The length of a rectangle is 5 inches longer than twice the width and the area is 12 inches squared. let l represent the length and w represent the width. what equation correctly models the situation?

Length = 8 inch , Width = , This equation models the situation

Step-by-step explanation:

Given as ;

The Area of Rectangle is 12 inches²

The Length of Rectangle is 5 inches longer than twice the width

Let The Length = L inches

      The Width   = W inches

According to question ,

L = 5 + (2 × w )

∵ The Area of Rectangle = Length × width

Or, 12  inches²  = [ 5 + (2 × w ) ] ×  w

Or,  12  inches² = 5 w +2 w²

Or,  2 w² + 5 w  - 12 = 0

Or,   2 w² + 8 w  -3 w - 12 = 0

Or,  2 w (w +4) - 3 (w + 4) = 0

I.e (w + 4) ( 2 w - 3) = 0

So ,  w = -4 , and w =

∴  Length = 5 + (2 × w ) =

   Length = 5 + (2 ×  )  

so ,  L = 5 + 3 = 8

Hence Length = 8 inch , Width =   This equation models the situation Answer

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.