Checking factors

[Probability]

I am asked to write;

[MATH]{6x}^{5}{y}^{8}[/MATH] in the form [MATH]{2x^2}{y}\times[/MATH] something

I thought;

[MATH]{2x^2}\times{3x^3}\times{y}\times{y^7}={6x^5}\times{y}\times{y^7}\implies[/MATH][MATH]{6x^5}\times{y^8}[/MATH]
But the solution given says;

[MATH]({2x^2}{y}\times{3x^3}\times{y^7})[/MATH]
If I multiply through by 1 I'll still get my same answer. Is this just another way of writing the solution or is it incomplete?

 

[Dr.Peterson]

So, what was your answer? You have all the pieces, but I don't see that you wrote anything in the form [MATH]2x^2y\times(\cdots)[/MATH].

Of course, I would have written the answer as [MATH](2x^2y)(3x^3y^7)[/MATH].

 

[Steven G]

I am asked to write;

[MATH]{6x}^{5}{y}^{8}[/MATH] in the form [MATH]{2x^2}{y}\times[/MATH] something

Let us see what you have. In 6x⁵y⁸ you have 3 factors. One is a number, the other is a power of x and the third is a power of y.
Now 2*3 = 6, x²*x³=x⁵ and y*y⁷=y⁸.

So what is the something?

 

[Probability]

I am asked to write;

[MATH]{6x}^{5}{y}^{8}[/MATH] in the form [MATH]{2x^2}{y}\times[/MATH] something

I thought;

[MATH]{2x^2}\times{3x^3}\times{y}\times{y^7}={6x^5}\times{y}\times{y^7}\implies[/MATH][MATH]{6x^5}\times{y^8}[/MATH]
I thought ^^this part was the solution

But the solution given says;

[MATH]({2x^2}{y}\times{3x^3}\times{y^7})[/MATH]
If I multiply through by 1 I'll still get my same answer. Is this just another way of writing the solution or is it incomplete?

So, what was your answer? You have all the pieces, but I don't see that you wrote anything in the form [MATH]2x^2y\times(\cdots)[/MATH].

Of course, I would have written the answer as [MATH](2x^2y)(3x^3y^7)[/MATH].

I thought this was the solution;
[MATH]{2x^2}\times{3x^3}\times{y}\times{y^7}[/MATH][MATH][/math]

 

[Probability]

Let us see what you have. In 6x⁵y⁸ you have 3 factors. One is a number, the other is a power of x and the third is a power of y.
Now 2*3 = 6, x²*x³=x⁵ and y*y⁷=y⁸.

So what is the something?

When you say I have 3 factors, one is a number. I see 2 numbers. Look first a number 6 in (6x^5). Factors of 6 are 2 and 3.

 

[HallsofIvy]

I thought this was the solution;
[MATH]{2x^2}\times{3x^3}\times{y}\times{y^7}[/MATH]

But that does not have "[math]2x^2y[/math]" in front so is not "[math]2x^2y\times[/math] something".

 

[Probability]

But that does not have "[math]2x^2y[/math]" in front so is not "[math]2x^2y\times[/math] something".

I see where you are coming from. I seem to have then two areas of confusion in this branch of mathematics. 1; keeping the like terms together is not a requirement, and 2; when looking for common factors I should be thinking about trying to reverse engineer the math so it ends up in brackets! At the moment I'm finding that breaking down the terms and arranging these into brackets is most difficult for me.

 

[Dr.Peterson]

I see where you are coming from. I seem to have then two areas of confusion in this branch of mathematics. 1; keeping the like terms together is not a requirement, and 2; when looking for common factors I should be thinking about trying to reverse engineer the math so it ends up in brackets! At the moment I'm finding that breaking down the terms and arranging these into brackets is most difficult for me.

As I said (post #2), you had all the pieces, but didn't answer the question.

All that you lacked was paying attention to the wording of the question. You just have to read what is asked in a specific question.

The point of this exercise is to prepare you for further factoring problems where this step is just part of the work. Splitting a product into two parts is a useful skill. What I typically do is to write [MATH]6x^5y^8 = (2x^2y)(.................)[/MATH] and think, "[MATH]6 = 2[/MATH] times what? [MATH]x^5 = x^2[/MATH] times what? [MATH]y^8 = y^1[/MATH] times what?" The answers go in the parentheses, and I'm done.

 

[Steven G]

When you say I have 3 factors, one is a number. I see 2 numbers. Look first a number 6 in (6x^5). Factors of 6 are 2 and 3.

6x^5 is made up of two factors, 6 and x^5. Sure you can break up 6 to 2*3 and you can also break up x^5 to x*x*x*x*x so what is your point?

 

[Probability]

As I said (post #2), you had all the pieces, but didn't answer the question.

All that you lacked was paying attention to the wording of the question. You just have to read what is asked in a specific question.

The point of this exercise is to prepare you for further factoring problems where this step is just part of the work. Splitting a product into two parts is a useful skill. What I typically do is to write [MATH]6x^5y^8 = (2x^2y)(.................)[/MATH] and think, "[MATH]6 = 2[/MATH] times what? [MATH]x^5 = x^2[/MATH] times what? [MATH]y^8 = y^1[/MATH] times what?" The answers go in the parentheses, and I'm done.

My problem seems to be understanding what I'm reading. It looks like I've been trying to separate 'y' from the [MATH]{2x^2}{y}[/MATH] and then asking myself why the author had not grouped the terms [MATH]{y}[/MATH] and [MATH]{y^7}[/MATH] in the solution given.

I'm leaning towards the idea that I've got a little understanding what is going on, but I still feel a long way from understanding this subject properly.

I think understanding what I know now I should have just said [MATH]{2x^2}{y}\times[/MATH] something means;

[MATH]{2x^2}{y}\times{3x^3}\times{y^7}[/MATH]

 

[Probability]

I'm still trying to get a good understanding of this subject. Look at this example;

[MATH]{6}{ab^7}{c^2}[/MATH] and [MATH]{9}{a^2}{b^5}[/MATH]
Looking for the highest common factors.

So I look at number 6 and 9 and I know that I can take 3 from each one.

I then ask to look at powers of 'a' and [MATH]{a^2}[/MATH] I'm told that the largest power of 'a' that divides both [MATH]{a}[/MATH] and [MATH]{a^2}[/MATH] is exactly 'a'. I see this as 'a' x 'a' = (a^2)

I'm then ask to consider powers of 'b', The largest power of 'b' that divides both (b^7) and (b^5) exactly is (b^5)

I don't know how that works?

Why not say the largest power of (b) is (b^7)

What am I not being told?

 

[Dr.Peterson]

I then ask to look at powers of 'a' and [MATH]{a^2}[/MATH] I'm told that the largest power of 'a' that divides both [MATH]{a}[/MATH] and [MATH]{a^2}[/MATH] is exactly 'a'. I see this as 'a' x 'a' = (a^2)

I'm then ask to consider powers of 'b', The largest power of 'b' that divides both (b^7) and (b^5) exactly is (b^5)

I don't know how that works?

Why not say the largest power of (b) is (b^7)

It's not "which is the largest power, [MATH]b^5[/MATH] or [MATH]b^7[/MATH]?", but "what is the largest power that evenly divides both [MATH]b^5[/MATH] and [MATH]b^7[/MATH]?" In other words, "what is the largest power that is no more than either [MATH]b^5[/MATH] or [MATH]b^7[/MATH]?" So you're looking for the smaller, not the larger, of the two powers.

It's the same as with numbers. You aren't asked which of 6 and 9 is greater, but what is the largest number that evenly divides both. Since 3 is a divisor of both 6 and 9, and nothing bigger does, that is the greatest common factor. This is a number that is no larger than the smaller of the two numbers.

Remember that the goal is to find a common factor. Is [MATH]b^7[/MATH] a factor of [MATH]b^5[/MATH]? No! Rather, [MATH]b^5[/MATH] is a factor of [MATH]b^7[/MATH], so it is a factor of both.

 

[Probability]

It's not "which is the largest power, [MATH]b^5[/MATH] or [MATH]b^7[/MATH]?", but "what is the largest power that evenly divides both [MATH]b^5[/MATH] and [MATH]b^7[/MATH]?" In other words, "what is the largest power that is no more than either [MATH]b^5[/MATH] or [MATH]b^7[/MATH]?" So you're looking for the smaller, not the larger, of the two powers.

It's the same as with numbers. You aren't asked which of 6 and 9 is greater, but what is the largest number that evenly divides both. Since 3 is a divisor of both 6 and 9, and nothing bigger does, that is the greatest common factor. This is a number that is no larger than the smaller of the two numbers.

Remember that the goal is to find a common factor. Is [MATH]b^7[/MATH] a factor of [MATH]b^5[/MATH]? No! Rather, [MATH]b^5[/MATH] is a factor of [MATH]b^7[/MATH], so it is a factor of both.

That is very well explained and thank you very much for your time and effort put in to write it. I've kept a copy for revision and understanding as I work through this topic.