How do I factor this?

Question

(x-5)^2-4?
Help?

Answer 1

Notice that it's the difference of 2 squares so it factors into
((x-5) - 2)((x-5) + 2) and simplifying
(x-7)(x-3)

Doug

Answer 2

Pay attention to the instructions. It says "factor", not "simplify".

To simplify (x-5)^2 - 4 you multiply (x-5) times itself, first outer inner last, then combine your like terms. But that's not what was asked.

To factor (x-5)^2 - 4 your goal is to come up with two expressions which when multiplied together would be equal to this one. This problem is tricky because it has the (x-5) where you would normally have just a variable. So let's pretend that it IS just a variable, and call it "a". If the question said to factor a^2 - 4, how would you do it? This is a quadratic, which factors into two binomials. It's a rather common type, called the difference of squares. a^2 is a square, and 4 is also a square. You factor it (a+2)(a-2).

You can check this by simplifying using the first-outer-inner-last method and you get (a+2)(a-2) = a^2 - 2a + 2a - 4 and you see the two middle terms cancel each other out so it is in fact equal to a^2 - 4.

Okay, so now we know a^2 - 4 factors to (a+2)(a-2).
Now put "x-5" in place of the "a".

Don't get rid of the parentheses. That's simplifying, which is the OPPOSITE of factoring.

Answer 3

(x - 5)^2 - 4
= (x - 5)^2 - (2)^2
= (x - 5 - 2) (x - 5 + 2)
=(x - 7)(x - 3)

Answer 4

(x-5)^2 - 4 = 0

Remember FOIL - first, outer, inner, last, to expand a squared term:

x^2 - 5x - 5x + 25 - 4 = 0
x^2 - 10x + 21

Now, we can factor the new equation as follows:
(x - 7)(x - 3) = 0
x = 7, 3