Factoring Difference of Two Squares online exercise for | Live ... - Free Printable

You're working on Factoring the Difference of Squares, which follows the formula:

> a² – b² = (a + b)(a – b)

This means if you have two perfect squares being subtracted, you can factor them into the product of a sum and a difference of their square roots.

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Let’s go through each problem one by one and factor them completely.

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1) \( a^2 - 49 \)

- \( 49 = 7^2 \)
- So: \( a^2 - 7^2 = (a + 7)(a - 7) \)

✔ Answer: \( (a + 7)(a - 7) \)

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2) \( a^2 - 64 \)

- \( 64 = 8^2 \)
- So: \( a^2 - 8^2 = (a + 8)(a - 8) \)

✔ Answer: \( (a + 8)(a - 8) \)

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3) \( p^2 - 144 \)

- \( 144 = 12^2 \)
- So: \( p^2 - 12^2 = (p + 12)(p - 12) \)

✔ Answer: \( (p + 12)(p - 12) \)

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4) \( b^2 - 25 \)

- \( 25 = 5^2 \)
- So: \( b^2 - 5^2 = (b + 5)(b - 5) \)

✔ Answer: \( (b + 5)(b - 5) \)

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5) \( x^2 - 9 \)

- \( 9 = 3^2 \)
- So: \( x^2 - 3^2 = (x + 3)(x - 3) \)

✔ Answer: \( (x + 3)(x - 3) \)

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6) \( x^2 - 4 \)

- \( 4 = 2^2 \)
- So: \( x^2 - 2^2 = (x + 2)(x - 2) \)

✔ Answer: \( (x + 2)(x - 2) \)

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7) \( k^2 - 121 \)

- \( 121 = 11^2 \)
- So: \( k^2 - 11^2 = (k + 11)(k - 11) \)

✔ Answer: \( (k + 11)(k - 11) \)

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8) \( k^2 - 36 \)

- \( 36 = 6^2 \)
- So: \( k^2 - 6^2 = (k + 6)(k - 6) \)

✔ Answer: \( (k + 6)(k - 6) \)

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9) \( n^2 - 289 \)

- \( 289 = 17^2 \) (since 17×17=289)
- So: \( n^2 - 17^2 = (n + 17)(n - 17) \)

✔ Answer: \( (n + 17)(n - 17) \)

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10) \( n^2 - 169 \)

- \( 169 = 13^2 \)
- So: \( n^2 - 13^2 = (n + 13)(n - 13) \)

✔ Answer: \( (n + 13)(n - 13) \)

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11) \( 4x^2 - 25 \)

- \( 4x^2 = (2x)^2 \), \( 25 = 5^2 \)
- So: \( (2x)^2 - 5^2 = (2x + 5)(2x - 5) \)

✔ Answer: \( (2x + 5)(2x - 5) \)

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12) \( 16b^2 - 1 \)

- \( 16b^2 = (4b)^2 \), \( 1 = 1^2 \)
- So: \( (4b)^2 - 1^2 = (4b + 1)(4b - 1) \)

✔ Answer: \( (4b + 1)(4b - 1) \)

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13) \( 9a^2 - 4 \)

- \( 9a^2 = (3a)^2 \), \( 4 = 2^2 \)
- So: \( (3a)^2 - 2^2 = (3a + 2)(3a - 2) \)

✔ Answer: \( (3a + 2)(3a - 2) \)

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14) \( n^2 - 16 \)

- \( 16 = 4^2 \)
- So: \( n^2 - 4^2 = (n + 4)(n - 4) \)

✔ Answer: \( (n + 4)(n - 4) \)

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15) \( 9b^2 - 25 \)

- \( 9b^2 = (3b)^2 \), \( 25 = 5^2 \)
- So: \( (3b)^2 - 5^2 = (3b + 5)(3b - 5) \)

✔ Answer: \( (3b + 5)(3b - 5) \)

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16) \( 1 - a^2 \)

- This is the same as \( -a^2 + 1 = -(a^2 - 1) \), but better to write as:
- \( 1^2 - a^2 = (1 + a)(1 - a) \) — or equivalently \( (a + 1)(-a + 1) \), but standard form is \( (1 + a)(1 - a) \)

✔ Answer: \( (1 + a)(1 - a) \)
*(Note: You may also see it written as \( (a + 1)(1 - a) \), but order doesn’t matter for multiplication.)*

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17) \( 16r^2 - 25 \)

- \( 16r^2 = (4r)^2 \), \( 25 = 5^2 \)
- So: \( (4r)^2 - 5^2 = (4r + 5)(4r - 5) \)

✔ Answer: \( (4r + 5)(4r - 5) \)

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18) \( m^2 - 9 \)

- \( 9 = 3^2 \)
- So: \( m^2 - 3^2 = (m + 3)(m - 3) \)

✔ Answer: \( (m + 3)(m - 3) \)

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19) \( 25m^2 - 9 \)

- \( 25m^2 = (5m)^2 \), \( 9 = 3^2 \)
- So: \( (5m)^2 - 3^2 = (5m + 3)(5m - 3) \)

✔ Answer: \( (5m + 3)(5m - 3) \)

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20) \( 16v^2 - 9 \)

- \( 16v^2 = (4v)^2 \), \( 9 = 3^2 \)
- So: \( (4v)^2 - 3^2 = (4v + 3)(4v - 3) \)

✔ Answer: \( (4v + 3)(4v - 3) \)

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## ✔ Final Answer Key:

1) \( (a + 7)(a - 7) \)
2) \( (a + 8)(a - 8) \)
3) \( (p + 12)(p - 12) \)
4) \( (b + 5)(b - 5) \)
5) \( (x + 3)(x - 3) \)
6) \( (x + 2)(x - 2) \)
7) \( (k + 11)(k - 11) \)
8) \( (k + 6)(k - 6) \)
9) \( (n + 17)(n - 17) \)
10) \( (n + 13)(n - 13) \)
11) \( (2x + 5)(2x - 5) \)
12) \( (4b + 1)(4b - 1) \)
13) \( (3a + 2)(3a - 2) \)
14) \( (n + 4)(n - 4) \)
15) \( (3b + 5)(3b - 5) \)
16) \( (1 + a)(1 - a) \)
17) \( (4r + 5)(4r - 5) \)
18) \( (m + 3)(m - 3) \)
19) \( (5m + 3)(5m - 3) \)
20) \( (4v + 3)(4v - 3) \)

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💡 Tip: Always check if both terms are perfect squares and that there’s a minus sign between them. Then apply \( a^2 - b^2 = (a + b)(a - b) \). If coefficients are involved (like 4x², 9a², etc.), take the square root of the coefficient too!

Let me know if you want to practice more or need explanations for similar problems!