Problem 28 - Algebra for College Students

Question

Factor completely each of the polynomials and indicate any that are not factorable using integers. $$4 n^{2}-19 n+21$$

Step-by-Step Solution

Answer

The polynomial \(4n^2 - 19n + 21\) factors to \((n - 3)(4n - 7)\).

Step 1: Identify the Polynomial Structure

The given polynomial is a quadratic in the form \( ax^2 + bx + c \), where \( a = 4 \), \( b = -19 \), and \( c = 21 \). Our task is to factor this expression completely.

Step 2: Search for Integer Factors of ac

Calculate \( ac = 4 \times 21 = 84 \). We need to find two integers whose product is 84 and whose sum is \(-19\).

Step 3: List Integer Pairs

Consider pairs of integers that multiply to 84: \( (1, 84), (2, 42), (3, 28), (4, 21), (6, 14), (7, 12) \). We are looking for such a pair where the sum is \(-19\).

Step 4: Identify Suitable Pair

Check the pair \( (-7, -12) \), as it meets both criteria: the product \((-7) \times (-12) = 84\) and the sum \(-7 + (-12) = -19\).

Step 5: Rewrite the Middle Term

Rewrite the middle term \(-19n\) of the quadratic as \(-7n - 12n\). The expression becomes \(4n^2 - 7n - 12n + 21\).

Step 6: Factor by Grouping

Group the terms: \((4n^2 - 7n) + (-12n + 21)\). Factor each group: \(n(4n - 7) - 3(4n - 7)\).

Step 7: Extract the Common Factor

Notice \((4n - 7)\) is a common factor: \((n - 3)(4n - 7)\). Thus, the factored form of the polynomial is \((n - 3)(4n - 7)\).

Key Concepts

Polynomial Expressions

Polynomial expressions are mathematical phrases made up of variables, coefficients, and exponents. A polynomial involves operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. A basic form of a polynomial expression is \( ax^n + bx^{n-1} + ... + k \), where \( a, b, ... \), and \( k \) are constants.

• Quadratic polynomials are a specific type of polynomial expression that can be written in the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \).
• The degree of a polynomial is the highest power of the variable in the expression. For a quadratic polynomial, the degree is \(2\).
• Quadratic polynomials often need to be factored into simple binomial expressions to find solutions or simplify expressions further.

Understanding polynomial expressions is crucial because they form the basis for many algebraic manipulations and are used in a variety of mathematical contexts.

Integer Factorization

Integer factorization involves writing an integer or a polynomial as a product of smaller, simpler terms. When dealing with polynomials, especially quadratics, the coefficients are often integers, and we aim to express these polynomials as a product of binomials with integer coefficients.

• The integer factorization of a polynomial means breaking it down into factors that multiply to give the original polynomial.
• To start, identify the coefficient \(ac\) from the quadratic polynomial \(ax^2 + bx + c\). Then, search for two integers that multiply to \(ac\) and add up to \(b\).
• This process can be likened to solving a puzzle where you're finding compatible pairs that meet defined criteria.
• Integer factorization simplifies a polynomial, making it easier to manipulate, solve equations, or integrate into more complex math problems.

Successfully factoring polynomials by finding integer factors is a fundamental skill in algebra that eases the way into more advanced topics.

Factor by Grouping

Factor by grouping is a technique used when a polynomial has four or more terms. In terms of a quadratic polynomial, factor by grouping is particularly applied after breaking down the middle term.

• Begin by rewriting the middle term of the quadratic expression, ensuring the sum of the new terms equals the original middle term. In our exercise, \(-19n\) was rewritten as \(-7n - 12n\).
• Arrange the expression into groups of terms. Each group should have a common factor so you can easily extract it.
• The objective here is to factor out the greatest common factor from each group. After grouping, such as \((4n^2 - 7n) + (-12n + 21)\), factor out \(n\) from the first group and \(-3\) from the second, resulting in \(n(4n - 7) - 3(4n - 7)\).
• Notice the common binomial in each group, which allows us to further factor the expression into \((n - 3)(4n - 7)\).
• All terms should multiply back to the original polynomial, verifying your factorization.

Factoring by grouping leverages the distributive property, enabling a structured approach to breaking down complex polynomials into more manageable parts.