Problem 24 - Elementary Algebra

Question

For the following problems, the first quantity represents the product and the second quantity a factor. Find the other factor. $$ 14 y^{2}-28 y+14, \quad 14 $$

Step-by-Step Solution

Answer

Answer: The other factor is \(y^2 - 2y + 1\).

Step 1: Identify the given product and factor

The given product is the polynomial expression: \(14y^2 - 28y + 14\), and the known factor is 14.

Step 2: Divide the polynomial expression by the known factor

To find the other factor, divide the polynomial expression by the known factor using polynomial long division or algebraic factorization. In this case, as the given factor is a constant (14), we can simply divide each term of the polynomial expression by 14: $$ \frac{14y^2}{14} - \frac{28y}{14} + \frac{14}{14} $$

Step 3: Perform the division

Divide each term by 14: $$ y^2 - 2y + 1 $$

Step 4: Write down the other factor

The other factor we were looking for is the polynomial expression obtained after the division: \(y^2 - 2y + 1\). Now we have the given product factored as: $$ 14y^2 - 28y + 14 = 14(y^2 - 2y + 1) $$

Key Concepts

Algebraic Division

Algebraic division is a crucial concept for tackling polynomial equations. It involves dividing one polynomial by another. In our exercise, we started by dividing a polynomial by a constant. This is similar to basic division but involves algebraic expressions. Algebraic division helps simplify expressions prominently in mathematics.

When you divide a polynomial by a constant, it simplifies to dividing each term of the polynomial individually. Assume we have the polynomial expression:

• \( 14y^2 - 28y + 14 \)
• and the constant or divisor is 14.

The division process involves taking each term of the polynomial and dividing it by 14. This process results in the expression \( y^2 - 2y + 1 \).This is basically breaking down a complex expression into a simpler one, which is the essence of algebraic division. It untangles the complexity of expressions, helping us solve equations more easily.

Factoring Polynomials

Factoring polynomials is like breaking down a large, complex problem into smaller, manageable parts. This powerful technique is used to find the ingredients, or 'factors,' of a given polynomial.

In our example, we needed to find factors of the polynomial \( 14y^2 - 28y + 14 \). The strategy here was to divide it by a known factor, which was 14. After performing algebraic division, we were left with the other factor: \( y^2 - 2y + 1 \). This means the original polynomial can be expressed as:

• \( 14(y^2 - 2y + 1) \)

Factoring polynomials is valuable because it transforms complex expressions into simpler ones. This is especially useful when solving polynomial equations or simplifying expressions. Once a polynomial is factored, it becomes easier to find roots or solutions, as we can set each factor equal to zero.

Polynomial Expressions

Polynomial expressions are mathematical phrases that can include numbers, variables, and exponents. Each component is called a 'term,' and the expression is formed by adding or subtracting these terms.

For instance, the polynomial expression we worked with is \( 14y^2 - 28y + 14 \). This expression consists of three terms:

• \( 14y^2 \)
• \( -28y \)
• \( 14 \)

Polynomials are foundational in algebra and are used in numerous applications. They help in modeling real-life situations, calculating areas, and understanding mathematical relationships. Understanding polynomial expressions is essential for succeeding in algebraic manipulation, solving equations, and comprehending more advanced mathematical concepts.