Algebraic identities: Questions and Answers

65 algebraic identities questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• Consider any three consecutive square numbers. For example, 1, 4, and 9. Add the smallest and the largest squares. Thus, 1 + 9 = 10. Then subtract twice the middle square from this sum. This leads to 10 − (2 × 4) = 10 − 8 = 2. Now try the same process with another set of three consecutive square numbers. Say 9, 16, 25.Answer: For 9, 16, 25: (9 + 25) − (2 × 16) = 34 − 32 = 2. The answer is always 2, because (n − 1)² + (n + 1)² − 2n² = 2 for every number n.
• Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.Answer: For any 4 consecutive squares, (smallest + largest) − (the two middle squares) = 4, always. Proof: n² + (n + 3)² − (n + 1)² − (n + 2)² = 4.
• Think of numbers a and b where a and b do not represent lengths of line segments. What if a and b are negative numbers? Let us check for some negative numbers and see if this equation still works.Answer: (i) (a + b)² = 25 and a² + 2ab + b² = 4 + 12 + 9 = 25 — equal. (ii) (a + b)² = 1144 and a² + 2ab + b² = 1144 — equal. So (a + b)² = a² + 2ab + b² works for negative numbers and fractions too.
• What is the difference between an equation and an identity?Answer: An identity is an equation that is true for all values of its variables, e.g. (x + y)² = x² + 2xy + y². An equation need not be true for all values, e.g. x² − 1 = 24 is true only for x = 5 or −5.
• By now you must have observed that (a + b)² ≠ a² + b². But can you find out which one of them will be greater? For example, if a = 10 and b = 2, what are the values of (a + b)² and a² + b²?Answer: (a + b)² = 12² = 144 and a² + b² = 100 + 4 = 104, so here (a + b)² is greater, by 2ab = 40. It is not greater for all numbers: it depends on the sign of 2ab.
• 1. What can you say about a and b if (a + b)² < a² + b²?
  2. What can you say about a and b if (a + b)² > a² + b²?
  3. When will (a + b)² be equal to a² + b²?
  Did you observe that (a + b)² and a² + b² are both positive? What term will decide which is larger? Use the expansion of (a + b)² to decide.Answer: The deciding term is 2ab, because (a + b)² − (a² + b²) = 2ab. 1. a and b have opposite signs (ab < 0). 2. a and b have the same sign and neither is 0 (ab > 0). 3. When a = 0 or b = 0 (ab = 0).
• Let us try to expand (5x + 2y)².Answer: (5x + 2y)² = 25x² + 20xy + 4y²
• To calculate 43², we can write it as (40 + 3)².Answer: 43² = (40 + 3)² = 1600 + 240 + 9 = 1849
• Using the identity (a + b)² = a² + 2ab + b², expand the following:Answer: (i) 49x² + 56xy + 16y² (ii) 4925x² + 215xy + 94y² (iii) 6.25p² + 7.5pq + 2.25q² (iv) 916s² + 12st + 64t² (v) x² + xy + 14y² (vi) 1x² + 2xy + 1y²
• Using the same identity, find the values of the following:Answer: (i) 64² = 4096 (ii) 105² = 11025 (iii) 205² = 42025
• The identity (a + b)² = a² + 2ab + b² can also be used to find factors of some algebraic expressions. Consider the algebraic expression x² + 4x + 4.Answer: x² + 4x + 4 = (x + 2)², so (x + 2) is a factor.
• Let us try to find factors of another algebraic expression: 36x² + 12x + 1.Answer: 36x² + 12x + 1 = (6x + 1)², so (6x + 1) is a factor.
• Let us try to factor 50p² + 60pq + 18q². What will a and b be in this case?Answer: Take out the common factor 2 first: 50p² + 60pq + 18q² = 2(25p² + 30pq + 9q²) = 2(5p + 3q)², with a = 5p and b = 3q.
• What if we replace b by −b in (a + b)² = a² + 2ab + b²?Answer: We get (a − b)² = a² − 2ab + b², which is also an identity.
• Suppose we have to calculate 29². We can express this as (30 − 1)².Answer: 29² = (30 − 1)² = 900 − 60 + 1 = 841
• Factor completely:Answer: (i) (3x + 4y)² (ii) (2s + 5t)² (iii) (7x + 2y)² (iv) (8p + 23q)² (v) 3(a + 23b)² (vi) 5(35s + v)²
• Find the values of the following using the identity (a − b)² = a² − 2ab + b².Answer: (i) 79² = 6241 (ii) 193² = 37249 (iii) 299² = 89401
• What will happen if we want to find the square of the sum of three numbers a, b and c, that is, (a + b + c)²?Answer: (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca: the three squares plus twice each of the three products of pairs.
• Label the squares and rectangles in the figure so that it represents the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.Answer: Each piece’s area = (its width) × (its height). Diagonal pieces: a², b², c². Off-diagonal pieces: two ab, two bc, two ca. Total = a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)².
• Let us use this identity to find the square of a number, say 119:Answer: 119² = (100 + 10 + 9)² = 10000 + 100 + 81 + 2000 + 1800 + 180 = 14161
• Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.Answer: (i) 117² = 13689 (ii) 78² = 6084 (iii) 198² = 39204 (iv) 214² = 45796 (v) 1104² = 1218816 (vi) 1120² = 1254400
• Factor using suitable identities:Answer: (i) (4y − 3)² (ii) (32s + 2t)² (iii) (m3 + k2 + 3n)² (iv) (p4 − 4p)² (v) (3a − 2b + c)²
• Expand the following using the identity
  (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca:Answer: (i) p² + 9q² + 49r² + 6pq + 42qr + 14pr (ii) 9x² + 4y² + 16z² − 12xy − 16yz + 24xz
• Is this an identity?
  (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c².Answer: No, it is not an identity. The left side simplifies to 3a² + 3b² + 3c² − 2ab − 2bc − 2ca, which is not 2a² + 2b² + 2c². For example, with a = 1, b = 0, c = 0 the left side is 3 but the right side is 2.
• Look at the following figure. Justify the identity a² = (a + b) (a − b) + b² for yourself.Answer: Cut a strip of width b off the bottom of the a × a square. Move its (a − b) × b part to the side of what is left: this makes a rectangle (a + b) × (a − b). Only a b × b square is left over. So a² = (a + b)(a − b) + b².
• 1. Try to evaluate the following using a suitable identity:
  (i) 35² (ii) 65² (iii) 85² (iv) 105²
  Do you observe any interesting pattern?
  2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.Answer: 1. 35² = 1225, 65² = 4225, 85² = 7225, 105² = 11025 (using a² = (a + 5)(a − 5) + 25). Pattern: for a number ending in 5, multiply the digits before the 5 by the next number, then write 25. 2. (a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = (2a)² + (2b)² + (2c)².
• Suppose 7x is split as 2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.Answer: No. With 2 x-tiles on one side and 5 below, the corner needs a 2 × 5 block of 10 unit tiles, but x² + 7x + 12 has 12. Checking every split (1 + 6, 2 + 5, 3 + 4) shows only 3x + 4x works, because only 3 × 4 = 12. So x² + 7x + 12 = (x + 3)(x + 4).
• Algebra tiles can be used to represent products and find factors.
  1. Figure out the product of x + 2 and x + 3 using algebra tiles.
  2. Lay out algebra tiles for x² + 11x + 30 in such a way that you will see its factors.Answer: 1. (x + 2)(x + 3) = x² + 5x + 6 (one x²-tile, 2 + 3 = 5 x-tiles, 2 × 3 = 6 unit tiles). 2. Put 5 x-tiles beside the x²-tile, 6 below, and the 30 unit tiles in a 5 × 6 block: the rectangle has sides x + 5 and x + 6, so x² + 11x + 30 = (x + 5)(x + 6).
• We have seen that (x + 3)(x + 4) = x² + 7x + 12.
  Also (x + 6)(x + 7) = x² + 13x + 42.
  Generalise the pattern to get an expression for (x + a) (x + b).Answer: (x + a)(x + b) = x² + (a + b)x + ab: the coefficient of x is the sum of the two numbers and the constant is their product.
• Now consider the case where we have a rectangle of sidelengths 2x + 3 and 3x + 1, as shown in the figure. What can you say about its area (2x + 3) (3x + 1)?Answer: Counting tiles: 6 x²-tiles, 9 + 2 = 11 x-tiles, 3 unit tiles. So the area is (2x + 3)(3x + 1) = 6x² + 11x + 3.
• Fill in the blanks with the appropriate expressions to make the equation true.
  (px + a) (qx + b) = (_____)x² + (_____)x + _____ .
  Also, verify your answer using the distributive property.Answer: (px + a)(qx + b) = (pq)x² + (pb + qa)x + ab
• Let us begin with x² + 7x + 12 = x² + (a + b)x + ab.Answer: a + b = 7 and ab = 12 give a = 3, b = 4, so x² + 7x + 12 = (x + 3)(x + 4).
• Let us try to factor x² + 11x + 30 in a similar manner.Answer: Need a + b = 11 and ab = 30: a = 5, b = 6. So x² + 11x + 30 = (x + 5)(x + 6).
• In order to factor x² − 5x + 6, we first note that the coefficient of x is negative.Answer: a + b = −5 and ab = 6 give a = −2, b = −3. So x² − 5x + 6 = (x − 2)(x − 3).
• Fill in the blanks to complete the following identities:Answer: (i) (s − 3)(s − 8) (ii) (3x − 7) (iii) (2x − 3)(5x + 2) (iv) (3x + 2)(2x + 1)
• Select and use the identity that will help you to find the following products without multiplying directly:Answer: (i) 1681 (ii) 729 (iii) 391 (iv) 18225 (v) 9409 (vi) 522 (vii) 1462 (viii) 42025
• Factor the following:Answer: (i) (3a − b + 2c)² (ii) (4s − 5t)² (iii) (r − 7)(r + 6) (iv) (7g + h)² (v) (8u − 11v − 2w)²
• James and Reshma were talking about algebraic identities they learnt in school.
  James: (a − b)² (a + b) = (a² − 2ab + b²)(a + b)
  Reshma: I have a different idea. (a − b)² (a + b) = (a − b) [(a − b) (a + b)] = (a − b)(a² − b²)
  I will find this product to get the answer.
  According to you, who is correct and why?
  Try to combine more such identities and find new results.Answer: Both are correct. Both ways give (a − b)²(a + b) = a³ − a²b − ab² + b³. They just group the same three brackets differently; Reshma’s way is a little shorter (4 products instead of 6).
• What do you think (a + b)³ will look like?Answer: (a + b)³ = a³ + 3a²b + 3ab² + b³
• What if we have a cube of edge a + b? Can we divide a cube of edge (a + b) into smaller cubes and cuboids and represent this new identity?Answer: Yes. Cutting each edge into a and b splits the cube into 8 pieces: one cube a³, one cube b³, three cuboids a × a × b (3a²b) and three cuboids a × b × b (3ab²). So (a + b)³ = a³ + 3a²b + 3ab² + b³.
• What happens when we replace b with −b in this new identity?Answer: We get (a − b)³ = a³ − 3a²b + 3ab² − b³. The signs alternate: +, −, +, −.
• What is the side of the cube whose volume is p³ + 6p²q + 12pq² + 8q³ cubic units?Answer: p³ + 6p²q + 12pq² + 8q³ = (p + 2q)³, so the side is p + 2q units.
• Now consider the expression 8n³ − 60n²m + 150nm² − 125m³. If you write it in the form (a − b)³, what will be a and b?Answer: 8n³ − 60n²m + 150nm² − 125m³ = (2n − 5m)³, so a = 2n, b = 5m.
• Now let us play with known identities to discover more identities. Try to multiply the following using the distributive property.Answer: 1. (x − y)(x² + xy + y²) = x³ − y³. 2. (x + y)(x² − xy + y²) = x³ + y³. Both are identities.
• 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?Answer: Yes to both. x⁴ − y⁴ = (x − y)(x + y)(x² + y²). x⁵ − y⁵ = (x − y)(x⁴ + x³y + x²y² + xy³ + y⁴). In fact x − y is a factor of xⁿ − yⁿ for every natural number n.
• Exploring further, let us multiply (x + y + z) and (x² + y² + z² − xy − xz − yz).Answer: (x + y + z)(x² + y² + z² − xy − xz − yz) = x³ + y³ + z³ − 3xyz
• 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.Answer: First xy + yz + zx = (100 − 38) ÷ 2 = 31. Then x³ + y³ + z³ = 10(38 − 31) + 3(25) = 70 + 75 = 145.
• Simplify the rational expression x² − 7x + 125x² + 5x − 100, assuming that 5x² + 5x − 100 ≠ 0.Answer: x² − 7x + 125x² + 5x − 100 = (x − 3)(x − 4)5(x − 4)(x + 5) = x − 35(x + 5)
• 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).Answer: Blanks: (t + 8s)(t + (−6s)), i.e. t² + 2ts − 48s² = (t + 8s)(t − 6s). Since (6s − t)² = (t − 6s)², the expression simplifies to t − 6st + 8s.
• Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:Answer: (i) 3(p − 3q)(p + 2q)(p + 5q)(p − 2q) (no common factor, so it does not reduce) (ii) n − m5 (iii) w² + v² + x² + wv + vx − wxw − v + x (iv) 5z − 2y5z + 2y (v) 1 (vi) (p + 2)(p² + 4)p − 2
• Saira has arranged a square of side x units, 8 rectangular strips of sides x units and width 1 unit, and 15 squares of side 1 unit to form a bigger rectangle. Find the length and breadth of the rectangle in terms of x.Answer: Area = x² + 8x + 15 = (x + 5)(x + 3). So length = x + 5 units and breadth = x + 3 units.
• A rectangular pool is such that its breadth is 4 metres less than its length and its area is 96 sq. metres. Find the length and breadth of the pool.Answer: x(x − 4) = 96 gives (x − 12)(x + 8) = 0, so x = 12 (length cannot be −8). Length = 12 m, breadth = 8 m.
• Use suitable identities to find the following products:Answer: (i) 9x² − 24x + 16 (ii) 4s² − 49 (iii) p⁴ − 14 (iv) 4n² − 49 (v) s³ − 8t³ (vi) 14r² − 4 + 16r² (vii) 9m² + 16k² + l² − 24mk − 8kl + 6ml (viii) x³ − x²y + 13xy² − 127y³ (ix) 3438k³ − 492k²m + 143km² − 827m³
• Find the values using suitable identities:Answer: (i) 357 (ii) 9984 (iii) 384 (iv) 3176523 (v) 7880599 (vi) 2048383 (vii) −1225043 (viii) −26730899
• Factor the following algebraic expressions: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
• Simplify the following:
  Note: Assume that the denominators are not equal to 0.Answer: (i) 2x + 12x − 1 (ii) 3(a² + 2ab + 4b²)a + 2b (iii) s² − 5st + 25t²s − 7t
• Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.Answer: (i) 25a² − 30ab + 9b² = (5a − 3b)², so length = breadth = (5a − 3b) units (the rectangle is a square). (ii) 36s² − 49t² = (6s + 7t)(6s − 7t), so length = (6s + 7t) units and breadth = (6s − 7t) units.
• Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.Answer: (i) 6a² − 24b² = 6(a − 2b)(a + 2b): dimensions 6, (a − 2b) and (a + 2b) units. (ii) 3ps² − 15ps + 12p = 3p(s − 1)(s − 4): dimensions 3p, (s − 1) and (s − 4) units.
• The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.Answer: Area of the path = (40 + 2s)² − 40² = 4s² + 160s = 4s(s + 40) m².
• If a number plus its reciprocal equals 103, find the number.Answer: The number is 3 or 13. Both work: 3 + 13 = 103 and 13 + 3 = 103.
• A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.Answer: 2x² + 7x + 3 = (2x + 1)(x + 3), so the length is (x + 3) hastas.
• If both x − 2 and x − 12 are factors of px² + 5x + r, show that p = r.Answer: Putting x = 2 and x = 12 gives 4p + r + 10 = 0 and p + 4r + 10 = 0. Subtracting, 3p − 3r = 0, so p = r (in fact p = r = −2).
• If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = −25.Answer: a³ + b³ + c³ − 3abc = (a + b + c)[(a + b + c)² − 3(ab + bc + ca)] = 5(25 − 30) = −25.
• By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.Answer: n³ − n = (n − 1)n(n + 1), a product of three consecutive integers. One of them is even and one is a multiple of 3, so the product is divisible by 2 × 3 = 6.
• Find the value ofAnswer: (i) 0 (ii) 0. Both are of the form a³ + b³ + c³ − 3abc with a + b + c = 0.

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