What are the possible dimensions of a rectangular room with an area given by the trinomial x² + 3x – 28?

To find the possible dimensions of the rectangle, we need to factor the trinomial x² + 3x – 28. The goal is to express this equation in the form of (x + a)(x + b) = 0, where a and b are constants.

First, we look for two numbers that multiply to -28 (the constant term) and add up to 3 (the coefficient of the linear term). The pair of numbers that satisfies these conditions is 7 and -4 because:

  • 7 * -4 = -28
  • 7 + (-4) = 3

Now we can rewrite the trinomial:

(x + 7)(x - 4) = 0

Next, we set each factor to zero to find the values of x:

x + 7 = 0  =>  x = -7
x - 4 = 0  =>  x = 4

Since dimensions cannot be negative, we discard x = -7. Therefore, the possible value for x in this context is 4. This means one possible dimension of the rectangle is 4 units.

To find the other dimension, we can substitute x back into one of our factor equations. If we take the value of x = 4 and plug it into either of the factor equations:

Length (x + 7) = 4 + 7 = 11

Thus, the dimensions of the rectangle can be expressed as:

  • Length: 11 units
  • Width: 4 units

In conclusion, the dimensions of the rectangular room can be 11 units and 4 units.

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