We already know that x2 − y2 = (x − y)(x + y).
Further, we have verified that x3 − y3 = (x − y)(x2 + xy + y2).
Observe that x − y is a common factor of x2 − y2 and x3 − y3.
Do you think x − y is also a factor of x4 − y4?
Note that x4 − y4 = (x2)2 − (y2)2 = (x2 − y2) (x2 + y2).
Can you see how x − y is a factor of x4 − y4?
How about x5 − y5? Does this also have x − y as a factor?
- x4 − y4 Do you think x − y is also a factor of x4 − y4?
- x5 − y5 How about x5 − y5? Does this also have x − y as a factor?
Step-by-step solution
Idea: For x4 − y4, use the difference of squares twice. For x5 − y5, copy the pattern of x3 − y3 = (x − y)(x2 + xy + y2): the second bracket has all the terms of degree one less, with powers of x going down and powers of y going up.
x4 − y4 Do you think x − y is also a factor of x4 − y4?
- x4 − y4 = (x2)2 − (y2)2 = (x2 − y2)(x2 + y2)½ mark
- Use the difference of squares again on the first bracket: x2 − y2 = (x − y)(x + y).½ mark
- So x4 − y4 = (x − y)(x + y)(x2 + y2). Yes, x − y is a factor.½ mark
x5 − y5 How about x5 − y5? Does this also have x − y as a factor?
- Follow the pattern: try (x − y)(x4 + x3y + x2y2 + xy3 + y4).½ mark
- x × (…) = x5 + x4y + x3y2 + x2y3 + xy4
−y × (…) = −x4y − x3y2 − x2y3 − xy4 − y5½ mark - All middle terms cancel, leaving x5 − y5. So yes, x − y is a factor.½ mark
Check: x = 3, y = 1: x5 − y5 = 243 − 1 = 242, and (x − y) = 2 with 81 + 27 + 9 + 3 + 1 = 121; 2 × 121 = 242 ✓. Also, putting x = y makes x5 − y5 = 0, which is what we expect when x − y is a factor.
Answer to write in the exam
x4 − y4
x4 − y4 = (x2)2 − (y2)2 = (x2 − y2)(x2 + y2) [a2 − b2 = (a + b)(a − b)]
= (x − y)(x + y)(x2 + y2)
∴ Yes, x − y is a factor of x4 − y4
x5 − y5
(x − y)(x4 + x3y + x2y2 + xy3 + y4)
= x5 + x4y + x3y2 + x2y3 + xy4 − x4y − x3y2 − x2y3 − xy4 − y5
= x5 − y5
∴ Yes, x5 − y5 = (x − y)(x4 + x3y + x2y2 + xy3 + y4), so x − y is a factor
Common mistakes that cost marks
- Writing x4 − y4 = (x2 − y2)2. That square equals x4 − 2x2y2 + y4, not x4 − y4.
- Trying to factor x2 + y2 as (x + y)(x − y). A sum of two squares does not factor this way.
- Leaving out a term in the second bracket for x5 − y5; it needs all five terms x4, x3y, x2y2, xy3, y4.
How this can come in the exam
Which of the following is NOT a factor of a4 − 16?
- a − 2
- a + 2
- a2 + 4
- a − 4
Show answer
(D) a − 4
a4 − 16 = (a2 − 4)(a2 + 4) = (a − 2)(a + 2)(a2 + 4). a − 4 is not a factor.
Assertion (A): 75 − 25 is divisible by 5.
Reason (R): x − y is a factor of x5 − y5.
- Both A and R are true, and R is the correct explanation of A.
- Both A and R are true, but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Show answer
(A) Both A and R are true, and R is the correct explanation of A.
With x = 7, y = 2, x − y = 5 divides 75 − 25 = 16807 − 32 = 16775 = 5 × 3355. R explains A.
Try one yourself
Factorise x4 − 81 completely.
Show answer
(x2 − 9)(x2 + 9) = (x − 3)(x + 3)(x2 + 9).
More questions like this
- Exploring further, let us multiply (x + y + z) and (x2 + y2 + z2 − xy − xz − yz).
- The sum of three numbers is 10 and their product is 25. The sum of their squares is 38. Try to use the previous identity to find the sum of the cubes of these three numbers.
- Simplify the rational expression x2 − 7x + 125x2 + 5x − 100, assuming that 5x2 + 5x − 100 ≠ 0.
- Try to simplify the following rational expression:
36s2 − 12st + t2t2 + 2ts − 48s2 = (6s − t)2(___ + ___)(___ + ___) .
(Hint: Factor t2 + 2ts − 48s2 and simplify the rational expressions assuming that t2 + 2ts − 48s2 ≠ 0). - Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero: