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Difficulty: MediumOperations on Polynomials

A closed cardboard box has a height of xx inches, a width of 3x−23x - 2 inches, and a length of 2x+52x + 5 inches. When the volume of the box, in cubic inches, is written as a polynomial in standard form, what is the coefficient of the x2x^2 term?

Answer: 11

Answer

The coefficient of the x2x^2 term is 11.
Expanding the volume expression V(x)=x(3x−2)(2x+5)V(x) = x(3x - 2)(2x + 5) yields 6x3+11x2−10x6x^3 + 11x^2 - 10x. The coefficient of the x2x^2 term is the numerical value associated with x2x^2, which is 11.

Step-by-Step Solution

1
Set up the polynomial expression for the volume.
V(x)=x(3x−2)(2x+5)V(x) = x(3x - 2)(2x + 5)
The volume of a rectangular prism is the product of its length, width, and height.
2
Multiply the binomials (3x−2)(3x - 2) and (2x+5)(2x + 5) by distributing terms.
(3x−2)(2x+5)=6x2+15x−4x−10=6x2+11x−10(3x - 2)(2x + 5) = 6x^2 + 15x - 4x - 10 = 6x^2 + 11x - 10
To find the product of two binomials, multiply each term of the first binomial by each term of the second binomial and combine like terms.
3
Distribute the monomial xx to each term in the simplified trinomial.
x(6x2+11x−10)=6x3+11x2−10xx(6x^2 + 11x - 10) = 6x^3 + 11x^2 - 10x
The height xx must scale the entire base area polynomial.
4
Identify the coefficient of the quadratic term x2x^2.
11
The coefficient of a term is the numerical factor multiplied by the variable part.

Key Concept

Multiplying polynomials and identifying coefficients of specific terms in the resulting standard form polynomial.
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Topics

Operations on Polynomials

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