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Difficult factorisation

  • Thread starter Heidegger
  • Start date
  • #1
7
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?
 

Answers and Replies

  • #2
90
1
Hello there.

What do you know about the factorization of a difference of squares?
 
  • #3
7
0
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 gonna try.
 
  • #4
33,646
5,313
Ok so the solution could easily be found by just applying the distributive law?

I'm gonna try.
More specifically, use the fact that a2 - b2 = (a + b)(a - b). That's what stringy was getting at.
 
  • #5
90
1
EDIT: Mark44 beat me to it. For future reference, the difference of squares refers to the identity that Mark44 wrote.
 
  • #6
7
0
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.
Thank you. I will return later with my answer.
 

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