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

Let's solve each of these trinomials by factoring completely. These are all quadratic trinomials in the form $ ax^2 + bx + c $, where $ a = 1 $. So we're looking for two numbers that:

- Multiply to give the constant term $ c $
- Add to give the coefficient of the middle term $ b $

We'll go through each one step-by-step.

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1) $ m^2 + m - 90 $

We need two numbers that:
- Multiply to $ -90 $
- Add to $ 1 $

Let’s list factor pairs of $ -90 $:
- $ 10 \times (-9) = -90 $, $ 10 + (-9) = 1 $ → ✔

So:
$$
m^2 + m - 90 = (m + 10)(m - 9)
$$

✔ Answer: $ (m + 10)(m - 9) $

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2) $ n^2 + 4n - 12 $

Need two numbers that:
- Multiply to $ -12 $
- Add to $ 4 $

Factor pairs of $ -12 $:
- $ 6 \times (-2) = -12 $, $ 6 + (-2) = 4 $ → ✔

So:
$$
n^2 + 4n - 12 = (n + 6)(n - 2)
$$

✔ Answer: $ (n + 6)(n - 2) $

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3) $ n^2 - 10n + 9 $

Need two numbers that:
- Multiply to $ 9 $
- Add to $ -10 $

Factor pairs of $ 9 $:
- $ -9 \times -1 = 9 $, $ -9 + (-1) = -10 $ → ✔

So:
$$
n^2 - 10n + 9 = (n - 9)(n - 1)
$$

✔ Answer: $ (n - 9)(n - 1) $

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4) $ b^2 + 16b + 64 $

This looks like a perfect square trinomial.

Check:
- $ 64 = 8^2 $
- $ 2 \times 8 = 16 $ → matches the middle term

So:
$$
b^2 + 16b + 64 = (b + 8)^2
$$

✔ Answer: $ (b + 8)(b + 8) $ or $ (b + 8)^2 $

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5) $ m^2 + 2m - 24 $

Need two numbers that:
- Multiply to $ -24 $
- Add to $ 2 $

Try:
- $ 6 \times (-4) = -24 $, $ 6 + (-4) = 2 $ → ✔

So:
$$
m^2 + 2m - 24 = (m + 6)(m - 4)
$$

✔ Answer: $ (m + 6)(m - 4) $

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6) $ x^2 + 4x + 32 $

Need two numbers that:
- Multiply to $ 32 $
- Add to $ 4 $

Factor pairs of 32:
- $ 1 \times 32 = 32 $, sum = 33
- $ 2 \times 16 = 32 $, sum = 18
- $ 4 \times 8 = 32 $, sum = 12
- All positive → no pair adds to 4

Wait — try negative? But both would have to be positive since product is positive and sum is positive.

But none add to 4.

Let’s check discriminant:
$ D = 4^2 - 4(1)(32) = 16 - 128 = -112 $ → negative

So this does not factor over real numbers.

But the problem says "factor completely", and if it can't be factored with integers, then it's prime.

⚠️ No integer factors exist, so it cannot be factored further.

✔ Answer: Prime (cannot be factored)

But wait — let’s double-check: maybe I made a mistake?

Try again:
Is there any pair of integers whose product is 32 and sum is 4?

- $ 8 \times 4 = 32 $, sum = 12
- $ 16 \times 2 = 32 $, sum = 18
- $ 32 \times 1 = 32 $, sum = 33
- $ -8 \times -4 = 32 $, sum = -12
- No pair gives sum = 4

So yes, it does not factor.

✔ Final answer: Cannot be factored (prime)

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7) $ k^2 - 13k + 40 $

Need two numbers that:
- Multiply to $ 40 $
- Add to $ -13 $

Factor pairs of 40:
- $ -8 \times -5 = 40 $, $ -8 + (-5) = -13 $ → ✔

So:
$$
k^2 - 13k + 40 = (k - 8)(k - 5)
$$

✔ Answer: $ (k - 8)(k - 5) $

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8) $ a^2 + 11a + 18 $

Need two numbers that:
- Multiply to $ 18 $
- Add to $ 11 $

Try:
- $ 9 \times 2 = 18 $, $ 9 + 2 = 11 $ → ✔

So:
$$
a^2 + 11a + 18 = (a + 9)(a + 2)
$$

✔ Answer: $ (a + 9)(a + 2) $

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11) $ n^2 - n - 56 $

Need two numbers that:
- Multiply to $ -56 $
- Add to $ -1 $

Factor pairs of $ -56 $:
- $ 7 \times (-8) = -56 $, $ 7 + (-8) = -1 $ → ✔

So:
$$
n^2 - n - 56 = (n + 7)(n - 8)
$$

✔ Answer: $ (n + 7)(n - 8) $

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12) $ n^2 - 5n + 6 $

Need two numbers that:
- Multiply to $ 6 $
- Add to $ -5 $

Try:
- $ -3 \times -2 = 6 $, $ -3 + (-2) = -5 $ → ✔

So:
$$
n^2 - 5n + 6 = (n - 3)(n - 2)
$$

✔ Answer: $ (n - 3)(n - 2) $

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## ✔ Final Answers:

1) $ (m + 10)(m - 9) $
2) $ (n + 6)(n - 2) $
3) $ (n - 9)(n - 1) $
4) $ (b + 8)(b + 8) $ or $ (b + 8)^2 $
5) $ (m + 6)(m - 4) $
6) Prime (cannot be factored)
7) $ (k - 8)(k - 5) $
8) $ (a + 9)(a + 2) $
11) $ (n + 7)(n - 8) $
12) $ (n - 3)(n - 2) $

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Let me know if you'd like a visual explanation or help with any specific step!