factor
(25-x^2)
difference of two squares
a^2-b^2 = (a+b)(a-b)
5^2-x^2 = (5+x)(5-x)
25-x^2=(5-x)*(5+x)
(5-x)*(5+x)=5*5-5*x+x*5-x^2=25-5x+5x-x^2=25-x^2
why is it not -(x-5)(x+5)
wouldnt you have to factor out a negative first so the x^2 isnt negative?
lou,
your answer of -(x-5)(x+5) is the same as the answer of (5-x)(5+x) that Damon and anonymous gave you
if you want to factor out the -1 at the start, you can do that
25 - x^2
= -(x^2 - 25)
= -(x+5)(x-5)
really makes no difference
To factor the expression (25 - x^2), we can use the difference of squares formula. The difference of squares formula states that for any two numbers a and b, the expression a^2 - b^2 can be factored as (a + b)(a - b).
In our case, we can see that 25 is a perfect square because it can be expressed as 5^2. Similarly, x^2 is also a perfect square because it can be expressed as (x)^2.
Now let's rewrite the expression (25 - x^2) as (5^2 - x^2) to apply the difference of squares formula.
(25 - x^2) = (5^2 - x^2)
Now, we can see that a = 5 and b = x in the difference of squares formula. Plugging these values into the formula, we get:
(5 + x)(5 - x)
Therefore, the factored form of the expression (25 - x^2) is (5 + x)(5 - x).