Algebra 1 Simplify problem. Please help! :)

Question

Simplify: 21y^5-42y^4+49y^3 over 7y^3

Answer 1

You have: (21y⁵-42y⁴+49y³)/7y³.

 

1. Do it one at a time. Start with the constants. You have a bunch of numbers- 21, 42, 49, over 7. Notice anything about those numbers- 21, 42, 49? They're all factors of 7. 

 

21/7 is 3, 42/7 is 6, 49/7 is 7. 

 

Simplify it just by dividing 21/7, 42/7, and 49/7!

 

 

2. Factor out Y - Now you should have (3y⁵-6y⁴+7y³)/y³. 

 

Notice there are "ys" in everything? There's a 3y, a 6y, a 7y- now it's time to take on the y's! "Pull out" as many y's as you can.

 

To do this, look at how many y's you have in the top: y⁵, y⁴,y³ You can only pull out 3 y's, since the term with the smallest amount of y's is y³.

 

Factor out or "pull out" y³ on the top. 

 

3. Finish simplifying You should have y³(3y²-6y+7)/y³

 

Cancel out y³ on the top and y³ on the bottom and you're left with 3y²-6y+7

 

 

= 3y² - 6y +7

 

 

Answer 2

I'm assuming that you mean that entire first part is over that last piece, correct?  

 

If so, you can think of this as 3 separate problems, at least to help you understand.  EVERY term in that numerator needs to be divided by the 7y³

 

21y⁵/7y³ - 42y⁴/7y³ +49y³/7y³

 

Dividing the number parts is straight forward.  21/7 for example is just 3.  To divide the y parts, you can use the exponential subtraction rule.  Remember that y⁵ is really y*y*y*y*y.  Therefore three of those y's will cancel with the y*y*y in the denominator, leaving y*y, or y², which you also get from 5-3.

 

The final answer then is 3y² - 6y + 7