Problem 14 - Algebra and Trigonometry

Question

\(13-20\) . Factor the trinomial. $$ x^{2}-6 x+5 $$

Step-by-Step Solution

Answer

The factored form is \((x-5)(x-1)\).

Step 1: Identify the trinomial structure

To factor the trinomial \(x^2 - 6x + 5\), we first determine its structure. This trinomial is in the form of \(ax^2 + bx + c\) where \(a = 1\), \(b = -6\), and \(c = 5\).

Step 2: Find two numbers that satisfy the conditions

We need to find two numbers whose product is \(c = 5\) and whose sum is \(b = -6\). Consider the factor pairs of 5: \((1, 5)\) and \((-1, -5)\). Only \((-1, -5)\) sums up to -6.

Step 3: Rewrite and factor the trinomial

Use the numbers from Step 2 to rewrite the middle term. The trinomial \(x^2 - 6x + 5\) can be split as \(x^2 - 1x - 5x + 5\). Group and factor by grouping: \((x^2 - 1x) + (-5x + 5)\) becomes \(x(x-1) - 5(x-1)\), which further factors to \((x-5)(x-1)\).

Step 4: Verify the factoring

Expand the factored form \((x-5)(x-1)\) to ensure it matches the original trinomial: \(x^2 - x - 5x + 5 = x^2 - 6x + 5\). Since it matches, the factoring is verified to be correct.

Key Concepts

Trinomial Structure

Understanding the structure of a trinomial is a crucial first step in factorization. A trinomial often appears in the form of \( ax^2 + bx + c \). It has three terms:

• Quadratic Term: This term, \( ax^2 \), includes the variable squared and is dictated by the coefficient \( a \).
• Linear Term: The middle term, \( bx \), contains the variable raised to the first power and the coefficient \( b \).
• Constant Term: A constant \( c \) which has no variable associated with it.

In our example, \( x^2 - 6x + 5 \), the trinomial has \( a = 1 \), \( b = -6 \), and \( c = 5 \). Recognizing that \( a = 1 \) simplifies our task since any number multiplied by 1 remains unchanged. This step helps us identify the key components we need for factoring.

Factoring by Grouping

Factoring by grouping is an approach used when a trinomial is rewritten to make it easier to factor. Here's how you can tackle this:Start by finding two numbers that multiply to the constant term \( c \), while also adding up to the linear coefficient \( b \). In the case of \( x^2 - 6x + 5 \):

• We need two numbers that multiply to 5 and add to -6.
• The pairs \((1, 5)\) and \((-1, -5)\) are considered.
• Only \((-1, -5)\) meets both conditions: the product is 5 and the sum is -6.

With these numbers, rewrite \(-6x\) as \(-1x - 5x\), yielding \(x^2 - 1x - 5x + 5\).Next, group and factor:

• Group terms: \((x^2 - 1x) + (-5x + 5)\).
• Factor within each group: \(x(x - 1) - 5(x - 1)\).
• Strikingly, both groups contain \((x - 1)\), allowing us to factor it out, giving \((x - 1)(x - 5)\).

This strategy of grouping confirms the arrangement, facilitating a smooth factorization process.

Verify Factoring

Verification of your factoring results is an essential step to ensure accuracy. Once you have factored the trinomial, it's important to expand the factored form to check if it matches the original trinomial.Let's verify \((x - 5)(x - 1)\):

• Start by applying the distributive property: Expand \((x - 5)(x - 1)\) as follows:
• First, multiply \(x\) by each term inside the second parenthesis, giving \(x^2 - x\).
• Then, multiply \(-5\) by each term, resulting in \(-5x + 5\).
• Combine the like terms: \(x^2 - x - 5x + 5\).
• Simplify further to arrive at \(x^2 - 6x + 5\).
• This matches the original trinomial exactly, confirming our factoring was done correctly.

The expansion back to the initial expression guarantees that every step of the process was executed properly, providing confidence in the solution.