This MCQ module is based on: Exploring Algebraic Identities — Exercises
TOPIC 12 OF 25
Exploring Algebraic Identities — Exercises
🎓 Class 9
Mathematics
CBSE
Theory
Ch 4 — Exploring Algebraic Identities
⏱ ~40 min
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This mathematics assessment will be based on: Exploring Algebraic Identities — Exercises
Targeting Class 9 level in Algebra, with Intermediate difficulty.
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4.9 Factorisation Using Algebraic Identities
Identities work in two directions. Reading from left to right, they let us expand products. Reading from right to left, they let us factorise polynomials. Mastery of both directions is the key to algebraic fluency.
| Pattern in Polynomial | Identity Used | Factored Form |
|---|---|---|
| \(a^2 + 2ab + b^2\) | I | \((a+b)^2\) |
| \(a^2 - 2ab + b^2\) | II | \((a-b)^2\) |
| \(a^2 - b^2\) | III | \((a+b)(a-b)\) |
| \(x^2 + (a+b)x + ab\) | IV | \((x+a)(x+b)\) |
| \(a^3 + b^3\) | VIII | \((a+b)(a^2 - ab + b^2)\) |
| \(a^3 - b^3\) | IX | \((a-b)(a^2 + ab + b^2)\) |
| \(a^3 + b^3 + c^3 - 3abc\) | X | \((a+b+c)(a^2+b^2+c^2-ab-bc-ca)\) |
Activity: Identity Detective
L4 AnalyseMaterials: The summary table above.
Predict: Given any quadratic or cubic polynomial, can you decide which identity to use just by looking at it?
- Look at how many terms the polynomial has.
- 2 terms: Difference of squares (Identity III) or sum/difference of cubes (VIII / IX).
- 3 terms: Probably \((a \pm b)^2\) (Identities I, II) — check whether the middle term is \(\pm 2\sqrt{\text{first} \cdot \text{last}}\). Or \((x+a)(x+b)\) form (Identity IV).
- 4 terms: Try grouping in pairs.
- Practice: classify which identity factorises (i) \(x^2 - 49\) (ii) \(x^2 + 12x + 36\) (iii) \(x^2 - x - 12\) (iv) \(8a^3 - 1\).
(i) Identity III → \((x-7)(x+7)\). (ii) Identity I → \((x+6)^2\). (iii) Identity IV with \(a=-4, b=3\) → \((x-4)(x+3)\). (iv) Identity IX → \((2a-1)(4a^2+2a+1)\).
Exercise 4.1 — Squaring & Difference Patterns
Q1. Use suitable identities to find the products:
(i) \((x+4)(x+10)\) (ii) \((x+8)(x-10)\) (iii) \((3x+4)(3x-5)\) (iv) \(\left(y^2 + \frac{3}{2}\right)\left(y^2 - \frac{3}{2}\right)\) (v) \((3-2x)(3+2x)\)
(i) \((x+4)(x+10)\) (ii) \((x+8)(x-10)\) (iii) \((3x+4)(3x-5)\) (iv) \(\left(y^2 + \frac{3}{2}\right)\left(y^2 - \frac{3}{2}\right)\) (v) \((3-2x)(3+2x)\)
(i) Identity IV: \(x^2 + 14x + 40\).
(ii) Identity IV: \(x^2 - 2x - 80\).
(iii) Use \((a+b)(a+c)\) form: \((3x)^2 + (4-5)(3x) + (4)(-5) = 9x^2 - 3x - 20\).
(iv) Identity III: \((y^2)^2 - \left(\frac{3}{2}\right)^2 = y^4 - \frac{9}{4}\).
(v) Identity III: \(9 - 4x^2\).
(ii) Identity IV: \(x^2 - 2x - 80\).
(iii) Use \((a+b)(a+c)\) form: \((3x)^2 + (4-5)(3x) + (4)(-5) = 9x^2 - 3x - 20\).
(iv) Identity III: \((y^2)^2 - \left(\frac{3}{2}\right)^2 = y^4 - \frac{9}{4}\).
(v) Identity III: \(9 - 4x^2\).
Q2. Evaluate using suitable identities:
(i) \(103 \times 107\) (ii) \(95 \times 96\) (iii) \(104 \times 96\)
(i) \(103 \times 107\) (ii) \(95 \times 96\) (iii) \(104 \times 96\)
(i) \((100+3)(100+7) = 10000 + 1000 + 21 = 11021\).
(ii) \((100-5)(100-4) = 10000 - 900 + 20 = 9120\).
(iii) \((100+4)(100-4) = 10000 - 16 = 9984\).
(ii) \((100-5)(100-4) = 10000 - 900 + 20 = 9120\).
(iii) \((100+4)(100-4) = 10000 - 16 = 9984\).
Q3. Factorise:
(i) \(9x^2 + 6xy + y^2\) (ii) \(36a^2 + 36a + 9\) (iii) \(\frac{x^2}{4} - \frac{y^2}{100}\)
(i) \(9x^2 + 6xy + y^2\) (ii) \(36a^2 + 36a + 9\) (iii) \(\frac{x^2}{4} - \frac{y^2}{100}\)
(i) \((3x)^2 + 2(3x)(y) + y^2 = (3x+y)^2\).
(ii) \((6a)^2 + 2(6a)(3) + 3^2 = (6a+3)^2 = 9(2a+1)^2\).
(iii) \(\left(\frac{x}{2}\right)^2 - \left(\frac{y}{10}\right)^2 = \left(\frac{x}{2}+\frac{y}{10}\right)\left(\frac{x}{2}-\frac{y}{10}\right)\).
(ii) \((6a)^2 + 2(6a)(3) + 3^2 = (6a+3)^2 = 9(2a+1)^2\).
(iii) \(\left(\frac{x}{2}\right)^2 - \left(\frac{y}{10}\right)^2 = \left(\frac{x}{2}+\frac{y}{10}\right)\left(\frac{x}{2}-\frac{y}{10}\right)\).
Q4. Expand using a suitable identity:
(i) \((x + 2y + 4z)^2\) (ii) \((2x - y + z)^2\) (iii) \((-2x + 3y + 2z)^2\)
(i) \((x + 2y + 4z)^2\) (ii) \((2x - y + z)^2\) (iii) \((-2x + 3y + 2z)^2\)
(i) Identity V: \(x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8zx\).
(ii) \(4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx\).
(iii) \(4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx\).
(ii) \(4x^2 + y^2 + z^2 - 4xy - 2yz + 4zx\).
(iii) \(4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8zx\).
Q5. Factorise: \(4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16xz\).
Match Identity V with \(a = 2x, b = 3y, c = -4z\): check \(2ab = 2(2x)(3y) = 12xy\) ✓, \(2bc = 2(3y)(-4z) = -24yz\) ✓, \(2ca = 2(-4z)(2x) = -16xz\) ✓. So the expression = \((2x + 3y - 4z)^2\).
Exercise 4.2 — Cube Identities and Factorisation
Q1. Write the following cubes in expanded form:
(i) \((2x + 1)^3\) (ii) \((2a - 3b)^3\) (iii) \(\left(\frac{3}{2}x + 1\right)^3\) (iv) \(\left(x - \frac{2}{3}y\right)^3\)
(i) \((2x + 1)^3\) (ii) \((2a - 3b)^3\) (iii) \(\left(\frac{3}{2}x + 1\right)^3\) (iv) \(\left(x - \frac{2}{3}y\right)^3\)
(i) \((2x+1)^3 = 8x^3 + 12x^2 + 6x + 1\).
(ii) \((2a)^3 - 3(2a)^2(3b) + 3(2a)(3b)^2 - (3b)^3 = 8a^3 - 36a^2b + 54ab^2 - 27b^3\).
(iii) \(\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1\).
(iv) \(x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3\).
(ii) \((2a)^3 - 3(2a)^2(3b) + 3(2a)(3b)^2 - (3b)^3 = 8a^3 - 36a^2b + 54ab^2 - 27b^3\).
(iii) \(\frac{27}{8}x^3 + \frac{27}{4}x^2 + \frac{9}{2}x + 1\).
(iv) \(x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3\).
Q2. Evaluate using suitable identities:
(i) \((99)^3\) (ii) \((102)^3\) (iii) \((998)^3\)
(i) \((99)^3\) (ii) \((102)^3\) (iii) \((998)^3\)
(i) \((100-1)^3 = 1000000 - 30000 + 300 - 1 = 970299\).
(ii) \((100+2)^3 = 1000000 + 60000 + 1200 + 8 = 1061208\).
(iii) \((1000-2)^3 = 10^9 - 6 \times 10^6 + 12000 - 8 = 994011992\).
(ii) \((100+2)^3 = 1000000 + 60000 + 1200 + 8 = 1061208\).
(iii) \((1000-2)^3 = 10^9 - 6 \times 10^6 + 12000 - 8 = 994011992\).
Q3. Factorise each of the following:
(i) \(8a^3 + b^3 + 12a^2b + 6ab^2\) (ii) \(8a^3 - b^3 - 12a^2b + 6ab^2\) (iii) \(27 - 125a^3 - 135a + 225a^2\) (iv) \(64a^3 - 27b^3 - 144a^2b + 108ab^2\)
(i) \(8a^3 + b^3 + 12a^2b + 6ab^2\) (ii) \(8a^3 - b^3 - 12a^2b + 6ab^2\) (iii) \(27 - 125a^3 - 135a + 225a^2\) (iv) \(64a^3 - 27b^3 - 144a^2b + 108ab^2\)
(i) Recognise Identity VI with \(2a, b\): \((2a+b)^3\).
(ii) Identity VII: \((2a - b)^3\).
(iii) Reorganise: \(27 - 135a + 225a^2 - 125a^3 = (3)^3 - 3(3)^2(5a) + 3(3)(5a)^2 - (5a)^3 = (3 - 5a)^3\).
(iv) Identity VII with \(4a, 3b\): \((4a - 3b)^3\).
(ii) Identity VII: \((2a - b)^3\).
(iii) Reorganise: \(27 - 135a + 225a^2 - 125a^3 = (3)^3 - 3(3)^2(5a) + 3(3)(5a)^2 - (5a)^3 = (3 - 5a)^3\).
(iv) Identity VII with \(4a, 3b\): \((4a - 3b)^3\).
Q4. Factorise: (i) \(27y^3 + 125z^3\) (ii) \(64m^3 - 343n^3\)
(i) Identity VIII: \((3y)^3 + (5z)^3 = (3y + 5z)(9y^2 - 15yz + 25z^2)\).
(ii) Identity IX: \((4m)^3 - (7n)^3 = (4m - 7n)(16m^2 + 28mn + 49n^2)\).
(ii) Identity IX: \((4m)^3 - (7n)^3 = (4m - 7n)(16m^2 + 28mn + 49n^2)\).
Q5. Verify that \(x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x+y+z)\left[(x-y)^2 + (y-z)^2 + (z-x)^2\right]\).
By Identity X: \(x^3+y^3+z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)\). Now multiply the bracket by 2:
\(2(x^2+y^2+z^2 - xy - yz - zx) = (x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2)\)
\( = (x-y)^2 + (y-z)^2 + (z-x)^2\). Hence the desired form. ✓
\(2(x^2+y^2+z^2 - xy - yz - zx) = (x^2 - 2xy + y^2) + (y^2 - 2yz + z^2) + (z^2 - 2zx + x^2)\)
\( = (x-y)^2 + (y-z)^2 + (z-x)^2\). Hence the desired form. ✓
Q6. If \(x + y + z = 0\), show that \(x^3 + y^3 + z^3 = 3xyz\).
By Identity X: \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(\ldots) = 0 \cdot (\ldots) = 0\). Therefore \(x^3 + y^3 + z^3 = 3xyz\). ✓
Q7. Without actually calculating, find the value of: (i) \((-12)^3 + (7)^3 + (5)^3\) (ii) \((28)^3 + (-15)^3 + (-13)^3\)
In each part the three numbers sum to 0.
(i) \(-12 + 7 + 5 = 0\), so the expression \(= 3(-12)(7)(5) = -1260\).
(ii) \(28 - 15 - 13 = 0\), so the expression \(= 3(28)(-15)(-13) = 16380\).
(i) \(-12 + 7 + 5 = 0\), so the expression \(= 3(-12)(7)(5) = -1260\).
(ii) \(28 - 15 - 13 = 0\), so the expression \(= 3(28)(-15)(-13) = 16380\).
Exercise 4.3 — Mixed Identities
Q1. Use suitable identities to expand and simplify:
(i) \(\left(x - \frac{1}{x}\right)^2\) (ii) \(\left(2x + \frac{1}{2x}\right)^2\) (iii) \((a + b - c)(a + b + c)\)
(i) \(\left(x - \frac{1}{x}\right)^2\) (ii) \(\left(2x + \frac{1}{2x}\right)^2\) (iii) \((a + b - c)(a + b + c)\)
(i) \(x^2 - 2 + \frac{1}{x^2}\). (ii) \(4x^2 + 2 + \frac{1}{4x^2}\). (iii) \(((a+b) - c)((a+b) + c) = (a+b)^2 - c^2 = a^2 + 2ab + b^2 - c^2\).
Q2. If \(a^2 + b^2 + c^2 = 50\) and \(ab + bc + ca = 47\), find \(a + b + c\).
By Identity V: \((a+b+c)^2 = 50 + 2(47) = 144\). Hence \(a+b+c = \pm 12\).
Q3. If \(a + b + c = 6\) and \(ab + bc + ca = 11\), find \(a^2 + b^2 + c^2\).
\(36 = a^2 + b^2 + c^2 + 2(11)\). So \(a^2 + b^2 + c^2 = 36 - 22 = 14\).
Q4. If \(x - \frac{1}{x} = 5\), find \(x^3 - \frac{1}{x^3}\).
Cube both sides: \(\left(x - \frac{1}{x}\right)^3 = x^3 - \frac{1}{x^3} - 3 \cdot 1 \cdot \left(x - \frac{1}{x}\right)\). So \(125 = x^3 - \frac{1}{x^3} - 3(5) = x^3 - \frac{1}{x^3} - 15\). Hence \(x^3 - \frac{1}{x^3} = 140\).
Chapter Summary
The Ten Algebraic Identities — Chapter 4
- I. \((a+b)^2 = a^2 + 2ab + b^2\)
- II. \((a-b)^2 = a^2 - 2ab + b^2\)
- III. \((a+b)(a-b) = a^2 - b^2\) — difference of squares
- IV. \((x+a)(x+b) = x^2 + (a+b)x + ab\) — quadratic factorisation
- V. \((a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\)
- VI. \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a+b)\)
- VII. \((a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a-b)\)
- VIII. \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) — sum of cubes
- IX. \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) — difference of cubes
- X. \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)\). Special case: if \(a+b+c = 0\), then \(a^3 + b^3 + c^3 = 3abc\)
Strategy Map for Factorisation
2 terms: Look for difference of squares (III) or sum/difference of cubes (VIII, IX).3 terms: Try perfect-square form \((a \pm b)^2\) (I, II); else use the product-sum technique (IV).
4 terms: Try grouping in pairs.
6 terms: Try Identity V — square of trinomial pattern.
Cubic, sum of three cubes: Identity X.
Competency-Based Questions
Scenario: A square garden of side \((p+q)\) metres is being designed by a landscape architect. Inside it, four equal flower beds will be made: two square beds (sides \(p\) and \(q\)) at opposite corners, and two rectangular paths (each \(p \times q\)) joining them — exactly modelling the area-decomposition of \((p+q)^2\).
Q1. Express the total area of the four sections, and verify it equals \((p+q)^2\).
L3 ApplyAnswer (b). \(p^2 + q^2 + 2 \cdot pq = (p+q)^2\). ✓ Identity I matches the geometry exactly.
Q2. Suppose the architect notices that doubling \(p\) (with \(q\) fixed at 4 m) makes the total garden area exactly four times the small square \(p^2\). Analyse: what value of \(p\) made this happen?
L4 AnalyseCondition: \((p+4)^2 = 4p^2\). Take square root (positive): \(p+4 = 2p\) → \(p = 4\). At \(p=4\): \((p+q)^2 = 64\) and \(4p^2 = 64\). ✓
Q3. Evaluate the cost of laying turf at ₹250 per m² over the rectangular paths only, when \(p = 8\) m and \(q = 5\) m.
L5 EvaluatePath area = \(2pq = 2(8)(5) = 80\) m². Cost = \(80 \times 250 = ₹20{,}000\).
Q4. Design a different garden of total area \(p^2 + 2pq + q^2\) using only THREE rectangular regions (not four). State dimensions and verify total area.
L6 CreateOne design: Split the side \((p+q)\) horizontally into two strips of heights \(p\) and \(q\). Top strip: \((p+q) \times p\). Bottom strip split into \(p \times q\) and \(q \times q\). Three regions: areas \(p(p+q), pq, q^2\). Sum: \(p^2 + pq + pq + q^2 = p^2 + 2pq + q^2 = (p+q)^2\) ✓. The same identity, three pieces instead of four.
Assertion–Reason Questions
Assertion (A): \(x^3 + 27 = (x + 3)(x^2 - 3x + 9)\).
Reason (R): \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\).
Reason (R): \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\).
(a) — Use \(a=x, b=3\). The factorisation in A directly applies the identity in R.
Assertion (A): If \(x + \frac{1}{x} = 3\), then \(x^3 + \frac{1}{x^3} = 18\).
Reason (R): \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\).
Reason (R): \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\).
(a) — Cubing: \(27 = x^3 + \frac{1}{x^3} + 3(1)(3)\), giving \(x^3 + \frac{1}{x^3} = 27 - 9 = 18\) ✓.
Assertion (A): The polynomial \(x^2 + 4x + 4\) has \((x+2)\) as a repeated factor.
Reason (R): \(x^2 + 4x + 4 = (x+2)^2\) by Identity I.
Reason (R): \(x^2 + 4x + 4 = (x+2)^2\) by Identity I.
(a) — \((x+2)^2 = (x+2)(x+2)\), so the factor \((x+2)\) appears twice. R explains A.
Frequently Asked Questions
How do you expand (3x + 2y)^2 using identities?
Use (a + b)^2 = a^2 + 2ab + b^2 with a = 3x and b = 2y: (3x + 2y)^2 = (3x)^2 + 2(3x)(2y) + (2y)^2 = 9x^2 + 12xy + 4y^2.
How do you factorise x^3 - 27 using cube identities?
Recognise 27 = 3^3, so x^3 - 27 = x^3 - 3^3. Apply a^3 - b^3 = (a - b)(a^2 + ab + b^2) with a = x and b = 3: x^3 - 27 = (x - 3)(x^2 + 3x + 9).
How do you compute 102 x 98 using algebraic identities?
Notice 102 = 100 + 2 and 98 = 100 - 2. Use (a + b)(a - b) = a^2 - b^2 with a = 100 and b = 2: 102 x 98 = (100 + 2)(100 - 2) = 100^2 - 2^2 = 10000 - 4 = 9996.
How do you factorise 8a^3 - 27b^3?
Write 8a^3 = (2a)^3 and 27b^3 = (3b)^3. Apply a^3 - b^3 = (a - b)(a^2 + ab + b^2) with a = 2a and b = 3b: 8a^3 - 27b^3 = (2a - 3b)((2a)^2 + (2a)(3b) + (3b)^2) = (2a - 3b)(4a^2 + 6ab + 9b^2).
If x + 1/x = 5, how do you find x^2 + 1/x^2?
Square both sides: (x + 1/x)^2 = 25. Expanding gives x^2 + 2 + 1/x^2 = 25, so x^2 + 1/x^2 = 23. The trick is to recognise the identity (a + b)^2 = a^2 + 2ab + b^2 with b = 1/a.
How do identities help in factorising polynomials in Class 9?
Identities convert recognisable patterns (squares, sums and differences of cubes) into ready-made factor forms. Once you spot a^2 - b^2, a^3 + b^3 or a^3 - b^3 inside a polynomial, factorisation becomes a single substitution.
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