(49w^5-28w^3)/7w

can you not compute

49/7 28/7 w^5/w and w^3/w?

Just divide each term by 7 and by w.

Hint: when dividing powers, subtract exponents.

@Steve How do I divide each term? and then subtract? Its Dividing Binomials by Monomials. It says you have to find the quotient.

Can you please help me? I tried to figure it out. Can you please give me the answer so I can put it on my homework.

My answer is 7w^6-4w^4 it was not correct.

(49w^5-28w^3)/7w=

49w^5/7w - 28w^3/7w
= 7w^4 - 4w^2

Why did you add exponents, when I suggested you need to subtract when dividing?

x^a/x^b = x^(a-b)

Thank you so much! Now I get Dividing Binomials by Monomials. Just with understanding how to do one of those math problems I got how to do the others. :)

To simplify the expression (49w^5 - 28w^3)/7w, we can divide each term individually.

First, let's simplify the numerator, 49w^5 - 28w^3. We notice that both terms have a common factor of 7w^3. Factoring out this common factor, we get:

49w^5 - 28w^3
= 7w^3(7w^2 - 4)

Now our expression becomes (7w^3(7w^2 - 4))/7w.

Next, we can simplify further by canceling out the common factor of 7w between the numerator and the denominator:

(7w^3(7w^2 - 4))/7w
= (7w^3 * (7w^2 - 4))/(7 * w)
= (7w^3 * (7w^2 - 4))/(7w)
= 7w^2 - 4

So, the simplified expression is 7w^2 - 4.