Problem 117 - Algebra and Trigonometry

Question

Exercises 115–117 will help you prepare for the material covered in the next section. In each exercise, replace the boxed question mark with an integer that results in the given product. Some trial and error may be necessary. $$ (4 x+1)(2 x-[?])=8 x^{2}-10 x-3 $$

Step-by-Step Solution

Answer

The integer that should replace the question mark is 3.

Step 1: Identify the structure

The given expression is of the form \((ax + b)(cx - d) = px^2 + qx + r\). This is the general form of multiplying two binomials to yield a quadratic form. The goal is to find the value of 'd' such that the given binomials multiply to yield \(8x^2 - 10x - 3\).

Step 2: Expand and analyze

Expanding out, we get the quadratic equation as \(8x^2 - 2dx + 2x - d = 8x^2 - 10x - 3\). Upon comparing, we observe that the coefficient of 'x' in the given equation is -10, while in our calculated equation, it is \(-2d + 2\). And also the constant term in the given equation is -3 which should be equal to -d in our calculated equation.

Step 3: Solve through system of equations

A system of two equations can be derived from the comparisons outlined in step 2. The equations are \(-2d + 2 = -10\) and \(-d = -3\). Solving the second equation immediately gives us d = 3. Substituting this into the first equation gives us \(-2*3 + 2 = -10\), which is a correct statement. Therefore, the missing integer is 3.

Key Concepts

Multiplying Binomials

Understanding how to multiply binomials is foundational in algebra. Binomials are expressions that contain two terms, such as \(4x + 1\) and \(2x - d\), where 'd' represents a constant or coefficient to be determined.

When we multiply binomials, we use the distributive property to multiply each term in the first binomial by each term in the second binomial. This process is also known as the FOIL method, representing the 'First', 'Outer', 'Inner', and 'Last' terms of the expressions being multiplied.

Steps To Multiply Binomials:

• Multiply the 'First' terms of each binomial.
• Multiply the 'Outer' terms of each binomial.
• Multiply the 'Inner' terms of each binomial.
• Multiply the 'Last' terms of each binomial.

Combining these products should give us our resultant quadratic expression.

Quadratic Expressions

Quadratic expressions are polynomials of the second degree, usually in the form of \(ax^2 + bx + c\). Here, 'a', 'b', and 'c' are coefficients, with 'a' being non-zero.

After multiplying binomials, we are often left with a quadratic expression. The coefficients of this expression provide key information: 'a' influences the width of the parabola when graphed, 'b' impacts the position of its vertex and direction of the graph, and 'c' represents the y-intercept.

What Can Quadratic Expressions Tell Us?

• \(ax^2\) term determines the opening direction and the width of the parabola.
• \(bx\) term helps in finding the axis of symmetry of the parabola.
• \(c\) gives the point where the parabola crosses the y-axis.

Having a solid grasp of quadratic expressions allows for better understanding and solving of quadratic equations.

Systems of Equations

A system of equations is a set that contains two or more equations with the same variables. The solution to such a system is the values of the variables that satisfy all equations simultaneously. There are multiple methods used to solve systems, including graphing, substitution, and elimination. In the context of factoring quadratic equations, we often deal with systems of equations that arise from setting each factor equal to zero.

When solving for the unknown in a quadratic equation derived from multiplying binomials, we equate the coefficients from both the expanded product and the given quadratic expression. This gives us a system of equations that, when solved, uncovers the unknown constants.

Benefits of Solving Systems of Equations:

• Find the values that satisfy multiple conditions at once.
• Understand the relationship between variables in different equations.
• Apply solutions to practical problems, such as optimization and prediction.

The ability to solve these systems is a powerful tool in algebra and beyond, aiding in the simplification and solution of complex problems.