*** Correction in the question: Area = [tex]x^3+12x^2+47x+60[/tex] ***
The correct answer is: Width of rectangle (expression) = w = [tex]x^2+7x+12[/tex]
Explanation:
The area of rectangle is given as follows:
Area.= Width * Length --- (A)
Where,
Area = A = [tex]x^3+12x^2+47x+60[/tex]
Length = l = x + 5
Width = w = ?
Plug in the values in equation (A) and solve for Width (w), as follows:
A = l * w
[tex]x^3+12x^2+47x+60 = (x+5) * w \\ By\thinspace using\thinspace synthetic \thinspace division \thinspace you\thinspace will\thinspace get \thinspace the\thinspace following:\\(x+3)(x+4)(x+5) = (x+5) * w \\ w = (x+3)(x+4) \\ w = x^2+7x+12[/tex]
Therefore, the answer is: width = [tex]x^2+7x+12[/tex]
EXTRA NOTE: I have done synthetic division directly above, but let me explain how I did that.
First you need to guess the value that will yeild the value 0 when you plug in that value in the following equation:
[tex]x^3+12x^2+47x+60[/tex]
As you can see in the above-mentioned expression that every sign is positive; therefore, the value must be negative so that the values can be cancelled out and give us the result 0.
Let us try the value -1 first:
[tex](-1)^3+12(-1)^2+47(-1)+60 \\ -1+12-47+60\\24\neq 0[/tex]
Let us try the value -2:
[tex](-2)^3+12(-2)^2+47(-2)+60 \\ -8+48-94+60\\6\neq 0[/tex]
Let us try the value -3:
[tex](-3)^3+12(-3)^2+47(-3)+60 \\ -27+108-141+60\\0=0[/tex]
Hence, the one root is x = -3.
Now apply the synthetic division.
-3 |1 12 47 60
-3 -27 -60
---------------------
1 9 20 0
The expression will now become the following:
[tex]1x^2+9x+20 \\ x^2 + 4x + 5x + 20 \\ x(x+4)+5(x+4) \\ (x+4)(x+5)[/tex]
And the other one was (x+3) since x = -3 was one root.
Hence [tex]x^3+12x^2+47x+60 = (x+3)(x+4)(x+5)[/tex]
