A. After the mod. One wall is “x” longer than original (horizontal in diagram) and the other is “x” shorter (vertical)
(30+x)(30-x)
b. He original area = 30 x 30 = 900 sq. ft.
The mod. Area is (30+6)(30-6) = 36x24= 764 sq. ft.
So the room was largest to begin with when it was a square
One stratagie you can use is called the big 7
Alright, as we can see the large rectangle's area is 4x² + 9x + 2.
You can also see that gray rectangle's area inside the larger rectangle is
x² + 4x + 3.
So if we want to find the white area in-between the two rectangles we will need to SUBTRACT the two formulas! (Subtract the large rectangle from the smaller one)
To do so we will need to subtract the like terms.
The like terms are 4x² and x², 9x and 4x, 2 and 3!
So first we do 4x² - x² and you get 3x²
Next, we do 9x - 4x and you get 5x
Lastly, we do 2 - 3 and you get -1
Now we put our solutions together and you get...
3x² + 5x - 1
This is the are for the white region of your problem. :D glad to help
Answer:
P(x)=(x-2)(x-4)(x+3)(x+6)
Step-by-step explanation:
Given: P(x)=x⁴+3x³-28x²-36x+144
It is a polynomial with degree 4.
It should maximum four factor.
Hit and trial error method.
Put x = 2 into P(x)
P(2)=2⁴+3×2³-28×2²-36×2+144
P(2) = 0
So, x-2 would be factor of P(x)
Now divide x⁴+3x³-28x²-36x+144 by x-2 to get another factors
Put x = 4
now divide 
by x-4
Now factor 
Complete factor of P(x)
P(x)=(x-2)(x-4)(x+3)(x+6)
