Algebra worksheet for factoring trinomials, featuring 20 quadratic expressions to factor completely.

Let's solve each trinomial factoring problem step by step. We'll factor each quadratic expression completely, starting by checking for a Greatest Common Factor (GCF) if needed, and then using the AC method or trial and error to factor.

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1. $ x^2 + 13x - 30 $

We need two numbers that multiply to $-30$ and add to $13$.
Try: $15$ and $-2$: $15 \times (-2) = -30$, $15 + (-2) = 13$ ✔
So:
$$
x^2 + 13x - 30 = (x + 15)(x - 2)
$$

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2. $ x^2 - 5x - 24 $

Need two numbers that multiply to $-24$ and add to $-5$.
Try: $-8$ and $3$: $-8 \times 3 = -24$, $-8 + 3 = -5$ ✔
$$
x^2 - 5x - 24 = (x - 8)(x + 3)
$$

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3. $ x^2 + 5x - 36 $

Need two numbers that multiply to $-36$ and add to $5$.
Try: $9$ and $-4$: $9 \times (-4) = -36$, $9 + (-4) = 5$ ✔
$$
x^2 + 5x - 36 = (x + 9)(x - 4)
$$

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4. $ x^2 + 15x + 56 $

Need two numbers that multiply to $56$ and add to $15$.
Try: $7$ and $8$: $7 \times 8 = 56$, $7 + 8 = 15$ ✔
$$
x^2 + 15x + 56 = (x + 7)(x + 8)
$$

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5. $ x^2 + 15x + 54 $

Need two numbers that multiply to $54$ and add to $15$.
Try: $6$ and $9$: $6 \times 9 = 54$, $6 + 9 = 15$ ✔
$$
x^2 + 15x + 54 = (x + 6)(x + 9)
$$

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6. $ x^2 - 8x - 20 $

Need two numbers that multiply to $-20$ and add to $-8$.
Try: $-10$ and $2$: $-10 \times 2 = -20$, $-10 + 2 = -8$ ✔
$$
x^2 - 8x - 20 = (x - 10)(x + 2)
$$

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7. $ x^2 + 4x - 32 $

Need two numbers that multiply to $-32$ and add to $4$.
Try: $8$ and $-4$: $8 \times (-4) = -32$, $8 + (-4) = 4$ ✔
$$
x^2 + 4x - 32 = (x + 8)(x - 4)
$$

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8. $ x^2 - x - 20 $

Need two numbers that multiply to $-20$ and add to $-1$.
Try: $-5$ and $4$: $-5 \times 4 = -20$, $-5 + 4 = -1$ ✔
$$
x^2 - x - 20 = (x - 5)(x + 4)
$$

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9. $ x^2 + 11x + 30 $

Need two numbers that multiply to $30$ and add to $11$.
Try: $5$ and $6$: $5 \times 6 = 30$, $5 + 6 = 11$ ✔
$$
x^2 + 11x + 30 = (x + 5)(x + 6)
$$

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10. $ x^2 + 14x + 49 $

This is a perfect square trinomial:
Check: $14/2 = 7$, $7^2 = 49$ ✔
$$
x^2 + 14x + 49 = (x + 7)^2
$$

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11. $ x^2 + 10x + 16 $

Need two numbers that multiply to $16$ and add to $10$.
Try: $8$ and $2$: $8 \times 2 = 16$, $8 + 2 = 10$ ✔
$$
x^2 + 10x + 16 = (x + 8)(x + 2)
$$

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12. $ x^2 + 3x + 2 $

Need two numbers that multiply to $2$ and add to $3$.
Try: $1$ and $2$: $1 \times 2 = 2$, $1 + 2 = 3$ ✔
$$
x^2 + 3x + 2 = (x + 1)(x + 2)
$$

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13. $ x^2 + 15x + 44 $

Need two numbers that multiply to $44$ and add to $15$.
Try: $11$ and $4$: $11 \times 4 = 44$, $11 + 4 = 15$ ✔
$$
x^2 + 15x + 44 = (x + 11)(x + 4)
$$

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15. $ x^2 + 6x + 5 $

Need two numbers that multiply to $5$ and add to $6$.
Try: $1$ and $5$: $1 \times 5 = 5$, $1 + 5 = 6$ ✔
$$
x^2 + 6x + 5 = (x + 1)(x + 5)
$$

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16. $ 2x^2 + 20x + 32 $

First, check for GCF: all terms divisible by 2
Factor out 2:
$$
2(x^2 + 10x + 16)
$$
Now factor inside: need two numbers that multiply to $16$ and add to $10$: $8$ and $2$
$$
= 2(x + 8)(x + 2)
$$

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18. $ 3x^2 - 15x + 18 $

Check GCF: all terms divisible by 3
Factor out 3:
$$
3(x^2 - 5x + 6)
$$
Now factor: need two numbers that multiply to $6$ and add to $-5$: $-3$ and $-2$
$$
= 3(x - 3)(x - 2)
$$

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19. $ 2x^2 + 8x - 24 $

Check GCF: all divisible by 2
Factor out 2:
$$
2(x^2 + 4x - 12)
$$
Now factor: need two numbers that multiply to $-12$ and add to $4$: $6$ and $-2$
$$
= 2(x + 6)(x - 2)
$$

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20. $ 2x^2 + 16x - 32 $

Check GCF: all divisible by 2
Factor out 2:
$$
2(x^2 + 8x - 16)
$$
Now factor $x^2 + 8x - 16$:
Need two numbers that multiply to $-16$ and add to $8$.
Try: $10$ and $-2$: $10 \times (-2) = -20$ ✘
Try: $8$ and $-2$: $8 \times (-2) = -16$, $8 + (-2) = 6$ ✘
Try: $16$ and $-1$: $16 + (-1) = 15$ ✘
No integers work → Not factorable over integers.

Wait — actually, let’s double-check:
Is $x^2 + 8x - 16$ factorable?
Discriminant: $8^2 - 4(1)(-16) = 64 + 64 = 128$, not a perfect square → Does not factor nicely.

But wait — maybe we missed something? Let's go back.

Actually, original: $2x^2 + 16x - 32$

We factored out 2: $2(x^2 + 8x - 16)$

But $x^2 + 8x - 16$ has no integer factors.

But let’s try again: maybe factor differently?

Wait — perhaps we can factor further?

Alternatively, maybe I made an error in earlier steps.

Wait — let’s recheck: $x^2 + 8x - 16$

We want two numbers that multiply to $-16$ and add to $8$:
Possible pairs:
(16, -1): sum = 15
(8, -2): sum = 6
(4, -4): sum = 0
(2, -8): sum = -6
(-16, 1): sum = -15
No pair gives sum of 8.

So it does not factor over integers.

Thus, the final answer is:
$$
2x^2 + 16x - 32 = 2(x^2 + 8x - 16)
$$
And this is not factorable with integer coefficients.

But wait — is there a mistake? Let's try factoring directly:

Use AC method: $a = 2$, $b = 16$, $c = -32$

$ac = 2 \times (-32) = -64$

Find two numbers that multiply to $-64$ and add to $16$:
Try: $32$ and $-2$: $32 \times (-2) = -64$, $32 + (-2) = 30$ ✘
Try: $16$ and $-4$: $16 \times (-4) = -64$, $16 + (-4) = 12$ ✘
Try: $8$ and $-8$: $8 \times (-8) = -64$, $8 + (-8) = 0$ ✘
Try: $-8$ and $8$: same
Try: $64$ and $-1$: $64 + (-1) = 63$ ✘
Try: $-64$ and $1$: sum = -63 ✘

No pair adds to 16.

So indeed, cannot be factored into binomials with integer coefficients.

But wait — let's try factoring out the GCF first, then see:

$2x^2 + 16x - 32 = 2(x^2 + 8x - 16)$

And $x^2 + 8x - 16$ does not factor over integers.

So the completely factored form is:
$$
2(x^2 + 8x - 16)
$$

But let's check if we can write it as a product of binomials with rational coefficients? Not required unless specified.

Since the worksheet asks to factor completely, and only integers are expected, this is fully factored.

Wait — but let’s double-check the original expression:
$2x^2 + 16x - 32$

Maybe we missed a GCF?
All terms divisible by 2: yes.

But $x^2 + 8x - 16$ — discriminant $64 + 64 = 128$, which is not a perfect square → irrational roots → not factorable over integers.

So final answer:
$$
2(x^2 + 8x - 16)
$$

But let’s confirm with a different approach.

Alternatively, could we have made a mistake in assuming it doesn’t factor?

Wait — what if we try to factor $2x^2 + 16x - 32$ directly?

Try grouping:
Look for two numbers that multiply to $2 \times (-32) = -64$, and add to $16$.

Try: $32$ and $-2$: $32 \times (-2) = -64$, $32 + (-2) = 30$ ✘
Try: $16$ and $-4$: $16 \times (-4) = -64$, $16 + (-4) = 12$ ✘
Try: $8$ and $-8$: $8 \times (-8) = -64$, $8 + (-8) = 0$ ✘
Try: $-16$ and $4$: $-16 + 4 = -12$ ✘
Try: $-64$ and $1$: $-64 + 1 = -63$ ✘

No pair works.

So no, cannot factor.

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## ✔ Final Answers:

1. $ (x + 15)(x - 2) $
2. $ (x - 8)(x + 3) $
3. $ (x + 9)(x - 4) $
4. $ (x + 7)(x + 8) $
5. $ (x + 6)(x + 9) $
6. $ (x - 10)(x + 2) $
7. $ (x + 8)(x - 4) $
8. $ (x - 5)(x + 4) $
9. $ (x + 5)(x + 6) $
10. $ (x + 7)^2 $
11. $ (x + 8)(x + 2) $
12. $ (x + 1)(x + 2) $
13. $ (x + 11)(x + 4) $
15. $ (x + 1)(x + 5) $
16. $ 2(x + 8)(x + 2) $
18. $ 3(x - 3)(x - 2) $
19. $ 2(x + 6)(x - 2) $
20. $ 2(x^2 + 8x - 16) $ ← Not factorable over integers

> Note: Problem #20 cannot be factored into binomials with integer coefficients. The expression $x^2 + 8x - 16$ has no integer factors.

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🔍 Summary of Strategy:

For trinomials of the form $ax^2 + bx + c$:
1. Always check for a GCF first.
2. If $a = 1$: Find two numbers that multiply to $c$ and add to $b$.
3. If $a \neq 1$: Use AC method or factoring by grouping.
- Multiply $a \cdot c$
- Find two numbers that multiply to $ac$ and add to $b$
- Rewrite middle term and factor by grouping
4. If no such integers exist → Not factorable over integers

Let me know if you'd like to see any specific one worked out in more detail!