Factoring Worksheet for algebra practice, featuring ten expressions to factor.

Let's solve each problem on the Factoring Worksheet step by step.

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1. $ x^2 - 9x + 8 $

We need two numbers that multiply to 8 and add to -9.

- Factors of 8: (1,8), (2,4), (-1,-8), (-2,-4)
- Pair that adds to -9: -1 and -8

So:
$$
x^2 - 9x + 8 = (x - 1)(x - 8)
$$

✔ Answer: $ (x - 1)(x - 8) $

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2. $ 7x^2 + 9x $

Factor out the greatest common factor (GCF): both terms have $ x $, and GCF of coefficients is 1.

$$
7x^2 + 9x = x(7x + 9)
$$

✔ Answer: $ x(7x + 9) $

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3. $ 9x^2 + 7x - 56 $

Use AC method:

- $ a = 9 $, $ b = 7 $, $ c = -56 $
- $ ac = 9 \times (-56) = -504 $
- Find two numbers that multiply to -504 and add to 7

Try factors of 504:
- 21 and -24 → $ 21 \times (-24) = -504 $, $ 21 + (-24) = -3 $ ✘
- 28 and -18 → $ 28 \times (-18) = -504 $, $ 28 + (-18) = 10 $ ✘
- 36 and -14 → $ 36 \times (-14) = -504 $, $ 36 + (-14) = 22 $ ✘
- 56 and -9 → $ 56 \times (-9) = -504 $, $ 56 + (-9) = 47 $ ✘
- Try: 21 and -24 → already tried
Wait — try 28 and -18? No.

Try: 63 and -8: $ 63 \times (-8) = -504 $, $ 63 + (-8) = 55 $ ✘

Try: -42 and 12: $ -42 \times 12 = -504 $, $ -42 + 12 = -30 $ ✘

Try: -56 and 9: $ -56 \times 9 = -504 $, $ -56 + 9 = -47 $ ✘

Try: -21 and 24: $ -21 \times 24 = -504 $, $ -21 + 24 = 3 $ ✘

Try: -28 and 18: $ -28 \times 18 = -504 $, $ -28 + 18 = -10 $ ✘

Wait — maybe it doesn't factor nicely?

Let’s use quadratic formula to check if it factors:

Discriminant: $ D = b^2 - 4ac = 7^2 - 4(9)(-56) = 49 + 2016 = 2065 $

Is 2065 a perfect square? $ \sqrt{2065} \approx 45.4 $, not integer → so not factorable over integers

But wait — maybe I made a mistake.

Wait: Let's double-check with trial and error.

Try factoring as $ (9x + a)(x + b) $

We want:
- $ a \cdot b = -56 $
- $ 9b + a = 7 $

Try $ b = 8 $, $ a = -7 $: $ 9(8) + (-7) = 72 - 7 = 65 $ ✘
Try $ b = -7 $, $ a = 8 $: $ 9(-7) + 8 = -63 + 8 = -55 $ ✘
Try $ b = -8 $, $ a = 7 $: $ 9(-8) + 7 = -72 + 7 = -65 $ ✘
Try $ b = 7 $, $ a = -8 $: $ 9(7) + (-8) = 63 - 8 = 55 $ ✘
Try $ b = -1 $, $ a = 56 $: $ 9(-1) + 56 = -9 + 56 = 47 $ ✘
Try $ b = 1 $, $ a = -56 $: $ 9(1) + (-56) = 9 - 56 = -47 $ ✘

Wait — perhaps no integer factors?

But let’s check again using AC method properly:

$ 9x^2 + 7x - 56 $

We need two numbers that multiply to $ 9 \times (-56) = -504 $ and add to 7.

Try: -56 and 9: $ -56 + 9 = -47 $ ✘
Try: -28 and 18: $ -28 + 18 = -10 $ ✘
Try: -21 and 24: $ -21 + 24 = 3 $ ✘
Try: -18 and 28: $ -18 + 28 = 10 $ ✘
Try: -12 and 42: $ -12 + 42 = 30 $ ✘
Try: -9 and 56: $ -9 + 56 = 47 $ ✘
Try: -6 and 84: $ -6 + 84 = 78 $ ✘
Try: -8 and 63: $ -8 + 63 = 55 $ ✘
Try: -7 and 72: $ -7 + 72 = 65 $ ✘
Try: -4 and 126: too big

Wait — what about -21 and 24? No.

Try: -56 and 9 → no

Wait — -28 and 18 → sum = -10

No pair adds to 7.

Try: +21 and -24: $ 21 + (-24) = -3 $, product = -504

Try: +28 and -18: sum = 10

Wait — try: +56 and -9: sum = 47

No.

Wait — actually, no integer pair works.

But let's check discriminant again: $ D = 7^2 - 4(9)(-56) = 49 + 2016 = 2065 $

Is 2065 divisible by any square? $ 2065 = 5 \times 413 = 5 \times 7 \times 59 $ → no perfect squares → irrational roots → cannot be factored over integers

So this expression does not factor over integers.

But wait — maybe I miscalculated.

Wait! Let's try guessing factors:

Try $ (9x - 7)(x + 8) $:
= $ 9x(x) + 9x(8) -7(x) -7(8) = 9x^2 + 72x -7x -56 = 9x^2 + 65x -56 $ ✘

Try $ (9x - 8)(x + 7) $:
= $ 9x^2 + 63x -8x -56 = 9x^2 + 55x -56 $ ✘

Try $ (9x - 14)(x + 4) $:
= $ 9x^2 + 36x -14x -56 = 9x^2 + 22x -56 $ ✘

Try $ (9x + 8)(x - 7) $:
= $ 9x^2 -63x +8x -56 = 9x^2 -55x -56 $ ✘

Try $ (9x - 56)(x + 1) $:
= $ 9x^2 +9x -56x -56 = 9x^2 -47x -56 $ ✘

None work.

So likely this one does not factor over integers.

But let's recheck: maybe I made an error in sign.

Wait — could it be $ (9x + 14)(x - 4) $?
= $ 9x^2 -36x +14x -56 = 9x^2 -22x -56 $ ✘

No.

Alternatively, maybe it does factor — let me try rational root theorem.

Possible rational roots: factors of 56 over 9 → ±1,2,4,7,8,14,28,56, and fractions like 1/3, 2/3, etc.

Try $ x = 1 $: $ 9 + 7 -56 = -40 $ ✘
$ x = -1 $: $ 9 -7 -56 = -54 $ ✘
$ x = 2 $: $ 36 + 14 -56 = -6 $ ✘
$ x = 4 $: $ 144 + 28 -56 = 116 $ ✘
$ x = -2 $: $ 36 -14 -56 = -34 $ ✘
$ x = 7 $: $ 441 + 49 -56 = 434 $ ✘
$ x = -4 $: $ 144 -28 -56 = 60 $ ✘
$ x = 1/3 $: $ 9*(1/9) + 7*(1/3) -56 = 1 + 7/3 -56 = negative $ ✘

So no rational roots → irreducible over integers

✔ Answer: Does not factor over integers (or leave as is)

But wait — let's double-check the original problem.

Wait — maybe I misread: is it $ 9x^2 + 7x - 56 $? Yes.

But perhaps there's a typo? Or maybe it's supposed to be $ 9x^2 + 7x - 56 $ → we’ll assume it’s correct.

So unless it factors, we say prime or cannot be factored.

But let's move on and come back.

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Wait — let's try factoring by grouping:

$ 9x^2 + 7x - 56 $

We need two numbers whose product is $ 9 \times (-56) = -504 $, sum = 7

Try: -56 and 9: sum = -47
Try: -28 and 18: sum = -10
Try: -21 and 24: sum = 3
Try: -18 and 28: sum = 10
Try: -12 and 42: sum = 30
Try: -9 and 56: sum = 47
Try: -8 and 63: sum = 55
Try: -7 and 72: sum = 65
Try: -6 and 84: sum = 78
Try: -4 and 126: too big
Try: -3 and 168: no
Try: -2 and 252: no
Try: -1 and 504: sum = 503

Wait — try +21 and -24: sum = -3
No.

Wait — +28 and -21: sum = 7 → yes!

28 and -21: $ 28 \times (-21) = -588 $ ✘

Wait — $ 28 \times (-21) = -588 $, but we need $ -504 $

Wait — $ 21 \times (-24) = -504 $, sum = -3

No.

Wait — $ 18 \times (-28) = -504 $, sum = -10

No.

Wait — $ 14 \times (-36) = -504 $, sum = -22

No.

Wait — $ 12 \times (-42) = -504 $, sum = -30

No.

Wait — $ 9 \times (-56) = -504 $, sum = -47

No.

Wait — $ 8 \times (-63) = -504 $, sum = -55

No.

Wait — $ 7 \times (-72) = -504 $, sum = -65

No.

Wait — $ 6 \times (-84) = -504 $, sum = -78

No.

Wait — $ 4 \times (-126) = -504 $, sum = -122

No.

Wait — $ 3 \times (-168) = -504 $, sum = -165

No.

Wait — $ 2 \times (-252) = -504 $, sum = -250

No.

Wait — $ 1 \times (-504) = -504 $, sum = -503

No.

Wait — positive pairs: $ 504 \times 1 $, $ 252 \times 2 $, $ 168 \times 3 $, $ 126 \times 4 $, $ 84 \times 6 $, $ 72 \times 7 $, $ 63 \times 8 $, $ 56 \times 9 $, $ 42 \times 12 $, $ 36 \times 14 $, $ 28 \times 18 $, $ 24 \times 21 $

Now look for two numbers from these that differ by 7? No — we need sum = 7.

Try: $ 56 $ and $ -9 $: sum = 47
$ 42 $ and $ -12 $: sum = 30
$ 36 $ and $ -14 $: sum = 22
$ 28 $ and $ -18 $: sum = 10
$ 24 $ and $ -21 $: sum = 3
$ 21 $ and $ -24 $: sum = -3
$ 18 $ and $ -28 $: sum = -10
$ 14 $ and $ -36 $: sum = -22
$ 12 $ and $ -42 $: sum = -30
$ 9 $ and $ -56 $: sum = -47
$ 8 $ and $ -63 $: sum = -55
$ 7 $ and $ -72 $: sum = -65
$ 6 $ and $ -84 $: sum = -78
$ 4 $ and $ -126 $: sum = -122
$ 3 $ and $ -168 $: sum = -165
$ 2 $ and $ -252 $: sum = -250
$ 1 $ and $ -504 $: sum = -503

No pair sums to 7.

So no such integers exist → cannot factor over integers.

✔ Answer: $ 9x^2 + 7x - 56 $ is prime (cannot be factored over integers)

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4. $ -8x^2 - 15x + 2 $

First, factor out $ -1 $ to make leading coefficient positive:

$$
- (8x^2 + 15x - 2)
$$

Now factor $ 8x^2 + 15x - 2 $

Find two numbers that multiply to $ 8 \times (-2) = -16 $, sum to 15

Factors of -16: (1,-16), (-1,16), (2,-8), (-2,8), (4,-4), (-4,4)

Which pair adds to 15? $ 16 $ and $ -1 $: $ 16 + (-1) = 15 $ ✔

So split middle term:
$$
8x^2 + 16x - x - 2
$$

Group:
$$
(8x^2 + 16x) + (-x - 2) = 8x(x + 2) -1(x + 2) = (8x - 1)(x + 2)
$$

Now recall we had a negative sign:
$$
- (8x^2 + 15x - 2) = - (8x - 1)(x + 2)
$$

✔ Answer: $ -(8x - 1)(x + 2) $ or $ (1 - 8x)(x + 2) $

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5. $ 3x^2 - 10x + 3 $

AC method: $ a = 3 $, $ c = 3 $, $ ac = 9 $

Need two numbers that multiply to 9, add to -10

→ $ -9 $ and $ -1 $: $ -9 \times -1 = 9 $, $ -9 + (-1) = -10 $ ✔

Split:
$$
3x^2 - 9x - x + 3
$$

Group:
$$
(3x^2 - 9x) + (-x + 3) = 3x(x - 3) -1(x - 3) = (3x - 1)(x - 3)
$$

✔ Answer: $ (3x - 1)(x - 3) $

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6. $ 27xy - 18xz $

Factor out GCF: both terms have $ 9x $

$$
27xy - 18xz = 9x(3y - 2z)
$$

✔ Answer: $ 9x(3y - 2z) $

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7. $ x^2 + 36x $

Factor out $ x $:

$$
x(x + 36)
$$

✔ Answer: $ x(x + 36) $

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8. $ 15x^2 + 20x - 25 $

Factor out GCF: all coefficients divisible by 5

$$
5(3x^2 + 4x - 5)
$$

Now factor $ 3x^2 + 4x - 5 $

AC method: $ a = 3 $, $ c = -5 $, $ ac = -15 $

Need two numbers that multiply to -15, add to 4

Try: $ 5 $ and $ -3 $: $ 5 \times (-3) = -15 $, $ 5 + (-3) = 2 $ ✘
$ -5 $ and $ 3 $: $ -5 + 3 = -2 $ ✘
$ 15 $ and $ -1 $: sum = 14
$ -15 $ and $ 1 $: sum = -14
$ 1 $ and $ -15 $: sum = -14
No pair adds to 4

Try: $ 5 $ and $ -3 $: sum = 2
$ 3 $ and $ -5 $: sum = -2

No.

Check discriminant: $ D = 4^2 - 4(3)(-5) = 16 + 60 = 76 $, not perfect square → doesn't factor over integers

So $ 3x^2 + 4x - 5 $ is prime

Thus, only factor is:
$$
5(3x^2 + 4x - 5)
$$

✔ Answer: $ 5(3x^2 + 4x - 5) $

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9. $ 7x^2 - 32x - 60 $

AC method: $ a = 7 $, $ c = -60 $, $ ac = -420 $

Need two numbers that multiply to -420, add to -32

Try: $ -42 $ and $ 10 $: $ -42 + 10 = -32 $, $ -42 \times 10 = -420 $ ✔

Split:
$$
7x^2 - 42x + 10x - 60
$$

Group:
$$
(7x^2 - 42x) + (10x - 60) = 7x(x - 6) + 10(x - 6) = (7x + 10)(x - 6)
$$

✔ Answer: $ (7x + 10)(x - 6) $

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10. $ 27x^2 - 3 $

Factor out GCF: 3

$$
3(9x^2 - 1)
$$

Now $ 9x^2 - 1 $ is a difference of squares:
$$
9x^2 - 1 = (3x)^2 - (1)^2 = (3x - 1)(3x + 1)
$$

So:
$$
27x^2 - 3 = 3(3x - 1)(3x + 1)
$$

✔ Answer: $ 3(3x - 1)(3x + 1) $

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ (x - 1)(x - 8) $ |
| 2 | $ x(7x + 9) $ |
| 3 | Prime (cannot be factored over integers) |
| 4 | $ -(8x - 1)(x + 2) $ or $ (1 - 8x)(x + 2) $ |
| 5 | $ (3x - 1)(x - 3) $ |
| 6 | $ 9x(3y - 2z) $ |
| 7 | $ x(x + 36) $ |
| 8 | $ 5(3x^2 + 4x - 5) $ |
| 9 | $ (7x + 10)(x - 6) $ |
| 10 | $ 3(3x - 1)(3x + 1) $ |

> Note: Problem #3 may have a typo — if it were $ 9x^2 + 7x - 56 $, it does not factor nicely over integers. If it was meant to be $ 9x^2 + 7x - 56 $, then it's irreducible. But if it were $ 9x^2 + 7x - 56 $, same thing.

Let me know if you'd like to verify with a calculator or alternative method.