Problem Solving with Factoring Polynomials

Solving Word Problems Involving Factoring Polynomials

The fundamental objective of these exercises is to translate verbal descriptions into algebraic expressions and solve for the variable by factoring the resulting polynomial equations. Each problem requires defining the variable $x$ based on the terms provided, setting up an equation usually in the form $x^2 + bx + c = 0$, and finding the roots of that equation to determine the solution.

Example 1: Consecutive Integers with a Specified Product

In this problem, the objective is to find the value of each integer given that the product of two consecutive integers is $272$. Consecutive integers are defined as integers that follow one another in order, such as $n$ and $n+1$.

First, we define the variables: let $x$ represent the first integer and $x + 1$ represent the second integer. According to the problem, their product is $272$, which gives the equation $x(x + 1) = 272$. By distributing the $x$, we obtain the quadratic expression $x^2 + x = 272$. To solve this by factoring, we must transpose the constant to the left side to set the equation to zero: $x^2 + x - 272 = 0$.

Next, we factor the polynomial $x^2 + x - 272$. We look for two factors of $-272$ that have a sum of $1$ (the coefficient of the middle term). The factors identified are $16$ and $17$. Specifically, $-16 \times 17 = -272$ and $-16 + 17 = 1$. Thus, the factored form of the equation is $(x - 16)(x + 17) = 0$. Setting each factor to zero provides two possible values for $x$:

$\begin{cases} x - 16 = 0 \rightarrow x = 16 \\ x + 17 = 0 \rightarrow x = -17 \end{cases}$

If we take $x = 16$ as the first integer, the second integer is $x + 1 = 17$. If we take $x = -17$ as the first integer, the second integer is $x + 1 = -16$. The product of both pairs ($16 \times 17$ and $-17 \times -16$) results in $272$. For this example, the integers are $16$ and $17$.

Example 2: Consecutive Even Integers and Their Product

This example focuses on finding two consecutive even integers whose product is $528$. Consecutive even integers are separated by an interval of $2$, such as $2, 4, 6$ or $x, x + 2$.

We define the variables as $x$ for the 1st even integer and $x + 2$ for the 2nd even integer. The mathematical representation of their product is $x(x + 2) = 528$. Expanding the left side results in $x^2 + 2x = 528$. Subtracting $528$ from both sides yields the standard quadratic form: $x^2 + 2x - 528 = 0$.

To solve this, we factor the trinomial $x^2 + 2x - 528$. We seek factors of $-528$ that sum to $2$. The factors identified are $24$ and $-22$. Thus, the equation is $(x + 24)(x - 22) = 0$. Solving for $x$ gives:

$\begin{cases} x + 24 = 0 \rightarrow x = -24 \\ x - 22 = 0 \rightarrow x = 22 \end{cases}$

If the first even integer is $22$, the second even integer is $x + 2 = 24$. Similarly, if the first even integer is $-24$, the second is $-22$. The primary solution pair provided is $22$ and $24$.

Example 3: Geometry Application - Floor Area and Classroom Perimeter

This problem involves calculating the perimeter of a classroom based on its floor area and the relationship between its length and width. The floor (area) of the classroom is given as $84\,sq.m$. The width of the classe is described as being $5\,meter\,ess$ than the length.

We begin by defining the dimensions: let the length $L = x$ and the width $W = x - 5$. The formula for the area $A$ of a rectangle is $A = L \times W$. Substituting the given values: $84 = x(x - 5)$. Expanding this results in the equation $84 = x^2 - 5x$. Rearranging into standard form gives $x^2 - 5x - 84 = 0$.

We factor the quadratic expression to find the value of $x$. We need factors of $-84$ that sum to $-5$. The appropriate factors are $-12$ and $7$, as $-12 \times 7 = -84$ and $-12 + 7 = -5$. This gives us $(x - 12)(x + 7) = 0$. Solving for $x$:

$\begin{cases} x - 12 = 0 \rightarrow x = 12 \\ x + 7 = 0 \rightarrow x = -7 \end{cases}$

Since a measurement of length cannot be negative, we reject $x = -7$. Therefore, the length $L = 12\,m$. Using our definition for width, $W = x - 5 = 12 - 5 = 7\,m$.

The final step is to find the perimeter of the classroom. The formula for the perimeter $P$ of a rectangle is $P = 2(L + W)$. Substituting the dimensions:

$P = 2(12 + 7)$

$P = 2(19)$

$P = 38\,m$

The perimeter of the classroom is $38\,meters$.