Factoring trinomials worksheet using the box method with examples and practice problems.

To solve the given trinomials using the box method, we need to follow these steps:

1. Identify the product and sum needed for factoring.
2. Split the middle term using the identified product and sum.
3. Use the box method to factor the trinomial.

Let's go through each problem step by step.

---

Problem 1: \( 4x^2 + 5x - 6 \)

#### Step 1: Identify the product and sum
- The product of the first and last terms: \( 4 \times (-6) = -24 \).
- We need two numbers that multiply to \(-24\) and add up to the coefficient of the middle term, \(5\).

The two numbers are \(8\) and \(-3\) because:
\[ 8 \times (-3) = -24 \]
\[ 8 + (-3) = 5 \]

#### Step 2: Split the middle term
Rewrite the trinomial by splitting the middle term:
\[ 4x^2 + 8x - 3x - 6 \]

#### Step 3: Use the box method
Set up the box and fill it in:
\[
\begin{array}{|c|c|}
\hline
4x^2 & 8x \\
\hline
-3x & -6 \\
\hline
\end{array}
\]

Factor out the greatest common factor (GCF) from each row and column:
- Top row: \( 4x^2 + 8x = 4x(x + 2) \)
- Bottom row: \( -3x - 6 = -3(x + 2) \)
- Left column: \( 4x^2 - 3x = x(4x - 3) \)
- Right column: \( 8x - 6 = 2(4x - 3) \)

The factors are:
\[ (4x - 3)(x + 2) \]

#### Final Answer:
\[
\boxed{(4x - 3)(x + 2)}
\]

---

Problem 2: \( 3x^2 + 16x + 5 \)

#### Step 1: Identify the product and sum
- The product of the first and last terms: \( 3 \times 5 = 15 \).
- We need two numbers that multiply to \(15\) and add up to the coefficient of the middle term, \(16\).

The two numbers are \(15\) and \(1\) because:
\[ 15 \times 1 = 15 \]
\[ 15 + 1 = 16 \]

#### Step 2: Split the middle term
Rewrite the trinomial by splitting the middle term:
\[ 3x^2 + 15x + x + 5 \]

#### Step 3: Use the box method
Set up the box and fill it in:
\[
\begin{array}{|c|c|}
\hline
3x^2 & 15x \\
\hline
x & 5 \\
\hline
\end{array}
\]

Factor out the GCF from each row and column:
- Top row: \( 3x^2 + 15x = 3x(x + 5) \)
- Bottom row: \( x + 5 = 1(x + 5) \)
- Left column: \( 3x^2 + x = x(3x + 1) \)
- Right column: \( 15x + 5 = 5(3x + 1) \)

The factors are:
\[ (3x + 1)(x + 5) \]

#### Final Answer:
\[
\boxed{(3x + 1)(x + 5)}
\]

---

Problem 3: \( x^2 + 2x - 48 \)

#### Step 1: Identify the product and sum
- The product of the first and last terms: \( 1 \times (-48) = -48 \).
- We need two numbers that multiply to \(-48\) and add up to the coefficient of the middle term, \(2\).

The two numbers are \(8\) and \(-6\) because:
\[ 8 \times (-6) = -48 \]
\[ 8 + (-6) = 2 \]

#### Step 2: Split the middle term
Rewrite the trinomial by splitting the middle term:
\[ x^2 + 8x - 6x - 48 \]

#### Step 3: Use the box method
Set up the box and fill it in:
\[
\begin{array}{|c|c|}
\hline
x^2 & 8x \\
\hline
-6x & -48 \\
\hline
\end{array}
\]

Factor out the GCF from each row and column:
- Top row: \( x^2 + 8x = x(x + 8) \)
- Bottom row: \( -6x - 48 = -6(x + 8) \)
- Left column: \( x^2 - 6x = x(x - 6) \)
- Right column: \( 8x - 48 = 8(x - 6) \)

The factors are:
\[ (x - 6)(x + 8) \]

#### Final Answer:
\[
\boxed{(x - 6)(x + 8)}
\]

---

Problem 4: \( x^2 + 2x - 24 \)

#### Step 1: Identify the product and sum
- The product of the first and last terms: \( 1 \times (-24) = -24 \).
- We need two numbers that multiply to \(-24\) and add up to the coefficient of the middle term, \(2\).

The two numbers are \(6\) and \(-4\) because:
\[ 6 \times (-4) = -24 \]
\[ 6 + (-4) = 2 \]

#### Step 2: Split the middle term
Rewrite the trinomial by splitting the middle term:
\[ x^2 + 6x - 4x - 24 \]

#### Step 3: Use the box method
Set up the box and fill it in:
\[
\begin{array}{|c|c|}
\hline
x^2 & 6x \\
\hline
-4x & -24 \\
\hline
\end{array}
\]

Factor out the GCF from each row and column:
- Top row: \( x^2 + 6x = x(x + 6) \)
- Bottom row: \( -4x - 24 = -4(x + 6) \)
- Left column: \( x^2 - 4x = x(x - 4) \)
- Right column: \( 6x - 24 = 6(x - 4) \)

The factors are:
\[ (x - 4)(x + 6) \]

#### Final Answer:
\[
\boxed{(x - 4)(x + 6)}
\]

---

Problem 5: \( 2x^2 + 9x + 10 \)

#### Step 1: Identify the product and sum
- The product of the first and last terms: \( 2 \times 10 = 20 \).
- We need two numbers that multiply to \(20\) and add up to the coefficient of the middle term, \(9\).

The two numbers are \(5\) and \(4\) because:
\[ 5 \times 4 = 20 \]
\[ 5 + 4 = 9 \]

#### Step 2: Split the middle term
Rewrite the trinomial by splitting the middle term:
\[ 2x^2 + 5x + 4x + 10 \]

#### Step 3: Use the box method
Set up the box and fill it in:
\[
\begin{array}{|c|c|}
\hline
2x^2 & 5x \\
\hline
4x & 10 \\
\hline
\end{array}
\]

Factor out the GCF from each row and column:
- Top row: \( 2x^2 + 5x = x(2x + 5) \)
- Bottom row: \( 4x + 10 = 2(2x + 5) \)
- Left column: \( 2x^2 + 4x = 2x(x + 2) \)
- Right column: \( 5x + 10 = 5(x + 2) \)

The factors are:
\[ (2x + 5)(x + 2) \]

#### Final Answer:
\[
\boxed{(2x + 5)(x + 2)}
\]

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Summary of Answers

1. \( 4x^2 + 5x - 6 \): \(\boxed{(4x - 3)(x + 2)}\)
2. \( 3x^2 + 16x + 5 \): \(\boxed{(3x + 1)(x + 5)}\)
3. \( x^2 + 2x - 48 \): \(\boxed{(x - 6)(x + 8)}\)
4. \( x^2 + 2x - 24 \): \(\boxed{(x - 4)(x + 6)}\)
5. \( 2x^2 + 9x + 10 \): \(\boxed{(2x + 5)(x + 2)}\)