This is giving me a lot of trouble.

Factoring Differences of Squares
Factor Completely
98-8w^2
and can you explain how you got the common factor if there is one please

just recall that a^2-b^2 = (a-b)(a+b)

So, first, make sure that both terms are perfect squares.

In this case, they are not. However, if you factor out a 2, you have

2(49-4w^2)
49 is 7^2 and 4w^2 = (2w)^2. So, we now have

2(7-2w)(7+2w)

To factor the expression 98-8w^2 completely, we need to recognize that it is a difference of squares. In general, a difference of squares is an expression of the form a^2 - b^2, where a and b are terms.

In this case, we can rewrite 98-8w^2 as (7^2) - (sqrt(8)w)^2.

Now, we can see that the expression fits the form a^2 - b^2, where a = 7 and b = sqrt(8)*w.

To factor a^2 - b^2, we can use the formula for the difference of squares: (a - b)(a + b).

Applying this to our expression, we have:
(7 - sqrt(8)w)(7 + sqrt(8)w)

So, the completely factored form of 98-8w^2 is (7 - sqrt(8)w)(7 + sqrt(8)w).