Problem 91 - Elementary Algebra

Question

The completed problem has one mistake. (a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: Use the \(a c\) method to factor \(6 c^{2}-5 c-4\). Incorrect Answer: \(6 c^{2}-5 c-4\) $$ \begin{aligned} &=6 c^{2}+8 c-3 c-4 \\ &=\left(6 c^{2}+8 c\right)+(-3 c-4) \\ &=2 c(-+\square)+-1(\square+\square) \\ &=2 c(3 c+4)-1(3 c+4) \\ &=(3 c+4)(2 c-1) \end{aligned} $$

Step-by-Step Solution

Answer

The correct factorization is \((2c + 1)(3c - 4)\). The mistake was in incorrectly identifying the terms to split \(-5c\).

Step 1: Identify the given polynomial

The given polynomial is \(6c^2 - 5c - 4\).

Step 2: Identify the mistake

The incorrect step is \(2c(-+\text{square}) + -1(\text{square}+\text{square})\), where the placeholders for terms were not correctly identified.

Step 3: Apply the ac method correctly

The ac method involves finding two numbers that multiply to \(a \times c\) and add up to \(b\). Here, \(a = 6\), \(b = -5\), and \(c = -4\). So, we need numbers that multiply to \(6 \times -4 = -24\) and add up to \(-5\). These numbers are \(3\) and \(-8\).

Step 4: Rewrite the polynomial

Rewrite \(-5c\) as \(3c - 8c\): \(6c^2 + 3c - 8c - 4\).

Step 5: Factor by grouping

Group the terms: \((6c^2 + 3c) + (-8c - 4)\). Factor out the common factors in each group: \(3c(2c + 1) - 4(2c + 1)\).

Step 6: Factor out the common binomial

Factor out the common binomial \((2c + 1)\): \((2c + 1)(3c - 4)\).

Key Concepts

Quadratic Equations

A quadratic equation is a second-order polynomial equation in a single variable. It has the standard form: \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). Quadratic equations have various methods of solution such as factoring, completing the square, and using the quadratic formula. Understanding how to factor quadratic equations is essential because it's directly related to finding the roots of the equation. In the exercise above, the goal was to factor a quadratic polynomial using a specific method called the 'ac method'.

ac Method

The 'ac method' is a useful technique for factoring quadratic polynomials, particularly when the leading coefficient (\(a\)) is not 1. Here’s how it works:

• First, identify the constants \(a\), \(b\), and \(c\) in the quadratic equation \(ax^2 + bx + c\).
• Calculate the product \(a \times c\).
• Find two numbers that multiply to \(a \times c\) and add up to \(b\).

For example, in the exercise \(6c^2 - 5c - 4\), the constants are \(a = 6\), \(b = -5\), and \(c = -4\). The product \(a \times c = 6 \times (-4) = -24\). The two numbers that fit are 3 and -8, because 3 \(\times -8 = -24\) and 3 \(+ -8 = -5\).

Factor by Grouping

After applying the 'ac method' and finding the two correct numbers, the next step is factor by grouping.

1. Rewrite the middle term of the polynomial using the two numbers found from the 'ac method'. For our example: \[ 6c^2 - 5c - 4 = 6c^2 + 3c - 8c - 4 \]
2. Group the terms in pairs: \[ (6c^2 + 3c) + (-8c - 4) \]
3. Factor out the greatest common factor from each pair: \[ 3c(2c + 1) - 4(2c + 1) \]

Note the common binomial factors \((2c + 1)\) in each group. We can then factor these out. This gives us the final factored form: \[ (2c + 1)(3c - 4) \].

Binomials

A binomial is a polynomial with exactly two terms. In the context of factoring, a binomial may appear after grouping terms in a quadratic equation. For instance, in the example \[ 6c^2 - 5c - 4 \], after performing grouping and factoring, we have two binomials: \[3c(2c + 1) - 4(2c + 1)\].Identifying and factoring these common binomials is crucial. The final step simplifies the quadratic equation into a product of two binomials: \[(2c + 1)(3c - 4)\].This form is essential for solving quadratic equations as it reduces the polynomial into simpler factors, facilitating the finding of its roots.