Problem 26

Question

For the following problems, fill in the missing term. $$ \frac{4 x-4}{x^{2}+2 x-3} $$

Step-by-Step Solution

Verified
Answer
The missing term in the given expression is \((x+3)(x-1)\).
1Step 1: Factor the numerator and the denominator
First, let's factor the numerator: $$ 4x - 4 = 4(x - 1) $$ Now, let's factor the denominator. To do this, we look for two numbers that multiply to -3 and add to 2. Those numbers are 3 and -1. So, the factored denominator will be: $$ x^2 + 2x - 3 = (x + 3)(x - 1) $$
2Step 2: Analyze the missing term
Now we have the expression factored: $$ \frac{4(x-1)}{(x+3)(x-1)} $$ Comparing the given expression to the factored expression, we find the missing term is not the factored form but rather the expanded form of the denominator. So, our original expression: $$ \frac{4 x-4}{x^{2}+2 x-3} $$ Is equivalent to the factored expression: $$ \frac{4(x-1)}{(x+3)(x-1)} $$
3Step 3: State the missing term
The missing term is simply the expanded denominator of the given expression, which we have found to be: \((x+3)(x-1)\).

Key Concepts

Quadratic EquationsRational ExpressionsPolynomial Factoring
Quadratic Equations
Quadratic equations are expressions of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). They are essential in algebra because they model parabolic patterns seen in physics, engineering, and many other fields. Solving a quadratic equation means finding the values of \( x \) that satisfy the equality.

To solve a quadratic equation, there are several methods:
  • Factoring: This method involves expressing the quadratic as a product of two binomials. For example, \( x^2 + 2x - 3 \) can be factored into \( (x+3)(x-1) \), which can directly yield the solutions \( x = -3 \) and \( x = 1 \).
  • Quadratic Formula: The solutions for any quadratic equation \( ax^2 + bx + c = 0 \) can be found using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula gives the roots directly without needing to factor.
  • Completing the Square: This approach modifies the equation to create a perfect square trinomial, making it simpler to solve.
Quadratic equations are foundational in understanding more complex algebraic expressions and become a stepping stone for higher mathematics. Remember, practice is key to mastering these equations.
Rational Expressions
Rational expressions are fractions where both the numerator and the denominator are polynomials. An example is \( \frac{2x + 3}{x - 1} \). Working with these expressions involves understanding how to simplify, multiply, divide, and solve them.

To simplify rational expressions, follow these steps:
  • Factor Completely: Factor both the numerator and the denominator as much as possible.
  • Cancel Common Factors: After factoring, cancel any common factors shared by the numerator and the denominator. This simplifies the expression.
When dealing with operations such as addition, subtraction, multiplication, and division:
  • Addition and Subtraction: Find a common denominator, similarly to adding and subtracting regular fractions.
  • Multiplication: Multiply the numerators together and the denominators together.
  • Division: Multiply by the reciprocal of the divisor.
The key with rational expressions is practice and familiarity with polynomial operations. They can model many real-world situations, making them vital for students to understand.
Polynomial Factoring
Polynomial factoring means expressing a polynomial as a product of its factors. This skill is essential for simplifying expressions, solving polynomial equations, and performing many algebraic manipulations.

When you factor polynomials, you should:
  • Look for Common Factors: Initially, identify and factor out any common elements from all terms.
  • Recognize Patterns: This includes differences of squares, perfect square trinomials, and sum/difference of cubes.
  • Apply the AC Method: For quadratics \( ax^2 + bx + c \), factor by finding two numbers that multiply to \( ac \) and add to \( b \).
  • Use Synthetic Division: For higher-degree polynomials, if you know one root, use synthetic division to reduce the polynomial's degree.
Factoring transforms complex expressions into simpler parts, making problem-solving easier. It's foundational for more advanced problems in calculus and higher algebra, solidifying its importance in mathematical studies.