6.5 Practice - Factoring Special Products - Apple - Free Printable

Let's solve the factoring special products practice problems step by step. These problems involve factoring expressions using special product patterns, such as:

- Difference of squares: $ a^2 - b^2 = (a - b)(a + b) $
- Perfect square trinomials: $ a^2 + 2ab + b^2 = (a + b)^2 $
- Sum and difference of cubes:
$ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $
$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $
- Difference of squares with higher powers (e.g., $ a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) $)

We’ll go through each problem one by one.

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1) $ x^2 - 16 $

This is a difference of squares:
$ x^2 - 4^2 = (x - 4)(x + 4) $

✔ Answer: $ (x - 4)(x + 4) $

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2) $ z^2 - 9 $

Difference of squares:
$ z^2 - 3^2 = (z - 3)(z + 3) $

✔ Answer: $ (z - 3)(z + 3) $

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3) $ x^2 - 25 $

Difference of squares:
$ x^2 - 5^2 = (x - 5)(x + 5) $

✔ Answer: $ (x - 5)(x + 5) $

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4) $ x^2 - 1 $

Difference of squares:
$ x^2 - 1^2 = (x - 1)(x + 1) $

✔ Answer: $ (x - 1)(x + 1) $

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5) $ y^2 - 4 $

Difference of squares:
$ y^2 - 2^2 = (y - 2)(y + 2) $

✔ Answer: $ (y - 2)(y + 2) $

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6) $ 4x^2 - 1 $

Factor out coefficient:
$ (2x)^2 - 1^2 = (2x - 1)(2x + 1) $

✔ Answer: $ (2x - 1)(2x + 1) $

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7) $ 9k^2 - 4 $

$ (3k)^2 - 2^2 = (3k - 2)(3k + 2) $

✔ Answer: $ (3k - 2)(3k + 2) $

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8) $ 9x^2 - 1 $

$ (3x)^2 - 1^2 = (3x - 1)(3x + 1) $

✔ Answer: $ (3x - 1)(3x + 1) $

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9) $ 3x^2 - 27 $

Factor out GCF first:
$ 3(x^2 - 9) = 3(x^2 - 3^2) = 3(x - 3)(x + 3) $

✔ Answer: $ 3(x - 3)(x + 3) $

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10) $ 5n^2 - 20 $

Factor GCF:
$ 5(n^2 - 4) = 5(n - 2)(n + 2) $

✔ Answer: $ 5(n - 2)(n + 2) $

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11) $ 16x^2 - 36 $

GCF: 4 → $ 4(4x^2 - 9) $
Now factor inside: $ 4x^2 - 9 = (2x)^2 - 3^2 = (2x - 3)(2x + 3) $
So total: $ 4(2x - 3)(2x + 3) $

✔ Answer: $ 4(2x - 3)(2x + 3) $

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12) $ 125x^2 + 45y^2 $

Wait — this is sum of squares? But sum of squares does not factor over real numbers unless it’s a perfect square trinomial or has common factors.

But here: $ 125x^2 + 45y^2 $. No subtraction, so not a difference of squares.

Check for GCF: both divisible by 5 → $ 5(25x^2 + 9y^2) $

But $ 25x^2 + 9y^2 $ is sum of squares, which doesn’t factor over reals.

⚠️ Note: This expression cannot be factored further over real numbers.

✔ Answer: $ 5(25x^2 + 9y^2) $ ← already simplified

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13) $ 18x^2 - 50y^2 $

GCF: 2 → $ 2(9x^2 - 25y^2) $
Now: $ 9x^2 - 25y^2 = (3x)^2 - (5y)^2 = (3x - 5y)(3x + 5y) $

So: $ 2(3x - 5y)(3x + 5y) $

✔ Answer: $ 2(3x - 5y)(3x + 5y) $

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14) $ 4m^2 + 64n^2 $

Sum of squares: $ 4m^2 + 64n^2 = 4(m^2 + 16n^2) $

But $ m^2 + 16n^2 $ is sum of squares — does not factor over reals.

✔ Answer: $ 4(m^2 + 16n^2) $ ← no further factoring

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15) $ x^2 - 2x + 1 $

Perfect square trinomial:
$ x^2 - 2x + 1 = (x - 1)^2 $

✔ Answer: $ (x - 1)^2 $

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16) $ k^2 + 4k + 4 $

Perfect square: $ k^2 + 2(k)(2) + 2^2 = (k + 2)^2 $

✔ Answer: $ (k + 2)^2 $

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17) $ x^2 + 6x + 9 $

$ x^2 + 2(3)x + 3^2 = (x + 3)^2 $

✔ Answer: $ (x + 3)^2 $

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18) $ n^2 - 8n + 16 $

$ (n - 4)^2 $

✔ Answer: $ (n - 4)^2 $

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19) $ x^2 - 6x + 9 $

$ (x - 3)^2 $

✔ Answer: $ (x - 3)^2 $

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20) $ k^2 - 4k + 4 $

$ (k - 2)^2 $

✔ Answer: $ (k - 2)^2 $

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21) $ 25y^2 - 10y + 1 $

Check if perfect square:
$ (5y)^2 - 2(5y)(1) + 1^2 = (5y - 1)^2 $

✔ Answer: $ (5y - 1)^2 $

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22) $ x^2 + 2x + 1 $

$ (x + 1)^2 $

✔ Answer: $ (x + 1)^2 $

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23) $ x^2 + 8xy + 16y^2 $

$ (x)^2 + 2(x)(4y) + (4y)^2 = (x + 4y)^2 $

✔ Answer: $ (x + 4y)^2 $

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24) $ 18m^2 - 24mn + 8n^2 $

GCF: 2 → $ 2(9m^2 - 12mn + 4n^2) $

Now check: $ 9m^2 - 12mn + 4n^2 = (3m)^2 - 2(3m)(2n) + (2n)^2 = (3m - 2n)^2 $

So: $ 2(3m - 2n)^2 $

✔ Answer: $ 2(3m - 2n)^2 $

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25) $ 4a^2 - 20ab + 25b^2 $

$ (2a)^2 - 2(2a)(5b) + (5b)^2 = (2a - 5b)^2 $

✔ Answer: $ (2a - 5b)^2 $

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26) $ 20x^2 + 20xy + 5y^2 $

GCF: 5 → $ 5(4x^2 + 4xy + y^2) $

Now: $ 4x^2 + 4xy + y^2 = (2x)^2 + 2(2x)(y) + y^2 = (2x + y)^2 $

So: $ 5(2x + y)^2 $

✔ Answer: $ 5(2x + y)^2 $

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27) $ 8x^2 - 24xy + 18y^2 $

GCF: 2 → $ 2(4x^2 - 12xy + 9y^2) $

Now: $ 4x^2 - 12xy + 9y^2 = (2x)^2 - 2(2x)(3y) + (3y)^2 = (2x - 3y)^2 $

So: $ 2(2x - 3y)^2 $

✔ Answer: $ 2(2x - 3y)^2 $

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28) $ x^2 + 64 $

Sum of squares: $ x^2 + 8^2 $ → does not factor over reals.

✔ Answer: $ x^2 + 64 $ ← prime over reals

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29) $ 8 - m^3 $

This is a difference of cubes:
$ 8 = 2^3 $, so $ 2^3 - m^3 = (2 - m)(4 + 2m + m^2) $

Use: $ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $

Here: $ a = 2, b = m $ →
$ (2 - m)(4 + 2m + m^2) $

✔ Answer: $ (2 - m)(4 + 2m + m^2) $

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30) $ x^4 + 8 $

Not a standard form. $ x^4 + 8 $ — sum of fourth power and constant.

No factoring pattern applies directly.

Note: $ x^4 + 8 $ is not a sum of squares in a factorable way.

✔ Answer: $ x^4 + 8 $ ← cannot be factored further over reals

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31) $ x^2 - 64 $

Difference of squares: $ x^2 - 8^2 = (x - 8)(x + 8) $

✔ Answer: $ (x - 8)(x + 8) $

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32) $ x^4 + 8 $

Same as #30 — no factoring

Wait, let’s double-check:
Is $ x^4 + 8 $ factorable?

Try writing as $ x^4 + 4 \cdot 2 = x^4 + 4 \cdot 2 $ — not helpful.

Alternatively, use sum of squares trick? Not really.

Actually, $ x^4 + 8 $ can be factored using sum of squares method via complex numbers, but not over reals.

So no real factoring.

✔ Answer: $ x^4 + 8 $ ← irreducible over reals

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Wait — there's a typo? Let's check problem 32 again.

Problem 32: $ x^4 + 8 $? But earlier we had $ x^4 + 8 $ in #30. Maybe it's a mistake.

Wait — actually, look at #32: $ x^4 + 8 $? That seems odd.

But let’s move on.

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33) $ 216 - x^3 $

$ 216 = 6^3 $, so $ 6^3 - x^3 = (6 - x)(36 + 6x + x^2) $

Using $ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $

So: $ (6 - x)(36 + 6x + x^2) $

✔ Answer: $ (6 - x)(36 + 6x + x^2) $

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34) $ 125x^3 - 216 $

$ 125 = 5^3 $, $ 216 = 6^3 $, so:
$ (5x)^3 - 6^3 = (5x - 6)(25x^2 + 30x + 36) $

✔ Answer: $ (5x - 6)(25x^2 + 30x + 36) $

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35) $ 125a^3 - 64 $

$ 125a^3 = (5a)^3 $, $ 64 = 4^3 $, so:
$ (5a)^3 - 4^3 = (5a - 4)(25a^2 + 20a + 16) $

✔ Answer: $ (5a - 4)(25a^2 + 20a + 16) $

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36) $ 64x^3 - 27 $

$ 64 = 4^3 $, $ 27 = 3^3 $, so:
$ (4x)^3 - 3^3 = (4x - 3)(16x^2 + 12x + 9) $

✔ Answer: $ (4x - 3)(16x^2 + 12x + 9) $

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37) $ 64x^3 + 27y^3 $

Sum of cubes: $ (4x)^3 + (3y)^3 = (4x + 3y)(16x^2 - 12xy + 9y^2) $

✔ Answer: $ (4x + 3y)(16x^2 - 12xy + 9y^2) $

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38) $ 32m^3 - 108n^3 $

GCF: 4 → $ 4(8m^3 - 27n^3) $

Now: $ 8m^3 = (2m)^3 $, $ 27n^3 = (3n)^3 $ → difference of cubes:

$ (2m)^3 - (3n)^3 = (2m - 3n)(4m^2 + 6mn + 9n^2) $

So: $ 4(2m - 3n)(4m^2 + 6mn + 9n^2) $

✔ Answer: $ 4(2m - 3n)(4m^2 + 6mn + 9n^2) $

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39) $ 54x^3 + 250y^3 $

GCF: 2 → $ 2(27x^3 + 125y^3) $

Now: $ 27x^3 = (3x)^3 $, $ 125y^3 = (5y)^3 $ → sum of cubes:

$ (3x)^3 + (5y)^3 = (3x + 5y)(9x^2 - 15xy + 25y^2) $

So: $ 2(3x + 5y)(9x^2 - 15xy + 25y^2) $

✔ Answer: $ 2(3x + 5y)(9x^2 - 15xy + 25y^2) $

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40) $ 375m^3 + 648n^3 $

GCF: 3 → $ 3(125m^3 + 216n^3) $

Now: $ 125m^3 = (5m)^3 $, $ 216n^3 = (6n)^3 $

Sum of cubes: $ (5m)^3 + (6n)^3 = (5m + 6n)(25m^2 - 30mn + 36n^2) $

So: $ 3(5m + 6n)(25m^2 - 30mn + 36n^2) $

✔ Answer: $ 3(5m + 6n)(25m^2 - 30mn + 36n^2) $

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41) $ x^4 - 81 $

This is a difference of squares:
$ x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9) $

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

So: $ (x - 3)(x + 3)(x^2 + 9) $

✔ Answer: $ (x - 3)(x + 3)(x^2 + 9) $

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42) $ x^4 - 256 $

$ 256 = 16^2 $, so $ x^4 - 16^2 = (x^2 - 16)(x^2 + 16) $

Then $ x^2 - 16 = (x - 4)(x + 4) $

So: $ (x - 4)(x + 4)(x^2 + 16) $

✔ Answer: $ (x - 4)(x + 4)(x^2 + 16) $

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43) $ x^4 - 1 $

$ x^4 - 1 = (x^2)^2 - 1^2 = (x^2 - 1)(x^2 + 1) $

Then $ x^2 - 1 = (x - 1)(x + 1) $

So: $ (x - 1)(x + 1)(x^2 + 1) $

✔ Answer: $ (x - 1)(x + 1)(x^2 + 1) $

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44) $ 16a^4 - b^4 $

$ (4a^2)^2 - (b^2)^2 = (4a^2 - b^2)(4a^2 + b^2) $

Now $ 4a^2 - b^2 = (2a - b)(2a + b) $

So: $ (2a - b)(2a + b)(4a^2 + b^2) $

✔ Answer: $ (2a - b)(2a + b)(4a^2 + b^2) $

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45) $ x^4 - y^4 $

$ (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2) $

Then $ x^2 - y^2 = (x - y)(x + y) $

So: $ (x - y)(x + y)(x^2 + y^2) $

✔ Answer: $ (x - y)(x + y)(x^2 + y^2) $

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46) $ m^4 - 81k^4 $

$ (m^2)^2 - (9k^2)^2 = (m^2 - 9k^2)(m^2 + 9k^2) $

Then $ m^2 - 9k^2 = (m - 3k)(m + 3k) $

So: $ (m - 3k)(m + 3k)(m^2 + 9k^2) $

✔ Answer: $ (m - 3k)(m + 3k)(m^2 + 9k^2) $

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47) $ m^4 - 81k^4 $

Wait — same as #46? Yes, probably typo.

But #47 says $ m^4 - 81k^4 $ — same as #46.

Possibly duplicate.

✔ Answer: $ (m - 3k)(m + 3k)(m^2 + 9k^2) $

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48) $ 81c^4 - 16d^4 $

$ (9c^2)^2 - (4d^2)^2 = (9c^2 - 4d^2)(9c^2 + 4d^2) $

Now $ 9c^2 - 4d^2 = (3c - 2d)(3c + 2d) $

So: $ (3c - 2d)(3c + 2d)(9c^2 + 4d^2) $

✔ Answer: $ (3c - 2d)(3c + 2d)(9c^2 + 4d^2) $

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## ✔ Final Answers Summary:

| Problem | Factored Form |
|--------|----------------|
| 1 | $ (x - 4)(x + 4) $ |
| 2 | $ (z - 3)(z + 3) $ |
| 3 | $ (x - 5)(x + 5) $ |
| 4 | $ (x - 1)(x + 1) $ |
| 5 | $ (y - 2)(y + 2) $ |
| 6 | $ (2x - 1)(2x + 1) $ |
| 7 | $ (3k - 2)(3k + 2) $ |
| 8 | $ (3x - 1)(3x + 1) $ |
| 9 | $ 3(x - 3)(x + 3) $ |
| 10 | $ 5(n - 2)(n + 2) $ |
| 11 | $ 4(2x - 3)(2x + 3) $ |
| 12 | $ 5(25x^2 + 9y^2) $ |
| 13 | $ 2(3x - 5y)(3x + 5y) $ |
| 14 | $ 4(m^2 + 16n^2) $ |
| 15 | $ (x - 1)^2 $ |
| 16 | $ (k + 2)^2 $ |
| 17 | $ (x + 3)^2 $ |
| 18 | $ (n - 4)^2 $ |
| 19 | $ (x - 3)^2 $ |
| 20 | $ (k - 2)^2 $ |
| 21 | $ (5y - 1)^2 $ |
| 22 | $ (x + 1)^2 $ |
| 23 | $ (x + 4y)^2 $ |
| 24 | $ 2(3m - 2n)^2 $ |
| 25 | $ (2a - 5b)^2 $ |
| 26 | $ 5(2x + y)^2 $ |
| 27 | $ 2(2x - 3y)^2 $ |
| 28 | $ x^2 + 64 $ (prime) |
| 29 | $ (2 - m)(4 + 2m + m^2) $ |
| 30 | $ x^4 + 8 $ (prime) |
| 31 | $ (x - 8)(x + 8) $ |
| 32 | $ x^4 + 8 $ (prime) |
| 33 | $ (6 - x)(36 + 6x + x^2) $ |
| 34 | $ (5x - 6)(25x^2 + 30x + 36) $ |
| 35 | $ (5a - 4)(25a^2 + 20a + 16) $ |
| 36 | $ (4x - 3)(16x^2 + 12x + 9) $ |
| 37 | $ (4x + 3y)(16x^2 - 12xy + 9y^2) $ |
| 38 | $ 4(2m - 3n)(4m^2 + 6mn + 9n^2) $ |
| 39 | $ 2(3x + 5y)(9x^2 - 15xy + 25y^2) $ |
| 40 | $ 3(5m + 6n)(25m^2 - 30mn + 36n^2) $ |
| 41 | $ (x - 3)(x + 3)(x^2 + 9) $ |
| 42 | $ (x - 4)(x + 4)(x^2 + 16) $ |
| 43 | $ (x - 1)(x + 1)(x^2 + 1) $ |
| 44 | $ (2a - b)(2a + b)(4a^2 + b^2) $ |
| 45 | $ (x - y)(x + y)(x^2 + y^2) $ |
| 46 | $ (m - 3k)(m + 3k)(m^2 + 9k^2) $ |
| 47 | $ (m - 3k)(m + 3k)(m^2 + 9k^2) $ |
| 48 | $ (3c - 2d)(3c + 2d)(9c^2 + 4d^2) $ |

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