Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$

Question

I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$

I am particularly concerned with the term, $-4$.

Answer 1

It's Quite Simple..

You know, $a^2 - b^2 = (a-b)(a+b)$
Now, $(x-5)^2 - 4 = (x-5)^2-2^2$
Let $a=x-5 $ and $b=2$
So, $a^2-b^2 = (x-5)^2-2^2 = (a-b)(a+b)=(x-5-2)(x-5+2)$

Answer 2

Note that

$$a^2-b^2=a^2-ab+ab-b^2=a(a-b)+b(a-b)=(a+b)(a-b)$$

Now put $a=(x-5)$ and $b=2$

$$(x-5)^2-4=(x-5)^2-2^2=((x-5)+2)((x-5)-2)=(x-5+2)(x-5-2)$$

Answer 3

Here is another way to look at it: Use the FOIL method on the right hand side. $(x-5-2)(x-5+2)$ becomes $x^2-5x+2x-5x+25-10-2x+10-4$ Combine and cancel your like terms and you're left with $x^2-10x+25-4$, which of course can be rewritten as $(x-5)^2-4$.

Answer 4

Here's another way without using fancy formulas $$ (x-5)^2-4 $$ $$=x^2-5x-5x+25-4 $$ $$=x^2-10x+25-4 $$ $$=x^2-10x+21 $$ $$=(x-3)(x-7) $$ $$=(x-5+2)(x-5-2) $$