This printable worksheet helps algebra students practice factoring quadratic trinomials where the leading coefficient is one.

To factor each of the given trinomials, we will use the method of factoring quadratic expressions of the form \( ax^2 + bx + c \) where \( a = 1 \). The general approach is to find two numbers that multiply to give the constant term \( c \) and add up to give the coefficient of the linear term \( b \).

Let's solve each problem step by step:

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1. \( b^2 + 8b + 7 \)

We need two numbers that multiply to \( 7 \) (the constant term) and add up to \( 8 \) (the coefficient of \( b \)).

- Factors of \( 7 \): \( (1, 7) \)
- Check: \( 1 + 7 = 8 \)

So, the numbers are \( 1 \) and \( 7 \).

Factorization:
\[
b^2 + 8b + 7 = (b + 1)(b + 7)
\]

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2. \( n^2 - 11n + 10 \)

We need two numbers that multiply to \( 10 \) (the constant term) and add up to \( -11 \) (the coefficient of \( n \)).

- Factors of \( 10 \): \( (-1, -10), (-2, -5) \)
- Check: \( -1 + (-10) = -11 \)

So, the numbers are \( -1 \) and \( -10 \).

Factorization:
\[
n^2 - 11n + 10 = (n - 1)(n - 10)
\]

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3. \( m^2 + m - 90 \)

We need two numbers that multiply to \( -90 \) (the constant term) and add up to \( 1 \) (the coefficient of \( m \)).

- Factors of \( -90 \): \( (10, -9), (9, -10), (15, -6), (6, -15), (18, -5), (5, -18), (30, -3), (3, -30) \)
- Check: \( 10 + (-9) = 1 \)

So, the numbers are \( 10 \) and \( -9 \).

Factorization:
\[
m^2 + m - 90 = (m + 10)(m - 9)
\]

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4. \( n^2 + 4n - 12 \)

We need two numbers that multiply to \( -12 \) (the constant term) and add up to \( 4 \) (the coefficient of \( n \)).

- Factors of \( -12 \): \( (6, -2), (4, -3), (3, -4), (2, -6) \)
- Check: \( 6 + (-2) = 4 \)

So, the numbers are \( 6 \) and \( -2 \).

Factorization:
\[
n^2 + 4n - 12 = (n + 6)(n - 2)
\]

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5. \( n^2 - 10n + 9 \)

We need two numbers that multiply to \( 9 \) (the constant term) and add up to \( -10 \) (the coefficient of \( n \)).

- Factors of \( 9 \): \( (-1, -9) \)
- Check: \( -1 + (-9) = -10 \)

So, the numbers are \( -1 \) and \( -9 \).

Factorization:
\[
n^2 - 10n + 9 = (n - 1)(n - 9)
\]

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6. \( b^2 + 16b + 64 \)

This is a perfect square trinomial because \( 64 = 8^2 \) and \( 16 = 2 \cdot 8 \).

Factorization:
\[
b^2 + 16b + 64 = (b + 8)^2
\]

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7. \( m^2 + 2m - 24 \)

We need two numbers that multiply to \( -24 \) (the constant term) and add up to \( 2 \) (the coefficient of \( m \)).

- Factors of \( -24 \): \( (6, -4), (4, -6), (8, -3), (3, -8), (12, -2), (2, -12), (24, -1), (1, -24) \)
- Check: \( 6 + (-4) = 2 \)

So, the numbers are \( 6 \) and \( -4 \).

Factorization:
\[
m^2 + 2m - 24 = (m + 6)(m - 4)
\]

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8. \( x^2 - 4x + 24 \)

We need two numbers that multiply to \( 24 \) (the constant term) and add up to \( -4 \) (the coefficient of \( x \)).

- Factors of \( 24 \): \( (-2, -12), (-3, -8), (-4, -6), (-6, -4), (-8, -3), (-12, -2) \)
- Check: None of these pairs add up to \( -4 \).

Since no such pair exists, this trinomial cannot be factored over the integers.

Factorization:
\[
x^2 - 4x + 24 \quad \text{(cannot be factored over integers)}
\]

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9. \( k^2 - 13k + 40 \)

We need two numbers that multiply to \( 40 \) (the constant term) and add up to \( -13 \) (the coefficient of \( k \)).

- Factors of \( 40 \): \( (-5, -8) \)
- Check: \( -5 + (-8) = -13 \)

So, the numbers are \( -5 \) and \( -8 \).

Factorization:
\[
k^2 - 13k + 40 = (k - 5)(k - 8)
\]

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10. \( a^2 + 11a + 18 \)

We need two numbers that multiply to \( 18 \) (the constant term) and add up to \( 11 \) (the coefficient of \( a \)).

- Factors of \( 18 \): \( (9, 2) \)
- Check: \( 9 + 2 = 11 \)

So, the numbers are \( 9 \) and \( 2 \).

Factorization:
\[
a^2 + 11a + 18 = (a + 9)(a + 2)
\]

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11. \( n^2 - n - 56 \)

We need two numbers that multiply to \( -56 \) (the constant term) and add up to \( -1 \) (the coefficient of \( n \)).

- Factors of \( -56 \): \( (-8, 7) \)
- Check: \( -8 + 7 = -1 \)

So, the numbers are \( -8 \) and \( 7 \).

Factorization:
\[
n^2 - n - 56 = (n - 8)(n + 7)
\]

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12. \( n^2 - 5n + 6 \)

We need two numbers that multiply to \( 6 \) (the constant term) and add up to \( -5 \) (the coefficient of \( n \)).

- Factors of \( 6 \): \( (-2, -3) \)
- Check: \( -2 + (-3) = -5 \)

So, the numbers are \( -2 \) and \( -3 \).

Factorization:
\[
n^2 - 5n + 6 = (n - 2)(n - 3)
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ (b + 1)(b + 7) \\
2. & \ (n - 1)(n - 10) \\
3. & \ (m + 10)(m - 9) \\
4. & \ (n + 6)(n - 2) \\
5. & \ (n - 1)(n - 9) \\
6. & \ (b + 8)^2 \\
7. & \ (m + 6)(m - 4) \\
8. & \ \text{(cannot be factored over integers)} \\
9. & \ (k - 5)(k - 8) \\
10. & \ (a + 9)(a + 2) \\
11. & \ (n - 8)(n + 7) \\
12. & \ (n - 2)(n - 3)
\end{aligned}
}
\]