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Hi everybody!

I hope you are doing well.

I really need help on these two algebra 1 questions. I've been stuck for a whole day and I would appreciate it if it could be explained..

1. Bradley has 3 square pieces of cardboard with each side equal to x units. For each piece, he does something different to it according to each part below:

Part A: Bradley pasted rectangular strips along two adjacent sides of the cardboard to increase its length and width by y units each. What will be the change in the area of the piece of cardboard? Show your work.

Part B: Bradley cut off rectangular strips from two adjacent sides of the cardboard to decrease its length and width by y units each. What will be the change in the area of the piece of cardboard? Show your work.

Part C: Bradley cut off a strip from one side of the cardboard and pasted the strip on an adjacent side of the cardboard to increase its length by y units and decrease its width by y units. What will be the change in the area of the cardboard? Show your work.

2. Part A: Divide (10x^4y^3 + 5x^3y^2 - 15x^2y - 25x^2y^4) by -5x^2y. Show your work, and justify each step.

Part B: How would your answer in Part A be affected if the x^2 variable in the denominator was just an x?

Part C: What is the degree and classification of the polynomial you got in Part A?

Even if you only have time to answer one of the questions, it would really help.

Thank you so much!

Respuesta :

NobleAlexander
These are some lengthy problems for a middle school student! 
Oh well, let me see what I can do. 

#1:
First off. Draw 3 Squares. 
Each side will have the length/width of X. 

First let's find the area of a normal square.
Because Area is Base * Height we will get this: 
x*x = x^2

Part A: 
Alright! So we have our square, but now they add 2 strips to both sides of this square. They tell us that the length/width has increased by y. Which means that our new side is equivalent to the following expression: 
X + Y

This is due to the fact that we are simply adding a new side. We haven't multiplied our length/width. So, let's examine the new area. 

(x+y)*(x+y) = (x+y)^2

Now we will actually square the expression by using the FOIL method. 
F - First
O -  Outer
I - Inner
L - Last

Multiply the First Terms: 
x*x = x^2

Multiply the Outer Terms: 
x*y = xy

Multiply the Inner Terms: 
y*x = xy (yx is the same as xy)

Multiply the Last Terms: 
y*y = y^2

Add them all together and we have our answer!

Area = x^2 + 2xy + y^2 
(The two xy's add together to form 2xy!) 
We are not done here! Now we must find the Change in the Area. To do so we must subtract the new area and the old area. So! Let's get right to it. (Keep in mind the "old area" is the area of a square which has the length/width of X) 

(New Area) - (Old Area)
(x^2 + 2xy + y^2) - (x^2)

Cancel out the like terms to get a final answer of... : 
Change in Area = 2xy + y^2

Part B: 
Same situation as the last problem! Except in this case we will be subtracting y. So our sides now look like this: 

X - Y

Area = Base*Height
Both are base and height equal x-y. So we just need to multiply them together like before. 

Area = (x - y)^2
Area = (x - y)(x - y)

Multiply the First Terms: 
x*x = x^2

Multiply the Outer Terms:
x*-y = -xy

Multiply the Inner Terms: 
-y*x = -xy

Multiply the Last Terms: 
-y*-y = y^2

Area = x^2 - 2xy + y^2 

Change in Area = x^2 -2xy + y^2 - x^2

Once again the x^2 will cancel out. 

Change in Area = -2xy + y^2

Part C: 
This is where it becomes a bit trickier. Now we have a rectangle instead of a square. 

So we have two different sides. 
I'll note that it doesn't matter if you say the x-y or x+y are the length or the width. As long as they are both represented it is fine. 

Area = L*W

Area = (x - y)(x + y) 

Multiply the First Terms: 
x*x = x^2

Multiply the Outer Terms: 
x*y = xy

Multiply the Inner Terms: 
-y*x = -xy

Multiply the Last Terms: 
-y*y = -y^2

Area = x^2 + xy - xy + y^2
The xy's cancel out. 

Change in Area = x^2 + y^2 - x^2

You can probably guess by now, but the x^2's will cancel out once again!

Change in Area = y^2

#2: Part A: 
(10x^4y^3 + 5x^3y^2 - 15x^2y - 25x^2y^4)/-5x^2y

We need to ask a few questions to get to our answer. 
Is the constant in front of the denominator (-5) divisible by all of the constants in front of all the terms in the numerator (10 + 5 - 15 - 25)? 

What variables are in the denominator? (x and y are) 

Do all of the terms in the numerator have x and y? 

If yes, then what is the lowest power of x? (x^2)

If yes, then what is the lowest power of y? (y)

What we have here is what we need to reduce the denominator by. 
So we reduce all the constants by -5. (Divide by -5)

10/-5 = -2
5/-5 = -1
-15/-5 = 3
-25/-5 = 5

Next up is the variables. 
For these we simply subtract the lowest power from all of them. (In reality we are dividing them by x^2, thus reducing 

x^4/x^2 = x^2
x^3/x^2 = x
x^2/x^2 = 1
x^2/x^2 = 1

Lastly the y terms.
Same deal with the x terms. 

y^3/y = y^2
y^2/y = y
y/y = 1
y^4/y = y^3

Now put them all together!

-2x^2 -xy +3 +5y^3

Part B: 
All the x terms would be increased by 1 due to being reduce one less power. 
So it'll look like this: 
-2x^3 - x^2y + 3 + 5xy^3

Part C: 
The Degree is 3, because the highest power is 3 (y^3). The classification for this is a Cubic Polynomial. 

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