How do I factorize this?

[Chikis]

How do I factorize?

[MATH]-1-p^2q^2[/MATH]

I would have used the idea of difference of two squares to handle this but I found it difficult because of the negative sign.

If I factor the -1, I have

[MATH]-1(1+p^2q^2)[/MATH] or [MATH]-1(p^2q^2+1)[/MATH] and that makes the terms inside the bracket unfactorizable?

Edited August 8, 2015 by Chikis

[Acme]

How do I factorize?

[math]-1-p^2q^2[/math]

I would have used the idea of difference of two squares to handle this but I found it difficult because of the negative sign.

If I factor the -1, I have

[math]-1(1+p^2q^2)[/math] or [math]-1(p^2q^2+1)[/math] and that makes the terms inside the bracket unfactorizable?

I'm a little rusty, but can you think of a way to rewrite -1 as a square?

 

[math]i^2-p^2q^2[/math]?

 

Edited August 8, 2015 by Acme

[Chikis]

The [MATH]i^2[/MATH] in the spoiler is an imaginery number and I don't think it is admmisible for this kind of work. I stand to be corrected.

Edited August 8, 2015 by Chikis

[Acme]

The [MATH]i^2 in the spoiler is an imaginery number and I don't think it is admmisible for this kind of work. I stand to be corrected.

Well, it's straightforward algebra. Try working it through. Have you covered imaginary numbers in your course? If so, maybe review your book to decide if it's admissible.

[ajb]

The [MATH]i^2 in the spoiler is an imaginery number and I don't think it is admmisible for this kind of work. I stand to be corrected.

 

You have to decide if you want to factorise this polynomial over the real numbers or complex numbers. I assume p and q are real here, and so in general you will need to extend to the complex numbers, but not to anything more exotic than that to factorise.

[Chikis]

what am saying is this, I know that the expression can be factorized so that it appears in complex number form. I also know that the expression can be factorized so that it appears in real number form.

I am not intrested in making it appear in complex number form. I want to factorize it such that the product appears in real number form. What I need now from you, is a guide to make that happen. Would you mind helping me actualize my determination?

[Acme]

what am saying is this, I know that the expression can be factorized so that it appears in complex number form. I also know that the expression can be factorized so that it appears in real number form.

I am not intrested in making it appear in complex number form. I want to factorize it such that the product appears in real number form. What I need now from you, is a guide to make that happen. Would you mind helping me actualize my determination?

I don't think it factorizes over the Reals; certainly not as the difference of squares. I put the expression to Wolfram Alpha and it gave no such factorization.(It doesn't give the factorization over complex numbers either.) It does give a derivative and an indefinite integral. ??

 

Derivative: (d)/(dp)(-1-p^2 q^2) = -2 p q^2

Indefinite Integral: integral (-1-p^2 q^2) dp = -1/3 p^3 q^2-p+constant

 

Edit: Wolfram Alpha does give complex roots for the expression.

p≠0, q = -i/p & p≠0, q = i/p

Edited August 25, 2015 by Acme

[ajb]

What I need now from you, is a guide to make that happen. Would you mind helping me actualize my determination?

To factorise means to write it as the product of two or more factors that are polynomials. You have a polynomial in two variables, p and q. Such things are almost never going to have a nice factorisation. Over the complex numbers I see that you can write

 

[math]-1 -(pq)^{2} = (i-pq)(i+pq)[/math].

 

Maybe you have some other expression in mind?

 

If so just write down the sort of expression you want and see if you can match the coefficients.

 

So, write

 

[math]-1 -(pq)^{2} = P(p,q)Q(p,q)[/math],

 

for some polynomials P and Q. An educated guess suggests that P and Q should at most be linear in pq. Set this up and match the terms and you will eventually get back to the solution we propose.

 

You could of course be looking for some other expression that is not really a proper factorisation. If you know the expression you are looking for, then show us and we maybe able to find out how to get it.

[Chikis]

Here is the factorization of [MATH]-1-p^2q^2-[/MATH]:

[MATH](1-pq)(pq-1)[/MATH]

[imatfaal]

[latex] (1-pq)(pq-1) [/latex]

forward outer inner last

[latex] pq -1 -p^2q^2 +pq[/latex]

the pq terms do not cancel.

 

One of the most famous non-factorizable (with real numbers) phrases is x2+1 - if it were factorizable then we would be able to find solutions to y=x^2+1 , which is equivalent to finding the roots ie t 0 = x^2+1 , x^2 = -1 .

 

Your expression is similar -1 -p2q2 = -1 * (1+p2q2) = -1 * (1+pq.pq) and if yu think abut it if pq has a solution then we can replace pq with x and there would be as solution to this simplified equation

[Chikis]

It cancels. See it here.

[MATH](1-pq)(pq-1)[/MATH]

= [MATH]1(pq)+(1)(-)-(pq)(pq)-(pq)(-1)[/MATH]. This gives

[MATH]pq-1-p^2q^2+pq[/MATH]

Study it carefully and you will see that the pq terms cancel to give the original expression.

Edited February 20, 2016 by Chikis

[imatfaal]

It cancels. See it here.

[MATH](1-pq)(pq-1)[/MATH]

= [MATH]1(pq)+(1)(-)-(pq)(pq)-(pq)(-1)[/MATH]

Study it carefully and you will see that the pq terms cancels to give the original expression.

 

do the next line and sort out the signs and you will see that it does not

 

[MATH]1(pq)+(1)(-1)-(pq)(pq)-(pq)(-1) = pq -1 -p^2q^2 + pq[/MATH]

 

your two pq terms are +1*(+pq) and -1*(-pq) they both equal +pq

[Chikis]

Yes, it is not factorizable. I did not notice the signs at first. I will rather have [MATH]2pq-1-p^2q^2[/MATH] which is not the original expression. But can the quadratic formular be used to solve this problem?

[imatfaal]

Yes, it is not factorizable. I did not notice the signs at first. I will rather have [MATH]2pq-1-p^2q^2[/MATH] which is not the original expression. But can the quadratic formular be used to solve this problem?

 

Not for Real number answers. The Imaginary answers have been given further up the thread.

[Chikis]

Ok! thank you.