The volume of a rectangular prism is $2 x^3+9 x^2-8 x-36$ with height $x+2$. Using synthetic division, what is the area of the base?
A. $2 x^3+13 x^2+18 x$
B. $2 x^3+5 x^2-18 x$
C. $2 x^2+13 x+18$
D. $2 x^2+5 x-18$
Answer (1)
Divide the volume of the rectangular prism by its height to find the area of the base.
Use synthetic division to divide the polynomial 2 x 3 + 9 x 2 − 8 x − 36 by x + 2 .
The result of the synthetic division is the quotient 2 x 2 + 5 x − 18 .
The area of the base is 2 x 2 + 5 x − 18 .
Explanation
Understanding the Problem We are given the volume of a rectangular prism as 2 x 3 + 9 x 2 − 8 x − 36 and its height as x + 2 . We need to find the area of the base. We know that the volume of a rectangular prism is given by the formula V = A × h , where V is the volume, A is the area of the base, and h is the height. Therefore, to find the area of the base, we need to divide the volume by the height: A = h V = x + 2 2 x 3 + 9 x 2 − 8 x − 36 . We can use synthetic division to perform this division.
Setting up Synthetic Division We set up the synthetic division with -2 as the divisor (since x + 2 = 0 implies x = − 2 ) and the coefficients of the polynomial 2 x 3 + 9 x 2 − 8 x − 36 as 2, 9, -8, -36.
Performing Synthetic Division Performing the synthetic division:
| -2 | 2 9 -8 -36
| | -4 -10 36 | ---------------------- | 2 5 -18 0
The result of the synthetic division gives us the coefficients of the quotient, which represents the area of the base. The quotient is 2 x 2 + 5 x − 18 . The remainder is 0, which confirms that x + 2 is a factor of the volume polynomial.
Finding the Area of the Base Therefore, the area of the base is 2 x 2 + 5 x − 18 .
Examples
Understanding the area of the base of a rectangular prism is crucial in various real-world applications. For example, when designing a building, knowing the volume and height allows architects to calculate the base area needed for the foundation. Similarly, in packaging and logistics, determining the base area helps in optimizing space utilization within containers or storage facilities. This calculation ensures efficient use of materials and resources, contributing to cost savings and structural integrity.
