Since the door is rectangular, its area is the multiplication of its width by its length:
[tex]A=w\cdot l[/tex]We know its area and width:
[tex]x^2+3x-28=(x-4)\cdot l[/tex]Since the area is a polynomial of second degree and the width is a polynomial of first degree, so it is expected that the length is also a first degree polynomial to complete the equation like this:
[tex]x^2+3x-28=(x-4)\cdot(x+b)[/tex]But, to make this, we can use polynomial synthetic division.
We will do the division:
[tex]\frac{x^2+3x-28}{x-4}=l[/tex]To do this, we look first to the denominator. Since it is x - 4, we pick the coefficient, change its sign and put it first:
4 |
|
|
Now, we take the coefficients of the numerator in order (from greater degree to lower) and put them. The coefficients are 1, 3 and -28, so:
4 | 1 3 -28
|
| 1
We also copy the first to the last row. Now, we will multiply the one in the last row by the "4" and put it under the next coefficient. 1 times 4 is 4, so:
4 | 1 3 -28
| 4
| 1
And now we add them and put the result in the last row:
4 | 1 3 -28
| 4
| 1 7
And now, we repeat:
4 | 1 3 -28
| 4 28
| 1 7
4 | 1 3 -28
| 4 28
| 1 7 0
The result is in the last row. the last is the remainder (which is 0, so there are no remainder), and the rest is the polynomial, the quotient:
[tex]\begin{gathered} 1\, \, \, \, \, \, \, \, 7 \\ x+7 \end{gathered}[/tex]So, this means that:
[tex]l=\frac{x^2+3x-28}{x-4}=x+7[/tex]Thus, the length is x + 7.
