Comprehensive worksheet titled 'Equations 13' designed to help students practice factorizing expressions using the difference of two squares method.

You're asked to factorise a set of equations using the difference of two squares.

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## 🔷 What is the Difference of Two Squares?

The identity is:

> a² – b² = (a – b)(a + b)

This works when you have two perfect squares separated by a minus sign.

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We’ll solve each section step-by-step.

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# ✔ SECTION A: Factorise using difference of two squares

All equations are in the form `x² - constant = 0`. We write the constant as a square, then apply the identity.

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1) x² – 25 = 0

→ 25 = 5²
→ x² – 5² = (x – 5)(x + 5)
✔ Answer: (x – 5)(x + 5)

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2) x² – 49 = 0

→ 49 = 7²
→ x² – 7² = (x – 7)(x + 7)
✔ Answer: (x – 7)(x + 7)

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3) x² – 100 = 0

→ 100 = 10²
→ x² – 10² = (x – 10)(x + 10)
✔ Answer: (x – 10)(x + 10)

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4) x² – 225 = 0

→ 225 = 15²
→ x² – 15² = (x – 15)(x + 15)
✔ Answer: (x – 15)(x + 15)

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5) 2x² – 32 = 0

⚠️ First, factor out the common factor 2:

→ 2(x² – 16) = 0
→ 16 = 4² → x² – 4² = (x – 4)(x + 4)
→ So overall: 2(x – 4)(x + 4)
✔ Answer: 2(x – 4)(x + 4)

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6) 5x² – 45 = 0

Factor out 5:

→ 5(x² – 9) = 0
→ 9 = 3² → x² – 3² = (x – 3)(x + 3)
→ So: 5(x – 3)(x + 3)
✔ Answer: 5(x – 3)(x + 3)

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7) 4x² – 144 = 0

Factor out 4:

→ 4(x² – 36) = 0
→ 36 = 6² → x² – 6² = (x – 6)(x + 6)
→ So: 4(x – 6)(x + 6)
✔ Answer: 4(x – 6)(x + 6)

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8) 7x² – 567 = 0

Factor out 7:

→ 7(x² – 81) = 0
→ 81 = 9² → x² – 9² = (x – 9)(x + 9)
→ So: 7(x – 9)(x + 9)
✔ Answer: 7(x – 9)(x + 9)

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# ✔ SECTION B: Factorise using difference of two squares

These involve different variables and coefficients. Again, look for perfect squares.

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1) 4a² – 9 = 0

→ 4a² = (2a)², 9 = 3²
→ (2a)² – 3² = (2a – 3)(2a + 3)
✔ Answer: (2a – 3)(2a + 3)

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2) 36s² – 121 = 0

→ 36s² = (6s)², 121 = 11²
→ (6s)² – 11² = (6s – 11)(6s + 11)
✔ Answer: (6s – 11)(6s + 11)

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3) 64 – p² = 0

→ 64 = 8², p² = p²
→ 8² – p² = (8 – p)(8 + p)
✔ Answer: (8 – p)(8 + p)

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4) 25 – 16c² = 0

→ 25 = 5², 16c² = (4c)²
→ 5² – (4c)² = (5 – 4c)(5 + 4c)
✔ Answer: (5 – 4c)(5 + 4c)

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5) x² – 9y² = 0

→ x² = x², 9y² = (3y)²
→ x² – (3y)² = (x – 3y)(x + 3y)
✔ Answer: (x – 3y)(x + 3y)

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6) 63 – 7q² = 0

⚠️ First, factor out 7:

→ 7(9 – q²) = 0
→ 9 = 3² → 3² – q² = (3 – q)(3 + q)
→ So: 7(3 – q)(3 + q)
✔ Answer: 7(3 – q)(3 + q)

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7) 196 – 4v² = 0

Factor out 4:

→ 4(49 – v²) = 0
→ 49 = 7² → 7² – v² = (7 – v)(7 + v)
→ So: 4(7 – v)(7 + v)
✔ Answer: 4(7 – v)(7 + v)

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8) 4x² – 121y² = 0

→ 4x² = (2x)², 121y² = (11y)²
→ (2x)² – (11y)² = (2x – 11y)(2x + 11y)
✔ Answer: (2x – 11y)(2x + 11y)

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# ✔ EXTENSION: Factorise 2a³b – 8ab³

This looks more complex — let’s break it down.

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Step 1: Look for a common factor

Both terms have:
- 2
- a
- b

So factor out 2ab:

→ 2ab(a² – 4b²)

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Step 2: Now factor the inside: a² – 4b²

→ 4b² = (2b)²
→ a² – (2b)² = (a – 2b)(a + 2b)

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Final Answer:

→ 2ab(a – 2b)(a + 2b)

✔ Answer: 2ab(a – 2b)(a + 2b)

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## 🧠 Summary

You used the difference of two squares formula:
> A² – B² = (A – B)(A + B)

Sometimes you had to factor out a common term first, then apply the identity.

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## 📝 All Answers:

Section A:

1) (x – 5)(x + 5)
2) (x – 7)(x + 7)
3) (x – 10)(x + 10)
4) (x – 15)(x + 15)
5) 2(x – 4)(x + 4)
6) 5(x – 3)(x + 3)
7) 4(x – 6)(x + 6)
8) 7(x – 9)(x + 9)

Section B:

1) (2a – 3)(2a + 3)
2) (6s – 11)(6s + 11)
3) (8 – p)(8 + p)
4) (5 – 4c)(5 + 4c)
5) (x – 3y)(x + 3y)
6) 7(3 – q)(3 + q)
7) 4(7 – v)(7 + v)
8) (2x – 11y)(2x + 11y)

Extension:

2ab(a – 2b)(a + 2b)

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Let me know if you want to solve for x or simplify further! 😊