Problem 64 - A Graphical Approach to College Algebra

Question

Completely factor each polynomial by substitution. $$7(3 k-1)^{2}+26(3 k-1)-8$$

Step-by-Step Solution

Answer

The polynomial factors to \(9(7k - 3)(k + 1)\).

Step 1: Identify the substitution

Notice that the expression has a repetitive factor \((3k-1)\). Let's say \(u = 3k - 1\). This substitution simplifies the polynomial, allowing us to rewrite it in terms of \(u\).

Step 2: Substitute and rewrite

Substitute \(u = 3k - 1\) into the polynomial. The expression becomes:\[7u^2 + 26u - 8\]

Step 3: Factor the quadratic polynomial

Now, we need to factor the quadratic polynomial \(7u^2 + 26u - 8\). To do this, look for two numbers that multiply to \(7 \times (-8) = -56\) and add up to \(26\). These numbers are \(28\) and \(-2\).

Step 4: Decompose the middle term

Using the numbers identified, rewrite the middle term:\[7u^2 + 28u - 2u - 8\]

Step 5: Factor by grouping

Group the terms and factor each group:\[7u(u + 4) - 2(u + 4)\]Notice that \((u + 4)\) is a common factor.

Step 6: Write the factored form

Factor out the common term \((u + 4)\):\[(7u - 2)(u + 4)\]

Step 7: Substitute back to original variable

Replace \(u\) back with \(3k - 1\):\[(7(3k - 1) - 2)((3k - 1) + 4)\]

Step 8: Simplify the expression

Simplify each factor:- For \(7(3k - 1) - 2\), it becomes \(21k - 9\)- For \((3k - 1) + 4\), it becomes \(3k + 3\)The fully factored form of the polynomial is:\[(21k - 9)(3k + 3)\]

Step 9: Factor out any common factor

Find and factor out any common factors from \(21k - 9\) and \(3k + 3\). Each factor has a common factor, \(3\):- From \(21k - 9\), factor out \(3\) to get \(3(7k - 3)\)- From \(3k + 3\), factor out \(3\) to get \(3(k + 1)\)Thus, the polynomial can be written as:\[3(7k - 3) \cdot 3(k + 1)\] which simplifies to \[9(7k - 3)(k + 1)\].

Key Concepts

Substitution Method

The substitution method is a clever algebraic tool that can simplify complex polynomial expressions by reducing them into simpler forms. In the given polynomial problem, there is a repeating expression, \(3k-1\), that makes substitution a helpful strategy. Here's how you use this method:

• Identify a part of the expression that appears multiple times, such as \(3k-1\).
• Introduce a new variable to represent this part; for instance, let \(u = 3k-1\).
• Replace every instance of the original expression with the new variable throughout the polynomial.

This transformation alters the polynomial into a more manageable form that makes further operations, like factoring, easier. Once the simplified polynomial is fully dealt with, remember to substitute back the original expression to get the solution back in the original variable's terms.

Quadratic Polynomial

A quadratic polynomial is a polynomial of degree two, typically having the form \(ax^2 + bx + c\). In the exercise, after using the substitution method, the polynomial reduces to a quadratic equation, specifically \(7u^2 + 26u - 8\).
Let's break down the structural elements of this quadratic:

• The term \(7u^2\) is the quadratic term, with a leading coefficient of 7.
• The middle term, \(26u\), is the linear part, affected during decomposition.
• The constant term is \(-8\), which remains unaffected by variable changes.

Factoring a quadratic polynomial often involves finding two numbers that multiply to the product of the quadratic term's coefficient and the constant term, \-56\ in this case, and add up to the middle term's coefficient, \26\. Once identified, these numbers guide the process of breaking down the polynomial for easier factoring.

Factoring by Grouping

Factoring by grouping is a method employed to factor polynomials of four or more terms. This technique
divides a polynomial into groups that can be factored individually. Our exercise involves factoring the transformed quadratic polynomial \(7u^2 + 26u - 8\).
The steps involve:

• Start by splitting the middle term based on two numbers that fit the sum and product rule identified.
• For example, you break down \(26u\) into \(28u\) and \(-2u\).
• Create two groups: \(7u^2 + 28u\) and \(-2u - 8\).
• Factor each group individually. The common factor in each group here is \(u+4\).

Once this common factor is identified for both groups, it is pulled out, leading to a clearly factored polynomial. This smart arrangement and factor extraction can be profoundly satisfying, revealing the elegance behind polynomial expressions.