The area of a rectangle is given by x^2-4x-21. What are the dimensions of the rectangle?
the area of the rectangle is given by the polynomial expression x^2+4x-23 ft.^2 and its length given by x-3 ft. find the width of the rectangle
To find the dimensions of the rectangle given the area, we need to factor the quadratic expression x^2-4x-21.
Step 1: Factor the quadratic expression x^2-4x-21.
The factored form can be written as (x+3)(x-7).
Step 2: Set each factor equal to zero.
x+3=0 or x-7=0
Step 3: Solve for x.
For x+3=0, we subtract 3 from both sides of the equation.
x=-3
For x-7=0, we add 7 to both sides of the equation.
x=7
Step 4: Determine the dimensions.
The dimensions of the rectangle are x+3 and x-7.
Substituting the values of x into the expressions:
- For x=-3, the dimensions are (-3+3) and (-3-7).
So, the dimensions are 0 and -10.
- For x=7, the dimensions are (7+3) and (7-7).
So, the dimensions are 10 and 0.
Therefore, the dimensions of the rectangle can be either 0 and -10 or 10 and 0.
To find the dimensions of a rectangle given the area, we need to factorize the quadratic equation x^2-4x-21 and examine the resulting factors.
The quadratic equation x^2-4x-21 can be factored as follows:
(x + 3)(x - 7)
So, the factors are (x + 3) and (x - 7).
In a rectangle, the length and width are represented by the factors of the area. Therefore, in this case, the dimensions of the rectangle are x + 3 and x - 7.
This question is nonsense.
You can pick any x >7 or <-3, solve for area (if positive), and then pick any length-width combination you want such that L*W = A
The teacher who assigned this seems to have confused algebra factoring with geometry