Yes. Can you factor (2p2−3pq+q2) and (q2−pq−2p2)?Is it possible to simplify further?
Maybe I can if only and if these are quadratic expressions.Yes. Can you factor (2p2−3pq+q2) and (q2−pq−2p2)?
They are quadratic expressions, but with 2 variables.Maybe I can if only and if these are quadratic expressions.
To OP.They are quadratic expressions, but with 2 variables.
I'll set up the first one for you. Think about what the ?'s should be.
2p2−3pq+q2=(2p− ?)(p− ?)
That is because I want that cancellation to happen for the expression to be reduced further.Yes, I am back again.
pq2−p2 q−2p32p2q−3pq2+q3=p(q2−pq−2p2)q(2p2−3pq+q2)=p(q−2p)(q+p)q(2p−q)(p−q)=p(−2p+q)(q+p)q(2p−q)(p−q)=−p(2p−q)(q+p)q(2p−q)(p−q)=−p(2p−q)(q+p)q(2p−q)(p−q)=−p(q+p)q(p−q)Thank you.
Reduced later? Are you saying that your last line is NOT your final answer?That is because I want that cancellation to happen for the expression to be reduced further.
How do you mean? I am not seeing it. Maybe you can just show me.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.
I did show you. Just read post #8.How do you mean? I am not seeing it. Maybe you can just show me.
I did show you. Just read post #8.
Please take a look atLooks 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)]
EX: 7-5 = 2 while 5-7 = -2Tell me whether
(2p−q) and (q−2p ) are the same. Please take note of the signs.
Now look at it. If I am to simplify further, say 5−77−5 I will not leave it as 5−77−5 I would rather write it as 5−77−5=−22=−1EX: 7-5 = 2 while 5-7 = -2
I am asking you this because it looks as if you're asking why should the negative sign show in −p(q+p)q(p−q) Remember, I was asked to simplify.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
No, it doesn't.Another way to look at it: Does subtraction have the commutative property?
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.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)]
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?I think Chijioke may be misunderstanding you, and thinking you just want to drop the negative sign, which would change the value.
Do you not see that he does NOT want to drop the negative? He wants to write it as I stated, making TWO changes: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?
That is, p(q+p)q(q−p)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)]
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.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, p(q+p)q(q−p)
This is equivalent to your answer.
Those two expressions are not equivalent. But no one is saying they are!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
p(q+p)q(q−p) is not equivalent to
−p(q+p)q(q−p)Please I stand to be corrected.
Steven said this:Yes, I am back again.
pq2−p2 q−2p32p2q−3pq2+q3=p(q2−pq−2p2)q(2p2−3pq+q2)=p(q−2p)(q+p)q(2p−q)(p−q)=p(−2p+q)(q+p)q(2p−q)(p−q)=−p(2p−q)(q+p)q(2p−q)(p−q)=−p(2p−q)(q+p)q(2p−q)(p−q)=−p(q+p)q(p−q)Thank you.
These are equivalent, because q-p = -(p-q).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)]