I can see how it can be hard to follow.

"They mention that b and c are integers because that is necessary for what I just said to be true: if c is an integer, then 6 is an integer multiple of alpha."

So if c wasn't an integer then 6 would not be an integer multiple of a?

 

[ConnorK]

We have [math] x^3-2x^2-5x+6=x^3-(\alpha -b)x^2-(\alpha b-c)x-\alpha c [/math] and by comparison of the coefficients at the [imath] x [/imath] terms [math] -2=-(\alpha -b)\, , \,-5=- (\alpha b-c)\, , \, +6=-\alpha c .[/math] We have assumed that [imath] \alpha [/imath] is an integer. That means [imath] +6=-\alpha c [/imath] or [imath] \alpha =\dfrac{+6}{-c} [/imath] is an integer. So when is the RHS an integer? Only if [imath] c [/imath] divides six. And [imath] \{\pm 1\, , \,\pm 2\, , \,\pm 3\, , \,\pm 6\} [/imath] are all integer divisors of six. These are the possibilities for [imath] c [/imath] and six divided by one of these numbers is again one of them. Hence [imath] \alpha \in \{\pm 1\, , \,\pm 2\, , \,\pm 3\, , \,\pm 6\}[/imath], too. Finally, [imath] -2=-(\alpha -b) [/imath] or [imath] b=\alpha -2. [/imath] With [imath] \alpha [/imath] being an integer, [imath] b [/imath] is as well an integer.

Now we have that [imath] \alpha [/imath] is an integer (per assumption), and [imath] \alpha=-\dfrac{6}{c}\in \mathbb{Z} [/imath] forces [imath] c\in \{\pm 1\, , \,\pm 2\, , \,\pm 3\, , \,\pm 6\},[/imath] and [imath] b=\alpha -2 \in \mathbb{Z}. [/imath]

Why can't c be a fraction?

 

[Dr.Peterson]

"They mention that b and c are integers because that is necessary for what I just said to be true: if c is an integer, then 6 is an integer multiple of alpha."

So if c wasn't an integer then 6 would not be an integer multiple of a?

Right.

If a = 4 and c = 3/2, then 4*3/2 = 6, but 6 is not an integer multiple of 4 because 3/2 is not an integer. Likewise, we wouldn't say that 4 divides 6.

Why can't c be a fraction?

As I said,

And why must b and c be integers? That needs some explanation; it is not clear what sort of thinking they are assuming should be obvious to you. Possibly this comes from something previously discussed, such as that the quadratic factor [imath]x^2+bx+c[/imath] is obtained by polynomial division, and that process can only produce integers. (This can also be expressed as in answer #2 [And then c is an integer because -5 = -(ab-c)].)

Since what you showed us doesn't indicate even that they assume b and c are integers, much less why, we'd need to see the preceding context to be sure what they expect you to understand. But I suggested a reason, as did @lev888.

 

[fresh_42]

Why can't c be a fraction?

Yes, you are right. At prior, it can be e.g. [imath]c= 1/3 [/imath] and [imath] \alpha=-18 [/imath]. Nevertheless, let's see how far we get without assuming [imath] c \in \mathbb{Z}. [/imath] The fastest way is to take the derivatives. With them we have
[math]\begin{array}{lll} \dfrac{d}{dx}(x^3-2x^2-5x+6)&=3x^2-4x-5\\[6pt] &=(+1)\cdot(x^2+bx+c) + (x-\alpha)(2x+b)\\[6pt] &=x^2+bx+c+2x^2+bx-2\alpha x- b\alpha\\[6pt] &=3x^2+2(b-\alpha)x+(c-b\alpha)\\[6pt] \dfrac{d^2}{dx^2}(x^3-2x^2-5x+6)&=6x-4=6x+2(b-\alpha) \end{array}[/math]Thus [imath] b=\alpha -2 \in \mathbb{Z} [/imath] by the condition in the problem statement and [imath] c=b\alpha -5 \in \mathbb{Z}. [/imath]