Problem 108 - Elementary Algebra

Question

(a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistake. (b) Do the problem correctly. Problem: Factor \(6 x^{2}+21 x+15\) completely. $$ \text { Incorrect Answer: } \begin{aligned} & 6 x^{2}+21 x+15 \\ &=3\left(2 x^{2}+7 x+5\right) \end{aligned} $$

Step-by-Step Solution

Answer

Correct polynomial factors into common multiples with proper check for accuracy

Step 1 - Identify and Describe the Mistake

The incorrect answer provided is: \[6x^2 + 21x + 15 = 3(2x^2 + 7x + 5)\]. The mistake is in the factoring. Specifically, the factor of 3 is not correctly applied to the entire trinomial. If 3 is a common factor, it should result in each term inside the parenthesis being correctly divided by 3.

Step 2 - Rewrite the Polynomial

Start by recognizing that the common factor for all terms is 3. This means that the original polynomial can be factored as follows: \[6x^2 + 21x + 15 = 3(2x^2 + 7x + 5)\]

Step 3 - Check Factoring

Verify that \(2x^2 + 7x + 5\) cannot be factored further. To confirm, use the quadratic formula setup or factorization if possible: \[2x^2 + 7x + 5 = 2x^2 + x + 6x + 5 = x(2x + 1) + 5(2x + 1) = (2x + 1)(x + 5)\]. Since this factorization is incorrect, consider simpler terms or double check the factor attempts.

Step 4 - Factor Properly

We should re-examine: \[2x^2 + 7x + 5 •= (2x + 5)(x + 1)\]. This does not yield correct roots as well. Rework: For verification, use factor methods or trial to get appropriate setup.

Key Concepts

Common Factors

Finding common factors is a key first step in factoring polynomials.
Identify the largest number that divides evenly into all the coefficients of the polynomial terms.
Let's consider the polynomial given in the problem: \(6x^2 + 21x + 15\).

We need to find the greatest common factor (GCF) of 6, 21, and 15. The largest number that evenly divides all three is 3.

• 6 divided by 3 = 2
• 21 divided by 3 = 7
• 15 divided by 3 = 5

Therefore, 3 is the common factor. We can factor out this number from the polynomial as follows: \(6x^2 + 21x + 15 = 3(2x^2 + 7x + 5)\). This step simplifies the equation and prepares it for further factoring.

Trinomial Factoring

After factoring out the common factor, the next step is to factor the resulting trinomial: \(2x^2 + 7x + 5\). This is a quadratic trinomial of the form \(ax^2 + bx + c\).

To factor trinomials, we look for two binomials that multiply to give the original trinomial. Since the leading coefficient is not 1, we need to find pairs of factors of each term that add up to the middle coefficient.
Here, we need two numbers that multiply to \(2 \times 5 = 10\) and add to 7.

• The pair 2 and 5 works: \(2 \times 5 = 10\) and \(2 + 5 = 7\).

Thus, we split the middle term using these pairs: \(2x^2 + 2x + 5x + 5\).
Then, we factor by grouping: \(2x(x+1) + 5(x+1)\).
This can be factored as: \( (2x+5)(x+1)\).

Quadratic Equations

Understanding quadratic equations is foundational for trinomial factoring. Quadratic equations are polynomials of degree 2, generally represented as \(ax^2 + bx + c = 0\).

The solutions to these equations are the roots or zeros of the polynomial. You can find these roots using factorization, completing the square, or the quadratic formula: \ \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. However, for factorable trinomials, we prefer factoring since it’s usually quicker and helps in fully understanding polynomials. In our exercise, \(2x^2 + 7x + 5\) was set up for factorization.

• We rewrote it as \(2x^2 + 2x + 5x + 5\)
• Factored by grouping to get \((2x+5)(x+1)\).

Factoring gives us an expression that can be solved by setting each binomial factor equal to zero. Solving \((2x+5)(x+1)=0\) provides you with the roots: \(x = -\frac{5}{2}\) and \(x = -1\).

Checking Factorization

After factoring a polynomial, it's important to check your factorization to ensure it's correct. You can do this by expanding the factors back out and verifying that you obtain the original polynomial.

For the given example, we factored \(6x^2 + 21x + 15\) into \(3(2x + 5)(x + 1)\). Now, let's check:
Expand \( (2x+5)(x+1)\):
\[= 2x(x+1) + 5(x+1)\]
\[= 2x^2 + 2x + 5x + 5\]
\[= 2x^2 + 7x + 5\].
Multiply everything by 3:
\[3(2x^2 + 7x + 5) = 6x^2 + 21x + 15\].

We return to the original polynomial, confirming our factorization is correct. Always make sure to perform this check to avoid mistakes and ensure accuracy.