how to solve this algebra problem?

Question

f(w)=w^2
find [f(w)-f(3)]/w-3
the answer is w+3, but i dont know how to get it.

thank u, except zachary efron lover, if its so easy, y couldnt u do it?

Answer 1

f(w) = w^2 & f(3) = 3^2 = 9

knowing this, we can get:

(w^2 - 9 ) / (w - 3)

= ((w+3)(w-3)) / (w - 3)

= w+3

Answer 2

f(w)=w^2.
f(3)=3^2=9

w^2-9 is a binonial that is derived from the multiplication of two factors both of which must have opposite second terms (for this results in the exclusion of a middle term.)

For example:
(x-1)x(x+1)= x^2 -1
-1 and 1 are the second terms. Because they are opposite a middle term is nonexistent in the reulting polynomial.

Now if they were the same, that is;
(x+1)x(x+1)= x^2+ 2x+1

(x-1)x(x-1)= x^2-2x+1

Resulting in a trinomial.
Now if this doesn't make sense try the FOIL (first, Inside, Outside, Last) method. Inwhich one multiplies the first terms then the inside terms, then the outside terms then the last terms.

You needed this background info. to solve the problem. (x^2-9)/x-3= x+3
This is because (x+3)x(x-3)= x^2-9

Answer 3

First step is to solve for f(3). Plugging 3 into w^2 gives you 9. You also know that f(w) is equal to w^2. So substitute what you now know into the equation. This would give you ((w^2)-9)/w-3. w^2-9 can be factored into (w+3)*(w-3). So now you have (w+3)*(w-3)/ (w-3). The w-3 can be cancelled and you are left with w+3.

Answer 4

f(w) - f(3) = w^2/(w-3) - 3^2/(w-3) =

(w^2 - 3^2)/ (w-3) =

(w + 3)(w-3)/(w-3) =

w +3

Answer 5

Plug it in:

[w^2 - (3)^2]/w - 3
[w^2 - 9]/w - 3 ==> factor w^2 - 9
[(w + 3)(w - 3)]/w - 3 ==> cancel w - 3 terms
w + 3

Answer 6

okay

[f(w) - f(3)]/(w-3)=

(w^2 -3^2)/(w-3)=

Factorize the numerator:

[(w-3)*(w+3)]/(w-3)=

w+3

Answer 7

f(w)-f(3)
is
w^2 - 3^2
If you ever have anything like a^2-b^2, this is (a+b)*(a-b).
So w^2-3^2 is (w+3)*(w-3)
From there the result follows directly.

Answer 8

w+3 its not that hard if u thimk its hard then ur a dumb ***