Chapter 3: Problem 44
Factor completely each of the polynomials and indicate any that are not factorable using integers. $$n^{2}-26 n+168$$
Short Answer
Expert verified
The polynomial factors to \((n - 14)(n - 12)\).
Step by step solution
01
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.
02
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 \( b = -26 \).
03
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 \(-14 + (-12) = -26\).
04
Rewrite the Middle Term
Express \( -26n \) as \( -14n - 12n \). The expression becomes: \( n^2 - 14n - 12n + 168 \).
05
Group the Terms
Group the terms into two pairs: \( (n^2 - 14n) + (-12n + 168) \).
06
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: \( n(n - 14) - 12(n - 14) \).
07
Complete the Factoring
Notice \( n - 14 \) is a common factor. Factor \( n - 14 \) from the expression: \( (n - 14)(n - 12) \).
08
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
- \( a = 1 \)
- \( b = -26 \)
- \( c = 168 \)
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:
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, \( n - 14 \).
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:
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 \(-12\).
- 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.
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:
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 \((n - 14)(n - 12)\).