Problem 94 - Algebra A Combined Function

Question

Factor each trinomial completely. \(3 x^{2}(a+3)^{3}-28 x(a+3)^{3}+25(a+3)^{3}\)

Step-by-Step Solution

Answer

The fully factored form is \((a+3)^3(3x-25)(x-1)\).

Step 1: Identify the Common Factor

First, observe that each term of the trinomial includes the factor \((a+3)^3\). Factor \((a+3)^3\) out of the expression: \[3x^2(a+3)^3 - 28x(a+3)^3 + 25(a+3)^3 = (a+3)^3(3x^2 - 28x + 25)\]

Step 2: Factor the Quadratic Trinomial

Now we need to focus on factoring the quadratic trinomial within the parentheses: \(3x^2 - 28x + 25\).We need to find two numbers that multiply to \(3 \times 25 = 75\) and add to \(-28\). These numbers are \(-25\) and \(-3\).

Step 3: Rewrite and Group the Terms

Rewrite the quadratic as:\[3x^2 - 3x - 25x + 25\].Next, group the terms:\[(3x^2 - 3x) + (-25x + 25)\].

Step 4: Factor by Grouping

Factor the common factors from each group:- From \(3x^2 - 3x\), factor out \(3x\) to get \(3x(x - 1)\).- From \(-25x + 25\), factor out \(-25\) to get \(-25(x - 1)\).The expression becomes:\[3x(x - 1) - 25(x - 1)\].

Step 5: Factor the Common Binomial

Now, notice that \((x - 1)\) is common in both terms. Factor \((x - 1)\) out:\[(3x - 25)(x - 1)\].Thus, the completely factored form is:\[(a+3)^3(3x-25)(x-1)\].

Key Concepts

Quadratic Trinomials

Quadratic trinomials are algebraic expressions of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(x\) is the variable. These are called trinomials because they consist of three terms. Understanding their structure is essential for factoring them correctly. To factor a quadratic trinomial, our goal is to express it as the product of two binomials.
A key step in the process is to identify two numbers that multiply to \(a \cdot c\) (the product of the leading coefficient and the constant term) and add to \(b\) (the coefficient of the middle term). For example, in \(3x^2 - 28x + 25\), we calculate \(3 \times 25 = 75\) and seek two numbers that multiply to 75 and add up to -28, which are -25 and -3. These numbers guide us in rearranging the trinomial into a form suitable for factoring.
Factoring quadratic trinomials can initially seem daunting, but with practice, it becomes a straightforward process. The key is to systematically look for pairs of numbers that satisfy both multiplication and addition conditions.

Factoring by Grouping

Factoring by grouping is a method used to simplify complex expressions by arranging terms into groups and factoring out common elements. This technique is particularly useful when dealing with quadratic trinomials where traditional methods are challenging to apply.
The strategy involves splitting the middle term based on two numbers identified earlier and organizing the expression into groups. Consider the example from our exercise, where \(3x^2 - 28x + 25\) was rewritten as \(3x^2 - 3x - 25x + 25\). This step sets the stage for grouping:

• First group: \(3x^2 - 3x\)
• Second group: \(-25x + 25\)

Next, factor out the greatest common factor from each group:

• From \(3x^2 - 3x\), factor out \(3x\), giving \(3x(x - 1)\).
• From \(-25x + 25\), factor out \(-25\), yielding \(-25(x - 1)\).

This reveals a common binomial \((x - 1)\) that can be factored further, combining the expression into \((3x - 25)(x - 1)\).
Grouping is a powerful tool that simplifies factoring by isolating parts of an expression where common factors emerge visibly.

Common Factors

Identifying common factors is a fundamental skill in algebra that helps simplify expressions and solve equations. A common factor is a number or expression that divides each term in the expression without leaving a remainder. Discovering these factors is often the first step in factoring trinomials.
For the given exercise, the common factor \((a+3)^3\) was present in every term of the expression \(3x^2(a+3)^3 - 28x(a+3)^3 + 25(a+3)^3\). By factoring out \((a+3)^3\), we simplified the trinomial to \((a+3)^3(3x^2 - 28x + 25)\).
Recognizing common factors can considerably reduce the complexity of an algebraic problem, making subsequent steps, such as applying factoring by grouping, much simpler.

• Look for terms with identical bases and exponents.
• Check for numerical common factors across coefficients.

By getting into the habit of quickly identifying common factors, you'll find that solving algebraic problems becomes far more efficient and less error-prone.