We are given that the area of a rectangle is given by the following function:
[tex]A=y^2+4y-5[/tex]
The area of a rectangle is the product of its width by its length:
[tex]A=wl[/tex]
We are given that the width is:
[tex]w=y-1[/tex]
Replacing in the formula for the area we get:
[tex]y^2+4y-5=(y-1)l[/tex]
Since the area is a quadratic equation this means that for the product of the given width by the length to yield a quadratic equation the length must be of the form:
[tex]l=y-b[/tex]
Replacing in the formula for the area:
[tex]y^2+4y-5=(y-1)(y-b)[/tex]
Now we need to determine the value of "b" to do that we will first solve the product on the right side.
[tex]y^2+4y-5=y^2-by-y+b[/tex]
Now we subtract "y squared" from both sides:
[tex]4y-5=-by-y+b[/tex]
Now we associate the terms that are multiplied by "y" on the right side:
[tex]4y-5=(-by-y)+b[/tex]
Now we take common factor on the associated terms;
[tex]4y-5=y(-b-1)+b[/tex]
Now each coefficient for the expression on the left side and the right side must be the same, therefore we have:
[tex]-b-1=4\text{ and -5=b}[/tex]
We get that b = -5. Therefore, the length of the floor must be equal to:
[tex]\begin{gathered} l=y-(-5) \\ l=y+5 \end{gathered}[/tex]
