Problem 21 - College Algebra

Question

For the following exercises, factor the polynomial. $$ 12 t^{2}+t-13 $$

Step-by-Step Solution

Answer

The polynomial factors to \((t - 1)(12t + 13)\).

Step 1: Identify the Coefficients

For the quadratic polynomial \(12t^2 + t - 13\), identify the coefficients: \(a = 12\), \(b = 1\), and \(c = -13\).

Step 2: Calculate the Product of a and c

Calculate the product of the coefficients \(a\) and \(c\). This is \(12 \times -13 = -156\).

Step 3: Find Two Numbers that Multiply to ac and Add to b

We need two numbers that multiply to \(-156\) and add to \(1\). These numbers are \(13\) and \(-12\).

Step 4: Rewrite the Middle Term

Rewrite the polynomial using the numbers found: \(12t^2 + 13t - 12t - 13\).

Step 5: Factor by Grouping

Group the terms: \((12t^2 + 13t) + (-12t - 13)\). Factor each group: \(t(12t + 13) - 1(12t + 13)\).

Step 6: Factor Out the Common Binomial

Since \(12t + 13\) is common in both terms, factor it out: \((t - 1)(12t + 13)\).

Key Concepts

Quadratic Polynomial

A quadratic polynomial is an algebraic expression of degree 2. It typically takes the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The highest power of the variable, usually denoted as \( x \) or \( t \), is squared.
Quadratic polynomials can describe various phenomena in algebra, including projectile motion in physics and the path of a parabolic curve in geometry. In the expression \( 12t^2 + t - 13 \), the term \( 12t^2 \) represents the quadratic part, \( t \) represents the linear part, and \(-13 \) is the constant. Understanding how to manipulate and factor these expressions is crucial in solving quadratic equations and simplifying complex algebraic problems.

Polynomial Coefficients

In algebra, the coefficients of a polynomial are the numerical components that multiply the variables. For a quadratic polynomial like \( 12t^2 + t - 13 \), the coefficients are:

• \( a = 12 \), which multiplies the quadratic term \( t^2 \)
• \( b = 1 \), which multiplies the linear term \( t \)
• \( c = -13 \), the constant term

These coefficients play a crucial role in factoring polynomials, as they influence the possible combinations of numbers used in techniques like the product-sum method. Recognizing the coefficients correctly allows you to accurately perform operations such as simplification, solving equations, and graphing.

Factor by Grouping

Factor by grouping is a method used for simplifying and factoring polynomials by grouping terms with common factors. This method involves rearranging the terms in a polynomial to create pairs (or groups) that can be factored easily.
For example, in \( 12t^2 + t - 13 \), first, the expression is rewritten, creating two groups: \((12t^2 + 13t) + (-12t - 13)\). Each group can then be factored separately.

• The first group \( 12t^2 + 13t \) factors to \( t(12t + 13) \)
• The second group \( -12t - 13 \) factors to \(-1(12t + 13) \)

Highlighting the common factor, which in this case is \( 12t + 13 \), allows you to factor the entire expression as \((t - 1)(12t + 13) \). This method is very helpful when handling more complex polynomials, as it breaks them down into simpler, more manageable parts.

Product-Sum Method

The product-sum method is a systematic approach to factoring quadratic polynomials, especially when the leading coefficient, \( a \), is not equal to 1. This method uses the relationship between the coefficients and their products and sums to simplify the expression.
Here's how it works:

• Multiply the leading coefficient \( a \) with the constant term \( c \), giving a product \(-156\) in our example
• Find two numbers that multiply to this product and add up to the linear coefficient \( b = 1 \)
• The identified numbers are used to split the middle term, rewriting \( t \) as \( 13t - 12t \)

Incorporating this step allows the polynomial to be grouped and factored more straightforwardly, as demonstrated in the factor by grouping method above. The product-sum method is essential for tackling polynomials where trial and error alone might not yield results efficiently.