Algebraic Simplification of (2p^2q - 3pq^2 + q^3)/(pq^2 - p^2q - 2p^3)

[chijioke]

I need help with this:
Simplify this algebraic expression
[math]\frac{ 2 p ^ { 2 } q - 3 p q^ { 2 } + q ^ { 3 }}{ p q ^ { 2 } - p ^ { 2 } \ q - 2 p ^ { 3 } }[/math]

Solution
[math]\frac{ 2 p ^ { 2 } q - 3 p q^ { 2 } + q ^ { 3 }}{ p q ^ { 2 } - p ^ { 2 } \ q - 2 p ^ { 3 } } = \frac{ q \left( 2 p ^ { 2 } - 3 p q + q^2 \right) }{ p \left( q ^ { 2 } - p q - 2 p^ { 2 } \right) }[/math]​

Is it possible to simplify further?

 

[BigBeachBanana]

Is it possible to simplify further?

Yes. Can you factor [imath] (2p^{2} - 3pq + q^2)[/imath] and [imath]\left( q^{ 2 } - pq - 2p^{ 2 } \right)[/imath]?

 

[chijioke]

Yes. Can you factor [imath] (2p^{2} - 3pq + q^2)[/imath] and [imath]\left( q^{ 2 } - pq - 2p^{ 2 } \right)[/imath]?

Maybe I can if only and if these are quadratic expressions.

 

[BigBeachBanana]

Maybe I can if only and if these are quadratic expressions.

They are quadratic expressions, but with 2 variables.
I'll set up the first one for you. Think about what the ?'s should be.
[math]2p^2 -3pq +q^2 = (2p - ~?)(p-~?)[/math]

 

[Steven G]

2p^2 -3pq +q^2 = 2p^2 -3qp +q^2 which is a quadratic in terms of p.
The a value = 2
The b value = -3p
The c value = q^2

2p^2 -3pq +q^2 = q^2 - 3pq + 2p^2 is also a quadratic in q

 

[Steven G]

They are quadratic expressions, but with 2 variables.
I'll set up the first one for you. Think about what the ?'s should be.
[math]2p^2 -3pq +q^2 = (2p - ~?)(p-~?)[/math]

To OP.
You want ?₁*?₂=q^2. So ?₁ and ?₂ might be equal to????

 

[chijioke]

Yes, I am back again.
[math]\frac{ 2 p ^ { 2 } q - 3 p q^ { 2 } + q ^ { 3 }}{ p q ^ { 2 } - p ^ { 2 } \ q - 2 p ^ { 3 } } = \frac{ q \left( 2 p ^ { 2 } - 3 p q + q^2 \right) }{ p \left( q ^ { 2 } - p q - 2 p^ { 2 } \right) }= \frac{ q \left( 2 p - q \right) \left( p - q\right) }{ p \left( q - 2 p \right) \left( q + p \right) } = \frac{ q \left( 2 p - q \right) \left( p - q \right) }{ p \left( - 2 p + q\right) \left( q + p\right) } = \frac{ q \left( 2 p - q \right) \left( p - q \right) }{ -p \left( 2 p - q\right) \left( { q + p}\right) } = - \frac{ q \left( \cancel{ 2 p - q } \right) \left( p - q \right) }{ p \left( \cancel {2 p - q}\right) \left( q + p\right) } = - \frac{ q \left( p - q \right) }{ p \left( q + p \right) }[/math]Thank you.

 

[Steven G]

Looks good to me.
I do have one comment to make. In mathematics, it is custom for final answers to have the least amount of negative signs. That is your final answer should be q(q-p)/[p(q+p)]

 

[chijioke]

Yes, I am back again.
[math]\frac{ 2 p ^ { 2 } q - 3 p q^ { 2 } + q ^ { 3 }}{ p q ^ { 2 } - p ^ { 2 } \ q - 2 p ^ { 3 } } = \frac{ q \left( 2 p ^ { 2 } - 3 p q + q^2 \right) }{ p \left( q ^ { 2 } - p q - 2 p^ { 2 } \right) }= \frac{ q \left( 2 p - q \right) \left( p - q\right) }{ p \left( q - 2 p \right) \left( q + p \right) } = \frac{ q \left( 2 p - q \right) \left( p - q \right) }{ p \left( - 2 p + q\right) \left( q + p\right) } = \frac{ q \left( 2 p - q \right) \left( p - q \right) }{ -p \left( 2 p - q\right) \left( { q + p}\right) } = - \frac{ q \left( \cancel{ 2 p - q } \right) \left( p - q \right) }{ p \left( \cancel {2 p - q}\right) \left( q + p\right) } = - \frac{ q \left( p - q \right) }{ p \left( q + p \right) }[/math]Thank you.

That is because I want that cancellation to happen for the expression to be reduced further.

 

[Steven G]

That is because I want that cancellation to happen for the expression to be reduced further.

Reduced later? Are you saying that your last line is NOT your final answer?
I never said that you cancelled wrong. Your 'final answer' has two negative signs but can be written with just one.

 

[chijioke]

Reduced later? Are you saying that your last line is NOT your final answer?
I never said that you cancelled wrong. Your 'final answer' has two negative signs but can be written with just one.

How do you mean? I am not seeing it. Maybe you can just show me.

 

[Steven G]

How do you mean? I am not seeing it. Maybe you can just show me.

I did show you. Just read post #8.

 

[chijioke]

I did show you. Just read post #8.

Looks good to me.
I do have one comment to make. In mathematics, it is custom for final answers to have the least amount of negative signs. That is your final answer should be q(q-p)/[p(q+p)]

Please take a look at
[math]\frac{ q \left( 2 p - q \right) \left( p - q \right) }{ p \left( q - 2 p \ \right) \left( q + p \right) }[/math]Tell me whether
[math]\left( 2 p - q \right)[/math] and [math]\left( q - 2 p \ \right)[/math] are the same. Please take note of the signs.

 

[Steven G]

Tell me whether
[math]\left( 2 p - q \right)[/math] and [math]\left( q - 2 p \ \right)[/math] are the same. Please take note of the signs.

EX: 7-5 = 2 while 5-7 = -2

2p is a place holder for a number, say 7 and q is a place holder for a number, say 5.
You tell us if you think 2p-q = q-2p

Another way to look at it: Does subtraction have the commutative property?

 

[chijioke]

EX: 7-5 = 2 while 5-7 = -2

Now look at it. If I am to simplify further, say [math]\frac{7-5}{5-7}[/math] I will not leave it as [math]\frac{7-5}{5-7}[/math] I would rather write it as [math]\frac{7-5}{5-7}=\frac{2}{-2}= -1[/math]

2p is a place holder for a number, say 7 and q is a place holder for a number, say 5.
You tell us if you think 2p-q = q-2p

I am asking you this because it looks as if you're asking why should the negative sign show in [math]- \frac{ q \left( p - q \right) }{ p \left( q + p \right)}[/math] Remember, I was asked to simplify.

Another way to look at it: Does subtraction have the commutative property?

No, it doesn't.

 

[Dr.Peterson]

In mathematics, it is custom for final answers to have the least amount of negative signs. That is your final answer should be q(q-p)/[p(q+p)]

This is simply not true. I have never judged an answer (mine or someone else's) based on whether there is a form with fewer negative signs.

You are making it sound as if it is wrong to leave the expression with two negatives (that is, a negative and a subtraction). No, it is not wrong, or even not customary. Both answers are equally good, and there can be good reasons to choose either over the other. You are just confusing Chijioke with your unnecessary comments.

I myself would probably write [math]- \frac{ q \left( p - q \right) }{ p \left( p + q \right)},[/math] prioritizing a different "custom" (or habit) of writing variables in alphabetical order. It's true that you could reduce the number of negatives by writing [math]\frac{ q \left( q - p \right) }{ p \left( q + p \right)},[/math] which I think is your intention; but to make it sound like there is something wrong with not doing that is bad pedagogy. In particular, I think Chijioke may be misunderstanding you, and thinking you just want to drop the negative sign, which would change the value.

 

[chijioke]

I think Chijioke may be misunderstanding you, and thinking you just want to drop the negative sign, which would change the value.

Exactly! You are just saying my mind. I feel that if the negative sign is dropped, the value would change? Sir or don't you think so?

 

[Dr.Peterson]

Exactly! You are just saying my mind. I feel that if the negative sign is dropped, the value would change? Sir or don't you think so?

Do you not see that he does NOT want to drop the negative? He wants to write it as I stated, making TWO changes:

Looks good to me.
I do have one comment to make. In mathematics, it is custom for final answers to have the least amount of negative signs. That is your final answer should be q(q-p)/[p(q+p)]

That is, [math]\frac{ q \left( q - p \right) }{ p \left( q + p \right)}[/math]
This is equivalent to your answer.

 

[chijioke]

Do you not see that he does NOT want to drop the negative? He wants to write it as I stated, making TWO changes:

That is, [math]\frac{ q \left( q - p \right) }{ p \left( q + p \right)}[/math]
This is equivalent to your answer.

I want to understand something here. Let take for instance; I believe -3 and 3 are not they same. So if they are not the same as I thought; I also believe that -3 is not equivalent to 3.
If that holds, I am also thinking that
[math]\frac{ q \left( q - p \right) }{ p \left( q + p \right)}[/math] is not equivalent to
[math]-\frac{ q \left( q - p \right) }{ p \left( q + p \right)}[/math]Please I stand to be corrected.

 

[Dr.Peterson]

I want to understand something here. Let take for instance; I believe -3 and 3 are not they same. So if they are not the same as I thought; I also believe that -3 is not equivalent to 3.
If that holds, I am also thinking that
[math]\frac{ q \left( q - p \right) }{ p \left( q + p \right)}[/math] is not equivalent to
[math]-\frac{ q \left( q - p \right) }{ p \left( q + p \right)}[/math]Please I stand to be corrected.

Those two expressions are not equivalent. But no one is saying they are!

I don't see that anyone has written the second expression you show here. Read carefully.

You said this:

Yes, I am back again.
[math]\frac{ 2 p ^ { 2 } q - 3 p q^ { 2 } + q ^ { 3 }}{ p q ^ { 2 } - p ^ { 2 } \ q - 2 p ^ { 3 } } = \frac{ q \left( 2 p ^ { 2 } - 3 p q + q^2 \right) }{ p \left( q ^ { 2 } - p q - 2 p^ { 2 } \right) }= \frac{ q \left( 2 p - q \right) \left( p - q\right) }{ p \left( q - 2 p \right) \left( q + p \right) } = \frac{ q \left( 2 p - q \right) \left( p - q \right) }{ p \left( - 2 p + q\right) \left( q + p\right) } = \frac{ q \left( 2 p - q \right) \left( p - q \right) }{ -p \left( 2 p - q\right) \left( { q + p}\right) } = - \frac{ q \left( \cancel{ 2 p - q } \right) \left( p - q \right) }{ p \left( \cancel {2 p - q}\right) \left( q + p\right) } = - \frac{ q \left( p - q \right) }{ p \left( q + p \right) }[/math]Thank you.

Steven said this:

Looks good to me.
I do have one comment to make. In mathematics, it is custom for final answers to have the least amount of negative signs. That is your final answer should be q(q-p)/[p(q+p)]

These are equivalent, because q-p = -(p-q).