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Algebraic identities · 5 marks

Factor the following algebraic expressions:

  1. (i) 4y2 + 1 + 116y2
  2. (ii) 9m2 − 125n2
  3. (iii) 27b3 − 164b3
  4. (iv) x2 + 5x6 + 16
  5. (v) 27u3 − 1125 − 27u25 + 9u25
  6. (vi) 64y3 + 1125z3
  7. (vii) p3 + 27q3 + r3 − 9pqr
  8. (viii) 9m2 − 12m + 4
  9. (ix) 9x3 − 83y3 + z33 + 6xyz
  10. (x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy
  11. (xi) 27u3 − 1216 − 9u22 + u4
Answer: (i) (2y + 14y)2 (ii) (3m − 15n)(3m + 15n) (iii) (3b − 14b)(9b2 + 34 + 116b2) (iv) (x + 12)(x + 13) (v) (3u − 15)3 (vi) (4y + z5)(16y2 − 4yz5 + z225) (vii) (p + 3q + r)(p2 + 9q2 + r2 − 3pq − 3qr − rp) (viii) (3m − 2)2 (ix) 13(3x − 2y + z)(9x2 + 4y2 + z2 + 6xy + 2yz − 3zx) (x) as printed it does not factor; with the cross terms 12xy + 24xz it would be (2x + 3y + 6z)2 (xi) (3u − 16)3

Step-by-step solution

Idea: Factorising is expanding in reverse. Count the terms and look at their shape: three terms with two perfect squares → (a ± b)2; two squares with a minus between them → a2 − b2; two cubes → a3 ± b3; four terms with two cubes → (a ± b)3; three cubes and a term in xyz → a3 + b3 + c3 − 3abc; six terms with three squares → (a + b + c)2. Always check the middle (cross) terms before you write the answer.

(i) 4y2 + 1 + 116y2

  1. 4y2 = (2y)2 and 116y2 = (14y)2 are both perfect squares, so try a2 + 2ab + b2 = (a + b)2 with a = 2y, b = 14y.
  2. Check the middle term: 2ab = 2 × 2y × 14y = 4y4y = 1. ✓ It matches the 1 in the middle.½ mark
  3. So 4y2 + 1 + 116y2 = (2y + 14y)2
(2y + 14y)2

(ii) 9m2 − 125n2

  1. Two perfect squares with a minus between them, so use a2 − b2 = (a − b)(a + b).
  2. 9m2 = (3m)2 and 125n2 = (15n)2, so a = 3m, b = 15n.½ mark
  3. 9m2 − 125n2 = (3m − 15n)(3m + 15n)
(3m − 15n)(3m + 15n)

(iii) 27b3 − 164b3

  1. Two perfect cubes with a minus between them: 27b3 = (3b)3 and 164b3 = (14b)3. Use the difference of cubes. (The letter b is already in the question, so write the identity with capital letters.) A3 − B3 = (A − B)(A2 + AB + B2) with A = 3b, B = 14b.
  2. A2 = 9b2; AB = 3b × 14b = 34 (the b cancels); B2 = 116b2.½ mark
  3. 27b3 − 164b3 = (3b − 14b)(9b2 + 34 + 116b2)
(3b − 14b)(9b2 + 34 + 116b2)

(iv) x2 + 5x6 + 16

  1. Use (x + a)(x + b) = x2 + (a + b)x + ab, because the expression is x2 + (number)x + (number). We need a + b = 56 and ab = 16.
  2. Try a = 12, b = 13: a + b = 36 + 26 = 56 ✓ and ab = 16 ✓.½ mark
  3. (Another way to find them: 16(6x2 + 5x + 1) = 16(2x + 1)(3x + 1), which is the same thing.)
  4. x2 + 5x6 + 16 = (x + 12)(x + 13)
(x + 12)(x + 13)

(v) 27u3 − 1125 − 27u25 + 9u25

  1. Four terms, two of them perfect cubes: 27u3 = (3u)3 and 1125 = (15)3. Signs alternate when rearranged, so try (a − b)3 = a3 − 3a2b + 3ab2 − b3 with a = 3u, b = 15.
  2. Rearrange in falling powers of u: 27u3 − 27u25 + 9u25 − 1125.
  3. Check the middle terms: 3a2b = 3 × 9u2 × 15 = 27u25 ✓; 3ab2 = 3 × 3u × 125 = 9u25 ✓.½ mark
  4. So the expression = (3u − 15)3
(3u − 15)3

(vi) 64y3 + 1125z3

  1. Sum of two cubes: 64y3 = (4y)3 and 1125z3 = (z5)3. Use a3 + b3 = (a + b)(a2 − ab + b2) with a = 4y, b = z5.
  2. a2 = 16y2; ab = 4y × z5 = 4yz5; b2 = z225.½ mark
  3. 64y3 + 1125z3 = (4y + z5)(16y2 − 4yz5 + z225)
(4y + z5)(16y2 − 4yz5 + z225)

(vii) p3 + 27q3 + r3 − 9pqr

  1. Three cubes and a term in pqr, so use a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca).
  2. a = p, b = 3q (since 27q3 = (3q)3), c = r. Check: 3abc = 3 × p × 3q × r = 9pqr ✓.½ mark
  3. a2 + b2 + c2 = p2 + 9q2 + r2; ab = 3pq, bc = 3qr, ca = rp.
  4. = (p + 3q + r)(p2 + 9q2 + r2 − 3pq − 3qr − rp)
(p + 3q + r)(p2 + 9q2 + r2 − 3pq − 3qr − rp)

(viii) 9m2 − 12m + 4

  1. 9m2 = (3m)2 and 4 = 22, with a minus middle term, so try a2 − 2ab + b2 = (a − b)2 with a = 3m, b = 2.
  2. Check: 2ab = 2 × 3m × 2 = 12m ✓.½ mark
  3. 9m2 − 12m + 4 = (3m − 2)2
(3m − 2)2

(ix) 9x3 − 83y3 + z33 + 6xyz

  1. The fractions with 3 at the bottom are awkward, so first take 13 out as a common factor:
    9x3 − 83y3 + z33 + 6xyz = 13(27x3 − 8y3 + z3 + 18xyz)
  2. Inside the bracket: three cubes and a term in xyz, so use a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca). Take a = 3x, b = −2y (because −8y3 = (−2y)3), c = z.
  3. Check: −3abc = −3 × 3x × (−2y) × z = +18xyz ✓.½ mark
  4. a2 + b2 + c2 = 9x2 + 4y2 + z2. −ab = −(3x)(−2y) = +6xy; −bc = −(−2y)(z) = +2yz; −ca = −(z)(3x) = −3zx.
  5. = 13(3x − 2y + z)(9x2 + 4y2 + z2 + 6xy + 2yz − 3zx)
13(3x − 2y + z)(9x2 + 4y2 + z2 + 6xy + 2yz − 3zx)

(x) 4x2 + 9y2 + 36z2 + 12xz + 36yz + 24xy

  1. Six terms: three perfect squares 4x2 = (2x)2, 9y2 = (3y)2, 36z2 = (6z)2, and three cross terms. So the natural identity to try is (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca with a = 2x, b = 3y, c = 6z.
  2. The cross terms this would give are:
    2ab = 2 × 2x × 3y = 12xy
    2bc = 2 × 3y × 6z = 36yz
    2ca = 2 × 6z × 2x = 24xz
  3. But the expression as printed has 24xy and 12xz: the numbers in front of xy and xz are swapped. The 36yz matches, the other two do not. So, as printed, the expression is not (2x + 3y + 6z)2.
  4. Could signs fix it? No. The x and y parts of any square must be ±2x and ±3y (to give 4x2 and 9y2), so the xy term can only be 2 × 2 × 3 = ±12xy, never 24xy. A quick test also shows it is not a square: put x = 1, y = −1, z = 0. The expression gives 4 + 9 − 24 = −11, and a square can never be negative.
  5. In fact, this printed expression cannot be split into factors with rational coefficients at all. It is almost certainly a printing slip for the expression with 12xy and 24xz.
  6. For the intended expression 4x2 + 9y2 + 36z2 + 12xy + 36yz + 24xz, every cross term matches, so it factors as (2x + 3y + 6z)2.
As printed (12xz and 24xy) it is not a perfect square and does not factor. The intended expression 4x2 + 9y2 + 36z2 + 12xy + 36yz + 24xz = (2x + 3y + 6z)2.

(xi) 27u3 − 1216 − 9u22 + u4

  1. Two perfect cubes, 27u3 = (3u)3 and 1216 = (16)3, plus two more terms: try (a − b)3 = a3 − 3a2b + 3ab2 − b3 with a = 3u, b = 16.
  2. Rearranged: 27u3 − 9u22 + u4 − 1216.
  3. Check: 3a2b = 3 × 9u2 × 16 = 27u26 = 9u22 ✓; 3ab2 = 3 × 3u × 136 = 9u36 = u4 ✓.½ mark
  4. So the expression = (3u − 16)3
(3u − 16)3
(i) (2y + 1/(4y))² (ii) (3m − 1/(5n))(3m + 1/(5n)) (iii) (3b − 1/(4b))(9b² + 3/4 + 1/(16b²)) (iv) (x + 1/2)(x + 1/3) (v) (3u − 1/5)³ (vi) (4y + z/5)(16y² − 4yz/5 + z²/25) (vii) (p + 3q + r)(p² + 9q² + r² − 3pq − 3qr − rp) (viii) (3m − 2)² (ix) (1/3)(3x − 2y + z)(9x² + 4y² + z² + 6xy + 2yz − 3zx) (x) as printed, not a perfect square and not factorable; the intended 4x² + 9y² + 36z² + 12xy + 36yz + 24xz = (2x + 3y + 6z)² (xi) (3u − 1/6)³

Check: Multiply back or put in numbers. (viii) m = 1: 9 − 12 + 4 = 1 and (3 − 2)2 = 1 ✓. (iv) x = 1: 1 + 56 + 16 = 2 and 32 × 43 = 2 ✓. (vii) p = q = r = 1: 1 + 27 + 1 − 9 = 20 and (5)(1 + 9 + 1 − 3 − 3 − 1) = 5 × 4 = 20 ✓. (v) u = 0: −1125 and (−15)3 = −1125 ✓.

Answer to write in the exam

(i)

4y2 + 1 + 116y2 = (2y)2 + 2(2y)(14y) + (14y)2 [a2 + 2ab + b2 = (a + b)2]

∴ 4y2 + 1 + 116y2 = (2y + 14y)2

(ii)

9m2 − 125n2 = (3m)2 − (15n)2 [a2 − b2 = (a + b)(a − b)]

∴ 9m2 − 125n2 = (3m − 15n)(3m + 15n)

(iii)

27b3 − 164b3 = (3b)3 − (14b)3 [A3 − B3 = (A − B)(A2 + AB + B2)]

= (3b − 14b)[(3b)2 + (3b)(14b) + (14b)2]

∴ 27b3 − 164b3 = (3b − 14b)(9b2 + 34 + 116b2)

(iv)

x2 + 5x6 + 16 = x2 + (12 + 13)x + (12)(13) [(x + a)(x + b) = x2 + (a + b)x + ab]

∴ x2 + 5x6 + 16 = (x + 12)(x + 13)

(v)

27u3 − 1125 − 27u25 + 9u25 = 27u3 − 27u25 + 9u25 − 1125

= (3u)3 − 3(3u)2(15) + 3(3u)(15)2 − (15)3 [a3 − 3a2b + 3ab2 − b3 = (a − b)3]

∴ 27u3 − 1125 − 27u25 + 9u25 = (3u − 15)3

(vi)

64y3 + 1125z3 = (4y)3 + (z5)3 [a3 + b3 = (a + b)(a2 − ab + b2)]

= (4y + z5)[(4y)2 − (4y)(z5) + (z5)2]

∴ 64y3 + 1125z3 = (4y + z5)(16y2 − 4yz5 + z225)

(vii)

p3 + 27q3 + r3 − 9pqr = p3 + (3q)3 + r3 − 3(p)(3q)(r) [a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)]

= (p + 3q + r)[p2 + (3q)2 + r2 − p(3q) − (3q)r − rp]

∴ p3 + 27q3 + r3 − 9pqr = (p + 3q + r)(p2 + 9q2 + r2 − 3pq − 3qr − rp)

(viii)

9m2 − 12m + 4 = (3m)2 − 2(3m)(2) + 22 [a2 − 2ab + b2 = (a − b)2]

∴ 9m2 − 12m + 4 = (3m − 2)2

(ix)

9x3 − 83y3 + z33 + 6xyz = 13(27x3 − 8y3 + z3 + 18xyz)

= 13[(3x)3 + (−2y)3 + z3 − 3(3x)(−2y)(z)] [a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)]

= 13(3x − 2y + z)[(3x)2 + (−2y)2 + z2 − (3x)(−2y) − (−2y)z − z(3x)]

∴ 9x3 − 83y3 + z33 + 6xyz = 13(3x − 2y + z)(9x2 + 4y2 + z2 + 6xy + 2yz − 3zx)

(x)

4x2 = (2x)2, 9y2 = (3y)2, 36z2 = (6z)2

(2x + 3y + 6z)2 = 4x2 + 9y2 + 36z2 + 12xy + 36yz + 24xz [(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

Given expression has 24xy and 12xz, so it is not of this form.

At x = 1, y = −1, z = 0: value = 4 + 9 − 24 = −11, negative, so it is not a perfect square.

∴ As printed it does not factorise; the intended 4x2 + 9y2 + 36z2 + 12xy + 36yz + 24xz = (2x + 3y + 6z)2

(xi)

27u3 − 1216 − 9u22 + u4 = 27u3 − 9u22 + u4 − 1216

= (3u)3 − 3(3u)2(16) + 3(3u)(16)2 − (16)3 [a3 − 3a2b + 3ab2 − b3 = (a − b)3]

∴ 27u3 − 1216 − 9u22 + u4 = (3u − 16)3

Common mistakes that cost marks

  • Writing a3 + b3 = (a + b)(a2 + ab + b2). For a sum of cubes the middle sign is minus: (a + b)(a2 − ab + b2). For a difference it is plus.
  • Matching the squares but not checking the cross terms. In (x) the three squares fit (2x + 3y + 6z)2, but the cross terms do not; always check every 2ab-type term before writing the answer.
  • In (ix), taking b = 2y instead of −2y. Then −3abc comes out as −18xyz, which does not match +18xyz. The minus sign of −8y3 belongs inside the cube: (−2y)3.

How this can come in the exam

MCQ (1 mark)

One factor of 8x3 − 27y3 is

  1. (2x + 3y)
  2. (2x − 3y)
  3. (4x − 9y)
  4. (8x − 27y)
Show answer

(B) (2x − 3y)
8x3 − 27y3 = (2x)3 − (3y)3 = (2x − 3y)(4x2 + 6xy + 9y2).

Assertion–Reason (1 mark)

Assertion (A): 4x2 + 9y2 + 36z2 + 12xy + 36yz + 24xz = (2x + 3y + 6z)2.
Reason (R): (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca.

  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Show answer

(A) Both A and R are true, and R is the correct explanation of A.
R is the standard identity. With a = 2x, b = 3y, c = 6z: 2ab = 12xy, 2bc = 36yz, 2ca = 24xz, which is exactly A.

Try one yourself

Factorise 8a3 + 127.

Show answer

(2a)3 + (13)3 = (2a + 13)(4a2 − 2a3 + 19), using a3 + b3 = (a + b)(a2 − ab + b2).

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