Factoring trinomials worksheet for algebra practice, focusing on expressions where the leading coefficient is 1.

You're working on Factoring Trinomials where the leading coefficient is 1 (i.e., trinomials of the form \( x^2 + bx + c \)).

The goal is to factor each trinomial into two binomials:
\[
(x + p)(x + q)
\]
where:
- \( p \cdot q = c \) (the constant term),
- \( p + q = b \) (the middle coefficient).

Let’s solve each one step by step.

---

1) \( m^2 + m - 90 \)

We need two numbers that multiply to -90 and add to +1.

Factors of -90:
- 10 and -9 → 10 × (-9) = -90, 10 + (-9) = 1 ✔

So:
\[
(m - 9)(m + 10)
\]

✔ Answer: \( (m - 9)(m + 10) \)

---

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

Find two numbers that multiply to -12, add to +4.

Factors:
- 6 and -2 → 6 × (-2) = -12, 6 + (-2) = 4 ✔

So:
\[
(n - 2)(n + 6)
\]

✔ Answer: \( (n - 2)(n + 6) \)

---

3) \( n^2 - 10n + 9 \)

Multiply to +9, add to -10.

Factors:
- -1 and -9 → (-1) × (-9) = 9, (-1) + (-9) = -10 ✔

So:
\[
(n - 1)(n - 9)
\]

✔ Answer: \( (n - 1)(n - 9) \)

---

4) \( b^2 + 16b + 64 \)

Multiply to +64, add to +16.

Factors:
- 8 and 8 → 8 × 8 = 64, 8 + 8 = 16 ✔

So:
\[
(b + 8)(b + 8) = (b + 8)^2
\]

✔ Answer: \( (b + 8)(b + 8) \)

---

5) \( m^2 + 2m - 24 \)

Multiply to -24, add to +2.

Factors:
- 6 and -4 → 6 × (-4) = -24, 6 + (-4) = 2 ✔

So:
\[
(m - 4)(m + 6)
\]

✔ Answer: \( (m - 4)(m + 6) \)

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6) \( x^2 + 4x + 32 \)

Multiply to +32, add to +4.

Check factor pairs of 32:
- 1 & 32 → sum=33
- 2 & 16 → sum=18
- 4 & 8 → sum=12

None add to 4. So this trinomial does NOT factor over the integers.

⚠️ But since the worksheet asks to “factor completely” and provides blank parentheses, it might expect you to write “prime” or leave it unfactored — however, looking at the format, maybe it's a trick question? Let’s double-check.

Wait — actually, no integer pair works. So this trinomial is prime (not factorable) over integers.

But since the problem expects filled parentheses, perhaps there’s a typo? Let me check if it was meant to be \( x^2 + 12x + 32 \) or something else.

As written: \( x^2 + 4x + 32 \) → cannot be factored with integer coefficients.

✔ Final answer: Prime / Not factorable

*(If forced to fill blanks, perhaps write “Not factorable” — but in context of worksheet, maybe an error.)*

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7) \( k^2 - 13k + 40 \)

Multiply to +40, add to -13.

Factors:
- -5 and -8 → (-5) × (-8) = 40, (-5) + (-8) = -13 ✔

So:
\[
(k - 5)(k - 8)
\]

✔ Answer: \( (k - 5)(k - 8) \)

---

8) \( a^2 + 11a + 18 \)

Multiply to +18, add to +11.

Factors:
- 2 and 9 → 2 × 9 = 18, 2 + 9 = 11 ✔

So:
\[
(a + 2)(a + 9)
\]

✔ Answer: \( (a + 2)(a + 9) \)

---

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

Multiply to -56, add to -1.

Factors:
- -8 and 7 → (-8) × 7 = -56, (-8) + 7 = -1 ✔

So:
\[
(n - 8)(n + 7)
\]

✔ Answer: \( (n - 8)(n + 7) \)

---

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

Multiply to +6, add to -5.

Factors:
- -2 and -3 → (-2) × (-3) = 6, (-2) + (-3) = -5 ✔

So:
\[
(n - 2)(n - 3)
\]

✔ Answer: \( (n - 2)(n - 3) \)

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## 📝 Final Answers:

1) \( (m - 9)(m + 10) \)
2) \( (n - 2)(n + 6) \)
3) \( (n - 1)(n - 9) \)
4) \( (b + 8)(b + 8) \)
5) \( (m - 4)(m + 6) \)
6) Cannot be factored over integers *(Prime)*
7) \( (k - 5)(k - 8) \)
8) \( (a + 2)(a + 9) \)
11) \( (n - 8)(n + 7) \)
12) \( (n - 2)(n - 3) \)

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📌 Note about #6: If this is from a standard curriculum worksheet, it’s possible there’s a typo — for example, if it were \( x^2 + 12x + 32 \), then factors would be (x+8)(x+4). As written, it doesn’t factor nicely. You may want to double-check the original problem.

Let me know if you’d like help checking #6 or any others!