Always x − y
x − y as a common factor
Two in a row
We know x² − y² = (x − y)(x + y) and x³ − y³ = (x − y)(x² + xy + y²). Both have the same first factor, x − y.
Does x⁴ − y⁴ also have x − y as a factor?
What this lesson covers
The idea
x – y is a common factor of x² – y² and x³ – y³, and also of x⁴ – y⁴ = (x² – y²)(x² + y²) and of x⁵ – y⁵.
Two in a row
We know x² − y² = (x − y)(x + y) and x³ − y³ = (x − y)(x² + xy + y²). Both have the same first factor, x − y.
Does x⁴ − y⁴ also have x − y as a factor?
- Yes
- No
- Only if x is bigger than y
Climb the ladder
Step up the ladder: n = 2, 3, 4, 5, 6. The second factor grows, but x − y stays. Then try your own x and y.
The factor that stays
x − y is a common factor of x² − y² and x³ − y³, and also of x⁴ − y⁴ = (x² − y²)(x² + y²) and of x⁵ − y⁵.
x⁴ − y⁴ = (x²)² − (y²)² = (x² − y²)(x² + y²), and x² − y² already has the factor x − y. In the same way, x⁵ − y⁵ also has x − y as a factor.
Notes
x − y is a common factor of x² − y², x³ − y³, x⁴ − y⁴ and x⁵ − y⁵.
Look at the ladder: the second factor of xⁿ − yⁿ has n terms.
Check yourself
Which of these has (x − y) as a factor?
Take x = 3 and y = 1. What is (x + y)(x² + y²)?
Answer: 40
4 × 10 = 40. Then x⁴ − y⁴ = (x − y)(x + y)(x² + y²) = 2 × 40 = 80, and 3⁴ − 1⁴ = 81 − 1 = 80.
Which is x⁴ − y⁴ written as a product?
x⁵ − y⁵ = (x − y)(…). How many terms does the second bracket have?
Answer: 5
The second bracket is x⁴ + x³y + x²y² + xy³ + y⁴: 5 terms.
- x² + y². This is a sum. Take x = y = 1: x − y = 0, but x² + y² = 2, and a multiple of 0 would be 0.
- x⁵ + y⁵. This is a sum too. Take x = y = 1: x − y = 0, but x⁵ + y⁵ = 2, and a multiple of 0 would be 0.
- x⁵ − y⁵ — correct. Yes. x − y is a factor of xⁿ − yⁿ for n = 2, 3, 4, 5.
- (x² − y²)(x² + y²) — correct. Yes: x⁴ − y⁴ = (x²)² − (y²)², and a² − b² = (a − b)(a + b).
- (x² − y²)(x² − y²). That is (x² − y²)² = x⁴ − 2x²y² + y⁴.
- (x² + y²)(x² + y²). That is (x² + y²)² = x⁴ + 2x²y² + y⁴.