Factorizing a Difficult Polynomial: Using the Difference of Squares Identity

  • Thread starter Heidegger
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In summary, the problem is to factorize the expression a^{2}-\left ( b+c \right )^{2} and the conversation involves discussing clues and techniques to simplify it. The key to solving it is using the difference of squares identity, a^{2}-b^{2}=(a+b)(a-b).
  • #1
Heidegger
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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?
 
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  • #2
Hello there.

What do you know about the factorization of a difference of squares?
 
  • #3
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.
 
  • #4
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.
 
  • #5
EDIT: Mark44 beat me to it. For future reference, the difference of squares refers to the identity that Mark44 wrote.
 
  • #6
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.
 

1. What is "difficult factorisation"?

"Difficult factorisation" refers to the process of finding the unique set of prime numbers that when multiplied together, equal a given number. This process can be challenging for larger numbers that have a large number of possible factors.

2. Why is factorisation important in science?

Factorisation is important in science because it allows us to break down complex numbers or equations into simpler components, making them easier to understand and manipulate. It is commonly used in fields such as physics and cryptography.

3. What are some strategies for tackling difficult factorisation problems?

Some strategies for tackling difficult factorisation problems include using trial and error, finding common factors, using prime factorisation, and using algebraic methods such as grouping and factoring by grouping.

4. How can technology aid in solving difficult factorisation problems?

Technology, such as calculators and computer programs, can help in solving difficult factorisation problems by quickly generating a list of factors and prime factors for a given number. This can save time and effort compared to manual calculations.

5. Are there any real-world applications of difficult factorisation?

Yes, there are many real-world applications of difficult factorisation. Some examples include breaking down large numbers into prime factors for cryptography and finding the least common multiple in scheduling algorithms.

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