Factorisation

**Benjy** · August 9th 2011, 01:11 PM

Factorize: $a^{2}-\left ( b+c \right )^{2}$

I’ve expanded it to see if I can find any solution:

$\left ( b+c \right )^{2}=b^{2}+2bc+c^{2}$

$a^{2}-\left ( b^{2}+2bc+c^{2} \right )$

$a^{2}- b^{2}-2bc-c^{2} \right )$

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?

**Also sprach Zarathustra** · August 9th 2011, 01:16 PM

Originally Posted by Benjy

Factorize: $a^{2}-\left ( b+c \right )^{2}$

I’ve expanded it to see if I can find any solution:

$\left ( b+c \right )^{2}=b^{2}+2bc+c^{2}$

$a^{2}-\left ( b^{2}+2bc+c^{2} \right )$

$a^{2}- b^{2}-2bc-c^{2} \right )$

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?

Answer: $(a-(b+c))(a+(b+c))$

**Plato** · August 9th 2011, 01:17 PM

Originally Posted by Benjy

Factorize: $a^{2}-\left ( b+c \right )^{2}$

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

**Benjy** · August 9th 2011, 01:38 PM

Originally Posted by Also sprach Zarathustra

Answer: $(a-(b+c))(a+(b+c))$

Originally Posted by Plato

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

Thanks! Could you please explain methodically how you should do to find this solution.

**Also sprach Zarathustra** · August 9th 2011, 01:41 PM

Originally Posted by Benjy

Thanks! Could you please explain methodically how you should do to find this solution.

Plato, he's yours.

**Plato** · August 9th 2011, 01:43 PM

Originally Posted by Also sprach Zarathustra

Plato, he's yours.

No I will let you have this one.

**Siron** · August 9th 2011, 01:46 PM

@ Benjy:
Notice:
$a^2-b^2=(a-b)\cdot (a+b)$

**Also sprach Zarathustra** · August 9th 2011, 01:54 PM

Originally Posted by Siron

@ Benjy:
Notice:
$a^2-b^2=(a-b)\cdot (a+b)$

"methodically" $a$ and $b$ are already taken, so you, Siron should say: $x^2-y^2=(x-y)(x+y)$

**Benjy** · August 9th 2011, 02:12 PM

Originally Posted by Siron

@ Benjy:
Notice:
$a^2-b^2=(a-b)\cdot (a+b)$

Originally Posted by Also sprach Zarathustra

"methodically" $a$ and $b$ are already taken, so you, Siron should say: $x^2-y^2=(x-y)(x+y)$

thanks. I'm actually familiar with this one. But the one that I posted was much more complicated I think.

**Siron** · August 9th 2011, 02:21 PM

In case of factorisation it's very important to know the (most important) special products and even more important if you can recognize them.

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