Problem 44 - Algebra for College Students

Question

Factor completely each of the polynomials and indicate any that are not factorable using integers. $$n^{2}-26 n+168$$

Step-by-Step Solution

Answer

The polynomial factors to \((n - 14)(n - 12)\).

Step 1: Understand the Expression

The given polynomial is a quadratic expression: \( n^2 - 26n + 168 \). Our goal is to factor it into the product of two binomials if possible.

Step 2: Identify the Factoring Method

For a quadratic polynomial \( an^2 + bn + c \), we often use factoring by splitting the middle term. Here, \( a = 1 \), \( b = -26 \), and \( c = 168 \). We need two numbers whose product is \( ac = 168 \) and sum is \( b = -26 \).

Step 3: Find Suitable Numbers

To factor the expression, find two numbers that multiply to 168 and add to -26. These numbers are -14 and -12, since \(-14 \times -12 = 168\) and \(-14 + (-12) = -26\).

Step 4: Rewrite the Middle Term

Express \( -26n \) as \( -14n - 12n \). The expression becomes: \( n^2 - 14n - 12n + 168 \).

Step 5: Group the Terms

Group the terms into two pairs: \( (n^2 - 14n) + (-12n + 168) \).

Step 6: Factor by Grouping

Factor out the common factor from each group. The first group is \( n(n - 14) \), and the second group is \(-12(n - 14)\). So, the expression becomes: \( n(n - 14) - 12(n - 14) \).

Step 7: Complete the Factoring

Notice \( n - 14 \) is a common factor. Factor \( n - 14 \) from the expression: \( (n - 14)(n - 12) \).

Step 8: Verify the Result

Multiply the factors \( (n - 14)(n - 12) \) to ensure it expands back to the original polynomial: \( n^2 - 12n - 14n + 168 = n^2 - 26n + 168 \). The factorization is correct.

Key Concepts

Quadratic Expressions

Quadratic expressions are polynomials of the form \( ax^2 + bx + c \) where \( a, b, \) and \( c \) are constants, and \( x \) represents the variable. In this particular expression, \( n^2 - 26n + 168 \), we have a quadratic equation because it is in the standard form where:

• \( a = 1 \)
• \( b = -26 \)
• \( c = 168 \)

Quadratic expressions are often solved by factoring because it can simplify solving equations that set the expression equal to zero. Understanding how to rewrite these expressions is essential, especially when working with roots. In this format, the highest exponent is 2, which is always true for a quadratic expression.

Quadratic equations often appear in various mathematical contexts, including physics, economics, and engineering, as they can represent trajectories, optimize operations, or solve real-world problems.

Factoring by Grouping

Factoring by grouping is a technique used to simplify polynomials like \( n^2 - 26n + 168 \) by rearranging and combining terms. This skill is crucial when direct factoring isn't evident.

The process generally involves:

• Splitting the middle term into two parts such that their sum equals the original middle term.
• Grouping terms into two pairs: here, we use \( n^2 - 14n \) and \(-12n + 168 \).
• Factoring out the greatest common factor from each group: for instance, \( n(n - 14) \) and \(-12(n - 14) \), respectively.
• Recognizing common factors between the groups, in this case, \( n - 14 \).

By mastering these steps, factoring by grouping becomes a powerful tool for revealing simpler polynomial forms that are much easier to handle or solve.

Factoring Techniques

Factoring techniques are the strategies mathematicians employ to break down polynomials into products of simpler expressions. A combination of methods can be applied depending on the nature of the polynomial.

In the example \( n^2 - 26n + 168 \), we used the combination of mental factoring and grouping:

• Determining that two numbers multiply to \( 168 \) and add to \(-26\): this is the crucial step that guides the correct factor selection.
• Application of basic arithmetic to identify these numbers as \(-14\) and \(-12\).

Additional quadratic factoring techniques may include the following:

• Using the quadratic formula when the expression does not factor neatly.
• Applying difference of squares for expressions like \( a^2 - b^2 \).
• Completing the square, particularly when seeking vertex forms or solving equations.

Each of these methods can be chosen based on the polynomial structure, solving context, or educational goals for practicing various algebraic manipulations.

Integer Factorization

Integer factorization involves breaking down a whole number into a product of smaller whole numbers, called factors. In algebra, it’s slightly adapted to factor polynomials with integer coefficients.

In this context, integer factorization entails:

• Identifying integers that multiply to yield another integer, here \( a \times c = 168 \).
• Ensuring these factors satisfy the equation's structural requirements, like adding to the desired middle term – this guides formulation into specific binomials such as \((n - 14)(n - 12)\).

Integer factorization is particularly important as it defines limitations where algebraic expressions might not neatly factor over the integers; hence you’ll need to verify whether a polynomial can be completely factored using integers or if it remains prime in that domain. This can help in understanding number properties and simplification solutions in advanced mathematics.