Factorizing a Difficult Polynomial: Using the Difference of Squares Identity

  • Thread starter Thread starter Heidegger
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
Heidegger
Messages
7
Reaction score
0
The problem

Factorize: [itex]a^{2}-\left ( b+c \right )^{2}[/itex]

expanded it to see if I can find any solution:

[itex]\left ( b+c \right )^{2}=b^{2}+2bc+c^{2}[/itex][itex]a^{2}-\left ( b^{2}+2bc+c^{2} \right )[/itex][itex]a^{2}- b^{2}-2bc-c^{2} \right )[/itex]


But I can’t get any further.
What should I do now to simplify it? Please explain and show me a couple of clues or something?
 
Physics news on Phys.org
Hello there.

What do you know about the factorization of a difference of squares?
 
stringy said:
Hello there.

What do you know about the factorization of a difference of squares?

Ok so the solution could easily be found by just applying the distributive law?

I'm going to try.
 
Heidegger said:
Ok so the solution could easily be found by just applying the distributive law?

I'm going to try.
More specifically, use the fact that a2 - b2 = (a + b)(a - b). That's what stringy was getting at.
 
EDIT: Mark44 beat me to it. For future reference, the difference of squares refers to the identity that Mark44 wrote.
 
Mark44 said:
More specifically, use the fact that a2 - b2 = (a + b)(a - b). That's what stringy was getting at.

stringy said:
EDIT: Mark44 beat me to it. For future reference, the difference of squares refers to the identity that Mark44 wrote.

Thank you. I will return later with my answer.