Problem 33 - Algebra 1: Concepts and Skills

Question

Factor the trinomial. \(4 x^{2}+27 x+35\)

Step-by-Step Solution

Answer

The factored form of the trinomial \(4x^{2}+27x+35\) is \((4x+5)(x+7)\).

Step 1: Identify the coefficients and constant

First, identify the coefficients of the terms. In the trinomial \(4x^{2}+27x+35\), the coefficient of \(x^{2}\) is 4, the coefficient of x is 27, and the constant term is 35.

Step 2: Calculate the product and identify the sum

Calculate the product of the coefficient of \(x^{2}\) (4) and the constant term (35), which equals 140. Then identify that we need two numbers that multiply together to give 140 and add together to give the coefficient of x, which is 27.

Step 3: Find the factors

Find two numbers that satisfy these requirements. These numbers are 5 and 20, since 5 times 20 equals 140 and 5 plus 20 equals 27.

Step 4: Rewrite the x-term using the found factors

Rewrite the trinomial by splitting the middle term using the factors found in the previous step: \(4x^{2}+5x+22x+35\).

Step 5: Factor pairs

Factor by grouping, pair off terms and factor out the common factor from each pair:\(x(4x+5)+7(4x+5)\). Both terms within parentheses are now identical, and can be factored out.

Step 6: Write the final factored form

The result from the grouping and factoring process will leave the final factored form of the trinomial:\((4x+5)(x+7)\).

Key Concepts

Coefficient Identification

In a trinomial, identifying the coefficients is crucial because these numbers dictate how you approach the factoring process. A trinomial of the form \( ax^2 + bx + c \) consists of three key parts:

• Quadratic coefficient (\( a \)): In our exercise, this is 4, associated with \( x^2 \).
• Linear coefficient (\( b \)): This is the number in front of \( x \), which here is 27.
• Constant term (\( c \)): The constant, or the part of the trinomial not tied to a variable, which in this problem is 35.

Recognizing these coefficients enables us to engage in more sophisticated factoring strategies.

Factor Pairing

Once the coefficients are identified, the next step is to find the factor pairs that will help break down the trinomial. To do this, we first calculate the product of the quadratic coefficient and the constant term: \\( 4 \times 35 = 140 \).
Then, the challenge is to find two numbers that multiply to 140 and add to give 27, the linear coefficient.

In this case, the numbers 5 and 20 satisfy both conditions:

• They multiply to 140: \( 5 \times 20 = 140 \)
• They add to 27: \( 5 + 20 = 27 \)

Obtaining the right pair is essential for rewriting the trinomial effectively.

Factoring by Grouping

After identifying factor pairs, we move on to factoring by grouping. This technique involves rewriting the middle term (27x) using the two numbers found in the factor pairing phase. In our trinomial, this becomes:
\( 4x^2 + 5x + 20x + 35 \).
By breaking it down this way, the expanded trinomial can be paired into two groups:

• \( (4x^2 + 5x) \)
• \( (20x + 35) \)

Each group is then factored separately:

• \( x(4x + 5) \) from the first pair
• \( 7(4x + 5) \) from the second pair

With identical expressions \(4x + 5\) arising in both groups, they can be factored out to complete the process.

Trinomial Factoring

Finally, assembling the components gives the completely factored form of the trinomial. In our solution, we take the common term \(4x + 5\) from each grouping and write the final expression. This step consolidates everything into the neat form:

• \((4x + 5)(x + 7)\)

Factoring trinomials can at first seem complex, but by methodically following each step from identifying coefficients, through pairing factors, and grouping them appropriately, the task becomes manageable.
This systematic approach is applicable to a wide range of similar problems and can simplify the learning journey for students.