Problem 45 - Algebra 1: Concepts and Skills

Question

Solve the equation by factoring. $$ 4 x^{2}-21 x+5=0 $$

Step-by-Step Solution

Answer

The solutions to the equation \(4x^{2}-21x+5=0\) by factoring are \(x = \frac{5}{4} \) and \(x = 4\)

Step 1: Find the product of a and c in the equation

In the quadratic equation \(ax^{2} + bx + c = 0\), identify a, b, and c. In this case, a is 4, b is -21, and c is 5. Calculate ac which will be \(4*5 = 20\).

Step 2: Find two numbers adding up to b and multiplying to ac

Find two numbers that add up to -21 and multiply to 20. After factoring, we find that these two numbers are -16 and -5 because they satisfy the condition (-16) + (-5) = -21 and (-16) * (-5) = 20.

Step 3: Rewrite the equation

Rewrite the original equation by breaking down the middle term. It becomes \(4x^{2} - 16x -5x + 5 = 0\).

Step 4: Factor by grouping

Group the terms to factor by grouping. The equation can be rewritten as \(4x(x - 4) -5(x - 4) = 0\) .

Step 5: Solve the equation

Since \(4x-5\) and \(x-4\) are the common factor, they become \( (4x - 5)(x - 4) = 0 \). Setting separate equal to zero gives the possible solutions for x as \(x = \frac{5}{4} \) and \(x = 4\).

Key Concepts

Solving Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). They often have two solutions due to their parabolic nature when graphed. Solving quadratic equations typically involves finding the values of \( x \) that make the equation true. To solve these equations, one can use several methods, such as factoring, completing the square, or using the quadratic formula.

Factoring is a common method when the equation can be rewritten as a product of two binomials. It is efficient and straightforward if the quadratic can be factored easily. Other methods, like the quadratic formula, are more general but may involve longer calculations. Each method has its own advantages and is better suited for different types of quadratic equations. In this article, we focus on solving by factoring, particularly using a technique called "factoring by grouping."

Factoring by Grouping

Factoring by grouping is a method used to factor quadratic expressions, especially when the leading coefficient \( a \) is not 1. This approach helps transform the quadratic into a simpler factorable form. First, you find two numbers that multiply to the product of \( a \) and \( c \) (from \( ax^2 + bx + c \)) and add up to \( b \). Once these numbers are found, the quadratic can be rewritten by splitting the middle term using these two values.

Next, the expression is grouped into pairs. Each pair is factored separately before identifying a common binomial factor from the groups. The quadratic equation \( 4x^2 - 21x + 5 = 0 \) becomes \( 4x^2 - 16x - 5x + 5 = 0 \) initially.

• Group the first pair of terms: \( 4x(x - 4) \).
• Group the second pair of terms: \( -5(x - 4) \).

Both groups now share a common factor \( (x - 4) \), allowing the equation to be rewritten as \((4x - 5)(x - 4) = 0 \). Factoring by grouping combines solving and factoring steps into one, often simplifying the process.

Quadratic Equation Solutions

Once a quadratic equation is successfully factored, finding its solutions involves setting each factor equal to zero. This is based on the zero product property, which states that if a product of two factors is zero, at least one of the factors must be zero.

In our example, \((4x - 5)(x - 4) = 0\) indicates two possible solutions:

• Solving \( 4x - 5 = 0 \):
  - Add 5 to both sides: \( 4x = 5 \)
  - Divide both sides by 4: \( x = \frac{5}{4} \)
• Solving \( x - 4 = 0 \):
  - Add 4 to both sides: \( x = 4 \)

These solutions, \( x = \frac{5}{4} \) and \( x = 4 \), are the points where the parabola crosses the x-axis. Understanding these solutions gives insight into the function's roots. Recognizing the need to solve by finding the zeros is crucial in realizing the full potential of factoring quadratic equations.