Trig question: Pythagorean Triples. ?

Question

I think I'm having a aproblem with the laws of exponents, but cansomeone explain this to me?

Okay so

a=2rs

b=r^2-s^2

c=r^2+s^2

Fine no prob.

So a^2+b^2= (2rs)^2 + (r^2-s^2)^2

I get that.

Now what I don't understand is what my book does next:

It says: That that last equasion above: (2rs)^2 + (r^2-s^2)^2

=4r^2s^2+r^4-2r^2s^2+s^4

I would have said it =4r^2s^2+r^4+s^4

Where did their 2r^2s^2 come from. It presents it as though this is obvious, but I totally don't follow. Please help.

I also understand that since a^2+b^2=c^2

Then 4r^2s^2+r^4+s^4 = r^2+s^2

But then what? 4r^2s^2+r^4+s^4-(r^2+s^2)=0 ?

What am I not understanding?

Answer 1

(2rs)² + (r² - s²)²

Remember that squaring is the same as multiplying by itself:

The first part is obvious:

(2rs)² = (2rs)(2rs) = 4r²s²

The second part requires that you use FOIL (First, Outer, Inner, Last):

(r² - s²)² = (r² - s²)(r² - s²)

First:

r² * r² = r^4

Outer:

r² * -s² = -r²s²

Inner:

-s² * r² = -r²s²

Last:

-s² * -s² = s^4

Second part:

r^4 - 2r²s² + s^4

All together:

4r²s² + r^4 - 2r²s² + s^4

Then you can continue from there.  Combine the like terms of 4r²s² and -2r²s²:

r^4 + 2r²s² + s^4

Finally, factor this as a perfect square:

(r² + s²)²

If you aren't sure of this step, do the FOIL again to confirm the answer.  Essentially work backwards from c² = (r² + s²)² and multiply it out.

(r² + s²)(r² + s²)

= r^4 + s²r² + s²r² + s^4

= r^4 + 2s²r² + s^4