Question

The room shown in the figure has a floor space of \(\left(2 x^{2}-x-3\right)\) square feet. If the width of the room is \((x+1)\) feet, what is the length?

Short answer

The length of the room is \(2x - 3\) feet, valid for \(x > -1\).

Step-by-step solution

Rewriting the formula

Given that the Area, \(A\), of a rectangle is the product of its length and width, we denote the length by \(L\) and express this relationship as \(A = L * W\). In this case \(A = (2x^2 - x - 3)\) and \(W = (x + 1)\). We're trying to solve for \(L\) so we can rearrange the formula: \(L = A/W\). Insert the given values: \(L = (2x^2 - x - 3) / (x + 1)\)

Simplifying the equation

We need to simplify the expression \(L = (2x^2 - x - 3) / (x + 1)\). The numerator (top part) of the fraction can be factored as \((2x^2 - x - 3) = (2x - 3)(x + 1)\). Now the equation looks like this: \(L = (2x - 3)(x + 1)/(x + 1)\).

Cancel out the common factor

In the equation \(L = (2x - 3)(x + 1)/(x + 1)\), we can see that \((x+1)\) appears in both the numerator and the denominator. As such, they can be cancelled by each other: \(L = 2x - 3\). So, the length of the room is \(2x - 3\) feet. However, we should specify that it's valid only for values of \(x\) that do not make the denominator zero, as division by zero is undefined. In this case, our width \(W = x + 1\) is always positive, so as long as \(x > -1\), we're okay.

Key concepts

Factoring Polynomials

Factoring polynomials is a crucial skill in algebra. It involves breaking down a polynomial into simpler components, multiplying to form the original polynomial. In our exercise, the polynomial given is \(2x^2 - x - 3\). This polynomial can be factored into two binomials: \((2x - 3)(x + 1)\).

To factorize, we look for two numbers that multiply to give the product of the leading coefficient (2 in our case) and the constant term (-3), while also combining to give the middle coefficient (-1). By trial, error, or through more systematic approaches like the AC method, these numbers are 2 and -3. Thus, splitting the middle term and grouping appropriately allows for the polynomial to be factored: \((2x - 3)(x + 1)\).

Factoring is particularly useful when simplifying expressions or solving equations, as it can reveal linear factors or other expressions which are easier to work with.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. In the problem, we are dealing with the rational expression \( \frac{2x^2 - x - 3}{x + 1} \).

The main goal when working with rational expressions is to simplify them as much as possible. This often involves factoring the numerator and the denominator and then canceling out any common factors. In our exercise, we factored the numerator \( (2x^2 - x - 3) = (2x - 3)(x + 1)\) and recognized \((x + 1)\) as a common factor in the denominator. Thus, these can be canceled out, simplifying the expression to \(2x - 3\).

It's important to note that when we cancel out terms, the solution is only valid as long as the original denominator is not zero. For \(x + 1 = 0\), the expression becomes undefined; hence, \(x\) must be greater than \(-1\) in this context.

Area of a Rectangle

The area of a rectangle is a basic yet essential concept in geometry. It is calculated by multiplying the length by the width. In the given exercise, the area is represented as a polynomial: \(2x^2 - x - 3\). Our task is to find the length when the width \(x + 1\) is given.

To find the length \(L\), we use the formula \(A = L \cdot W\). Rearranging gives \(L = \frac{A}{W}\). By substituting \(A = 2x^2 - x - 3\) and \(W = x + 1\) into this formula, we obtain our rational expression to simplify.

Upon simplifying, the length turns out to be \(2x - 3\). This clear division helps illustrate the relationship between area, length, and width in polynomial form.

Understanding these concepts of area also helps in practical situations, like determining a room's dimensions based on known parameters.