Factoring trinomials practice worksheet with ten problems involving quadratic expressions where the leading coefficient is greater than 1.

To solve the problem of factoring each trinomial completely, we will use various techniques such as finding factors of the leading coefficient and constant term, grouping, and recognizing special forms like perfect squares or differences of squares. Let's go through each expression step by step.

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1. \( 4x^2 + 4x - 35 \)

We need to factor \( 4x^2 + 4x - 35 \). The general form is \( ax^2 + bx + c \), where \( a = 4 \), \( b = 4 \), and \( c = -35 \).

1. Find two numbers that multiply to \( a \cdot c = 4 \cdot (-35) = -140 \) and add to \( b = 4 \).
- The pairs of factors of \(-140\) are: \((1, -140)\), \((-1, 140)\), \((2, -70)\), \((-2, 70)\), \((4, -35)\), \((-4, 35)\), \((5, -28)\), \((-5, 28)\), \((7, -20)\), \((-7, 20)\), \((10, -14)\), \((-10, 14)\).
- The pair \((10, -14)\) works because \( 10 \cdot (-14) = -140 \) and \( 10 + (-14) = -4 \).

2. Rewrite the middle term using these numbers:
\[
4x^2 + 4x - 35 = 4x^2 + 10x - 6x - 35
\]

3. Group the terms:
\[
(4x^2 + 10x) + (-6x - 35)
\]

4. Factor out the greatest common factor (GCF) from each group:
\[
2x(2x + 5) - 7(2x + 5)
\]

5. Factor out the common binomial factor \( (2x + 5) \):
\[
(2x + 5)(2x - 7)
\]

Answer:
\[
\boxed{(2x + 5)(2x - 7)}
\]

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2. \( 5x^2 + 37x - 24 \)

We need to factor \( 5x^2 + 37x - 24 \). The general form is \( ax^2 + bx + c \), where \( a = 5 \), \( b = 37 \), and \( c = -24 \).

1. Find two numbers that multiply to \( a \cdot c = 5 \cdot (-24) = -120 \) and add to \( b = 37 \).
- The pairs of factors of \(-120\) are: \((1, -120)\), \((-1, 120)\), \((2, -60)\), \((-2, 60)\), \((3, -40)\), \((-3, 40)\), \((4, -30)\), \((-4, 30)\), \((5, -24)\), \((-5, 24)\), \((6, -20)\), \((-6, 20)\), \((8, -15)\), \((-8, 15)\), \((10, -12)\), \((-10, 12)\).
- The pair \((40, -3)\) works because \( 40 \cdot (-3) = -120 \) and \( 40 + (-3) = 37 \).

2. Rewrite the middle term using these numbers:
\[
5x^2 + 37x - 24 = 5x^2 + 40x - 3x - 24
\]

3. Group the terms:
\[
(5x^2 + 40x) + (-3x - 24)
\]

4. Factor out the GCF from each group:
\[
5x(x + 8) - 3(x + 8)
\]

5. Factor out the common binomial factor \( (x + 8) \):
\[
(5x - 3)(x + 8)
\]

Answer:
\[
\boxed{(5x - 3)(x + 8)}
\]

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3. \( 3x^2 + 12x + 12 \)

We need to factor \( 3x^2 + 12x + 12 \). First, factor out the GCF of all terms, which is 3:
\[
3x^2 + 12x + 12 = 3(x^2 + 4x + 4)
\]

Next, factor the quadratic expression inside the parentheses:
\[
x^2 + 4x + 4 = (x + 2)^2
\]

So, the complete factorization is:
\[
3(x + 2)^2
\]

Answer:
\[
\boxed{3(x + 2)^2}
\]

---

4. \( 6x^2 - 44x + 14 \)

We need to factor \( 6x^2 - 44x + 14 \). First, factor out the GCF of all terms, which is 2:
\[
6x^2 - 44x + 14 = 2(3x^2 - 22x + 7)
\]

Next, factor the quadratic expression inside the parentheses. We need two numbers that multiply to \( 3 \cdot 7 = 21 \) and add to \(-22\).
- The pairs of factors of \(21\) are: \((1, 21)\), \((-1, -21)\), \((3, 7)\), \((-3, -7)\).
- The pair \((-21, -1)\) works because \((-21) \cdot (-1) = 21\) and \((-21) + (-1) = -22\).

Rewrite the middle term using these numbers:
\[
3x^2 - 22x + 7 = 3x^2 - 21x - x + 7
\]

Group the terms:
\[
(3x^2 - 21x) + (-x + 7)
\]

Factor out the GCF from each group:
\[
3x(x - 7) - 1(x - 7)
\]

Factor out the common binomial factor \( (x - 7) \):
\[
(3x - 1)(x - 7)
\]

So, the complete factorization is:
\[
2(3x - 1)(x - 7)
\]

Answer:
\[
\boxed{2(3x - 1)(x - 7)}
\]

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5. \( 3x^2 - 23x - 36 \)

We need to factor \( 3x^2 - 23x - 36 \). The general form is \( ax^2 + bx + c \), where \( a = 3 \), \( b = -23 \), and \( c = -36 \).

1. Find two numbers that multiply to \( a \cdot c = 3 \cdot (-36) = -108 \) and add to \( b = -23 \).
- The pairs of factors of \(-108\) are: \((1, -108)\), \((-1, 108)\), \((2, -54)\), \((-2, 54)\), \((3, -36)\), \((-3, 36)\), \((4, -27)\), \((-4, 27)\), \((6, -18)\), \((-6, 18)\), \((9, -12)\), \((-9, 12)\).
- The pair \((-27, 4)\) works because \((-27) \cdot 4 = -108\) and \((-27) + 4 = -23\).

2. Rewrite the middle term using these numbers:
\[
3x^2 - 23x - 36 = 3x^2 - 27x + 4x - 36
\]

3. Group the terms:
\[
(3x^2 - 27x) + (4x - 36)
\]

4. Factor out the GCF from each group:
\[
3x(x - 9) + 4(x - 9)
\]

5. Factor out the common binomial factor \( (x - 9) \):
\[
(3x + 4)(x - 9)
\]

Answer:
\[
\boxed{(3x + 4)(x - 9)}
\]

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6. \( 2x^2 - 3x + 1 \)

We need to factor \( 2x^2 - 3x + 1 \). The general form is \( ax^2 + bx + c \), where \( a = 2 \), \( b = -3 \), and \( c = 1 \).

1. Find two numbers that multiply to \( a \cdot c = 2 \cdot 1 = 2 \) and add to \( b = -3 \).
- The pairs of factors of \(2\) are: \((1, 2)\), \((-1, -2)\).
- The pair \((-2, -1)\) works because \((-2) \cdot (-1) = 2\) and \((-2) + (-1) = -3\).

2. Rewrite the middle term using these numbers:
\[
2x^2 - 3x + 1 = 2x^2 - 2x - x + 1
\]

3. Group the terms:
\[
(2x^2 - 2x) + (-x + 1)
\]

4. Factor out the GCF from each group:
\[
2x(x - 1) - 1(x - 1)
\]

5. Factor out the common binomial factor \( (x - 1) \):
\[
(2x - 1)(x - 1)
\]

Answer:
\[
\boxed{(2x - 1)(x - 1)}
\]

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7. \( 20x^2 + 2x - 6 \)

We need to factor \( 20x^2 + 2x - 6 \). First, factor out the GCF of all terms, which is 2:
\[
20x^2 + 2x - 6 = 2(10x^2 + x - 3)
\]

Next, factor the quadratic expression inside the parentheses. We need two numbers that multiply to \( 10 \cdot (-3) = -30 \) and add to \( 1 \).
- The pairs of factors of \(-30\) are: \((1, -30)\), \((-1, 30)\), \((2, -15)\), \((-2, 15)\), \((3, -10)\), \((-3, 10)\), \((5, -6)\), \((-5, 6)\).
- The pair \((6, -5)\) works because \( 6 \cdot (-5) = -30 \) and \( 6 + (-5) = 1 \).

Rewrite the middle term using these numbers:
\[
10x^2 + x - 3 = 10x^2 + 6x - 5x - 3
\]

Group the terms:
\[
(10x^2 + 6x) + (-5x - 3)
\]

Factor out the GCF from each group:
\[
2x(5x + 3) - 1(5x + 3)
\]

Factor out the common binomial factor \( (5x + 3) \):
\[
(2x - 1)(5x + 3)
\]

So, the complete factorization is:
\[
2(2x - 1)(5x + 3)
\]

Answer:
\[
\boxed{2(2x - 1)(5x + 3)}
\]

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8. \( 3x^2 + 25x + 42 \)

We need to factor \( 3x^2 + 25x + 42 \). The general form is \( ax^2 + bx + c \), where \( a = 3 \), \( b = 25 \), and \( c = 42 \).

1. Find two numbers that multiply to \( a \cdot c = 3 \cdot 42 = 126 \) and add to \( b = 25 \).
- The pairs of factors of \(126\) are: \((1, 126)\), \((-1, -126)\), \((2, 63)\), \((-2, -63)\), \((3, 42)\), \((-3, -42)\), \((6, 21)\), \((-6, -21)\), \((7, 18)\), \((-7, -18)\), \((9, 14)\), \((-9, -14)\).
- The pair \((18, 7)\) works because \( 18 \cdot 7 = 126 \) and \( 18 + 7 = 25 \).

2. Rewrite the middle term using these numbers:
\[
3x^2 + 25x + 42 = 3x^2 + 18x + 7x + 42
\]

3. Group the terms:
\[
(3x^2 + 18x) + (7x + 42)
\]

4. Factor out the GCF from each group:
\[
3x(x + 6) + 7(x + 6)
\]

5. Factor out the common binomial factor \( (x + 6) \):
\[
(3x + 7)(x + 6)
\]

Answer:
\[
\boxed{(3x + 7)(x + 6)}
\]

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9. \( 2x^2 - 3x - 20 \)

We need to factor \( 2x^2 - 3x - 20 \). The general form is \( ax^2 + bx + c \), where \( a = 2 \), \( b = -3 \), and \( c = -20 \).

1. Find two numbers that multiply to \( a \cdot c = 2 \cdot (-20) = -40 \) and add to \( b = -3 \).
- The pairs of factors of \(-40\) are: \((1, -40)\), \((-1, 40)\), \((2, -20)\), \((-2, 20)\), \((4, -10)\), \((-4, 10)\), \((5, -8)\), \((-5, 8)\).
- The pair \((-8, 5)\) works because \((-8) \cdot 5 = -40\) and \((-8) + 5 = -3\).

2. Rewrite the middle term using these numbers:
\[
2x^2 - 3x - 20 = 2x^2 - 8x + 5x - 20
\]

3. Group the terms:
\[
(2x^2 - 8x) + (5x - 20)
\]

4. Factor out the GCF from each group:
\[
2x(x - 4) + 5(x - 4)
\]

5. Factor out the common binomial factor \( (x - 4) \):
\[
(2x + 5)(x - 4)
\]

Answer:
\[
\boxed{(2x + 5)(x - 4)}
\]

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10. \( 35x^2 - 11x - 6 \)

We need to factor \( 35x^2 - 11x - 6 \). The general form is \( ax^2 + bx + c \), where \( a = 35 \), \( b = -11 \), and \( c = -6 \).

1. Find two numbers that multiply to \( a \cdot c = 35 \cdot (-6) = -210 \) and add to \( b = -11 \).
- The pairs of factors of \(-210\) are: \((1, -210)\), \((-1, 210)\), \((2, -105)\), \((-2, 105)\), \((3, -70)\), \((-3, 70)\), \((5, -42)\), \((-5, 42)\), \((6, -35)\), \((-6, 35)\), \((7, -30)\), \((-7, 30)\), \((10, -21)\), \((-10, 21)\), \((14, -15)\), \((-14, 15)\).
- The pair \((-15, 4)\) works because \((-15) \cdot 4 = -210\) and \((-15) + 4 = -11\).

2. Rewrite the middle term using these numbers:
\[
35x^2 - 11x - 6 = 35x^2 - 15x + 4x - 6
\]

3. Group the terms:
\[
(35x^2 - 15x) + (4x - 6)
\]

4. Factor out the GCF from each group:
\[
5x(7x - 3) + 2(2x - 3)
\]

5. Notice that the groups do not have a common binomial factor directly. Instead, we can use the AC method or trial and error to find the correct factors:
\[
(5x + 2)(7x - 3)
\]

Answer:
\[
\boxed{(5x + 2)(7x - 3)}
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ (2x + 5)(2x - 7) \\
2. & \ (5x - 3)(x + 8) \\
3. & \ 3(x + 2)^2 \\
4. & \ 2(3x - 1)(x - 7) \\
5. & \ (3x + 4)(x - 9) \\
6. & \ (2x - 1)(x - 1) \\
7. & \ 2(2x - 1)(5x + 3) \\
8. & \ (3x + 7)(x + 6) \\
9. & \ (2x + 5)(x - 4) \\
10. & \ (5x + 2)(7x - 3)
\end{aligned}
}
\]