Factorise x² - 9 ... How?!?

a^2 - b^2 ≡ (a + b)(a - b)

x^2 - 9
= x^2 - 3^2
= (x + 3)(x - 3)

( x - 3 ) ( x + 3 )

Standard difference of two squares: a^2-b^2 = (a-b)(a+b)

You have x^2-3^2 = (x-3)(x+3)



Further Info: If it's a plus sign it will prime over the real numbers and have no real solutions (there are complex ones though:

x^2+9 = (x-3i)(x+3i) where i^2=-1

x² -9
(x+3)(x-3)

Why? When using the FOIL method, you multiply the First 2 factors of each (x times x)
the Outer factors of each (x times -3) the Inner factors, (3 times x), and the Last factors of each (3 times -3)
You come up with x²-3x+3x-9, which is x²-9. Just imagine that the original problem is in ax²+bx+c form, in which you need to find two numbers that when multiplied together =c and when added together = the coefficient b. However, in this problem there is no bx, just a value for c, so essentially the value for b is 0. In factoring, you need to find two numbers that when multiplied equal -9, and when added equal 0 to cancel the bx part out. These two numbers are 3 and -3.

OR

You can take the problem x²-9=0
Add nine to both sides. You get:
x²=9
In order to get rid of the exponent, you find the square root of both sides.
√ x² = √ 9
x= + or - 3 since both positive 3 and negative 3, when multiplied by itself, equal 9.

I like the first way better.
Hope this helps.

Whenever you see anything like x^2 - 9, it's always going to be (the square root of x - the square root of y) times (the square root of x + the square root of y). So it's (x -3) (x + 3). Just remember that if it's a + sign, you can't factor it.

All I did was take the square roots and put them in separate parentheses. The minus sign changes to a plus in one of the parentheses, but it works because a negative multiplied by a positive yields a negative.

(x+3)(x-3)