1
Basics — Building Blocks
What is a Variable vs a Constant?
A variable (x, y, z, t…) can take any real value — it changes. A constant (a, b, 2, π…) stays fixed in a given problem. When we write ax, 'a' is the unknown constant and 'x' is the variable.
Polynomial in one variable: An algebraic expression of the form
anxⁿ + an−1xⁿ⁻¹ + … + a₁x + a₀
where all exponents of x are whole numbers (0, 1, 2, 3, …) and coefficients are constants with an ≠ 0.
anxⁿ + an−1xⁿ⁻¹ + … + a₁x + a₀
where all exponents of x are whole numbers (0, 1, 2, 3, …) and coefficients are constants with an ≠ 0.
Why some expressions are NOT polynomials
| Expression | Reason NOT a polynomial |
|---|---|
x + 1/x | x⁻¹ has exponent −1 (not a whole number) |
√x + 3 | x^(1/2) has exponent 1/2 (not a whole number) |
∛y + y² | y^(1/3) has exponent 1/3 (not a whole number) |
y + 2/y | 2y⁻¹ has negative exponent |
x¹⁰ + y³ + t⁵⁰ | THREE variables — not a polynomial in one variable |
Key Terminology
Terms and Coefficients
In −x³ + 4x² + 7x − 2:
Terms → −x³, 4x², 7x, −2
Coefficient of x³ = −1 | Coefficient of x² = 4 | Coefficient of x = 7 | Coefficient of x⁰ = −2
Degree of a Polynomial
The degree is the highest power of the variable in the polynomial.
e.g., degree of 3x⁷ − 4x⁶ + x + 9 is 7.
The degree of a non-zero constant is 0.
The degree of the zero polynomial is not defined.
e.g., degree of 3x⁷ − 4x⁶ + x + 9 is 7.
The degree of a non-zero constant is 0.
The degree of the zero polynomial is not defined.
Classification by Number of Terms
Monomial
1 term
2x, −5x², 7
Binomial
2 terms
x+1, x²−x
Trinomial
3 terms
x²+x+π
Classification by Degree
Linear
Degree 1
ax+b (a≠0)
Quadratic
Degree 2
ax²+bx+c (a≠0)
Cubic
Degree 3
ax³+bx²+cx+d (a≠0)
Constant
Degree 0
5, −3, π
Max terms rule: Linear → at most 2 terms | Quadratic → at most 3 terms | Cubic → at most 4 terms | Degree n → at most (n+1) terms.
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All Concepts
2.1 Zeroes of a Polynomial
A real number c is a zero of the polynomial p(x) if p(c) = 0.
Equivalently, c is a root of the equation p(x) = 0.
Equivalently, c is a root of the equation p(x) = 0.
Key Facts About Zeroes
- A zero of a polynomial need not be 0 itself.
- 0 may be a zero of a polynomial (e.g., p(x) = x² − 2x → p(0) = 0).
- Every linear polynomial ax + b has exactly one zero: x = −b/a.
- A polynomial can have more than one zero (e.g., x² − 2x has zeroes 0 and 2).
- A non-zero constant polynomial has NO zero.
- Every real number is a zero of the zero polynomial.
Zero of linear polynomial ax + b → x = −b/a
2.2 Remainder Theorem
If a polynomial p(x) of degree ≥ 1 is divided by a linear polynomial (x − a), then the remainder = p(a).
How to Apply
- Identify the linear divisor (x − a). Find a (zero of divisor).
- Substitute x = a in p(x).
- The value p(a) is the remainder — no long division needed!
Division by (x − a): p(x) = (x − a)·q(x) + p(a). If p(a) = 0, then (x − a) divides p(x) exactly.
2.3 Factor Theorem
(x − a) is a factor of p(x) ⟺ p(a) = 0
Both directions: (i) if p(a) = 0 then (x − a) is a factor; (ii) if (x − a) is a factor then p(a) = 0.
Both directions: (i) if p(a) = 0 then (x − a) is a factor; (ii) if (x − a) is a factor then p(a) = 0.
Proof Outline
By the Remainder Theorem: p(x) = (x − a)·q(x) + p(a).
If p(a) = 0 → p(x) = (x − a)·q(x) → (x − a) is a factor. ✓
If (x − a) is a factor → p(x) = (x − a)·g(x) → p(a) = (a−a)·g(a) = 0. ✓
2.4 Factorisation Strategy
For Quadratic ax² + bx + c
- Splitting middle term: Find p, q such that p + q = b and p × q = ac. Split bx = px + qx, then group and factor.
- Factor Theorem method: Try rational roots ±(factors of c)/(factors of a). If p(a) = 0, then (x − a) is a factor. Divide to get the other factor.
For Cubic p(x)
- Try ±1, ±2, ±3, … (factors of constant term) until you find one root 'a'.
- Extract (x − a) by grouping or synthetic/long division.
- Factorise the resulting quadratic.
2.5 Algebraic Identities (All 8)
Identity I
(x + y)² = x² + 2xy + y²
Identity II
(x − y)² = x² − 2xy + y²
Identity III
x² − y² = (x + y)(x − y)
Identity IV
(x+a)(x+b) = x²+(a+b)x+ab
Identity V
(x+y+z)² = x²+y²+z²+2xy+2yz+2zx
Identity VI
(x+y)³ = x³+y³+3xy(x+y)
Identity VII
(x−y)³ = x³−y³−3xy(x−y)
Identity VIII
x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²−xy−yz−zx)
Special cases of Identity VIII:
If x + y + z = 0, then x³ + y³ + z³ = 3xyz.
Also: x³ + y³ = (x + y)(x² − xy + y²) and x³ − y³ = (x − y)(x² + xy + y²).
If x + y + z = 0, then x³ + y³ + z³ = 3xyz.
Also: x³ + y³ = (x + y)(x² − xy + y²) and x³ − y³ = (x − y)(x² + xy + y²).
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Examples — Step by Step
Example 1 — Degree of a polynomial
Find the degree of: (i) x⁵ − x⁴ + 3 (ii) 2 − y² − y³ + 2y⁸ (iii) 2
Solution
(i) Terms: x⁵, −x⁴, 3. Highest exponent = 5. Degree = 5.
(ii) Terms: 2, −y², −y³, 2y⁸. Highest exponent = 8. Degree = 8.
(iii) 2 = 2·x⁰. Exponent of x is 0. Degree = 0.
Example 2 — Value of a polynomial
(i) p(x) = 5x² − 3x + 7 at x = 1 (ii) q(y) = 3y³ − 4y + √11 at y = 2 (iii) p(t) = 4t⁴ + 5t³ − t² + 6 at t = a
Solution
p(1) = 5(1)² − 3(1) + 7 = 5 − 3 + 7 = 9
q(2) = 3(8) − 4(2) + √11 = 24 − 8 + √11 = 16 + √11
p(a) = 4a⁴ + 5a³ − a² + 6 (substitute t = a directly)
Example 3 — Verify zeroes
Check whether −2 and 2 are zeroes of p(x) = x + 2.
Solution
p(2) = 2 + 2 = 4 ≠ 0 → 2 is NOT a zero.
p(−2) = −2 + 2 = 0 → −2 IS a zero.
Example 4 — Find zero of linear polynomial
Find the zero of p(x) = 2x + 1.
Solution
Set p(x) = 0 → 2x + 1 = 0
2x = −1 → x = −1/2. Zero is −1/2.
Example 5 — Verify two zeroes of a quadratic
Verify whether 2 and 0 are zeroes of p(x) = x² − 2x.
Solution
p(2) = 4 − 4 = 0 ✓ → 2 is a zero.
p(0) = 0 − 0 = 0 ✓ → 0 is a zero.
Both 2 and 0 are zeroes of x² − 2x.
Example 6 — Factor Theorem check
Is (x + 2) a factor of x³ + 3x² + 5x + 6? Also of 2x + 4?
Solution
Zero of (x + 2) is −2. Compute p(−2) = (−2)³ + 3(4) + 5(−2) + 6 = −8 + 12 − 10 + 6 = 0.
Since p(−2) = 0, by Factor Theorem, (x + 2) is a factor of x³ + 3x² + 5x + 6.
s(−2) = 2(−2) + 4 = 0 → (x + 2) is also a factor of 2x + 4. [Also: 2x+4 = 2(x+2)]
Example 7 — Find k using Factor Theorem
Find k if (x − 1) is a factor of 4x³ + 3x² − 4x + k.
Solution
If (x − 1) is a factor, then p(1) = 0.
p(1) = 4(1) + 3(1) − 4(1) + k = 4 + 3 − 4 + k = 3 + k = 0.
k = −3.
Example 8 — Factorise quadratic (two methods)
Factorise 6x² + 17x + 5.
Method 1: Splitting the Middle Term
Need p, q: p + q = 17 and p × q = 6 × 5 = 30. Pair: 2 and 15 (2 + 15 = 17, 2 × 15 = 30).
6x² + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1)(2x + 5).
Method 2: Factor Theorem
Try p(−1/3): = 6/9 − 17/3 + 5 = 0 ✓. So (x + 1/3) → (3x + 1) is a factor.
Divide → other factor is (2x + 5). Answer: (3x + 1)(2x + 5).
Example 9 — Factorise quadratic using Factor Theorem
Factorise y² − 5y + 6.
Solution
p(2) = 4 − 10 + 6 = 0 → (y − 2) is a factor.
p(3) = 9 − 15 + 6 = 0 → (y − 3) is a factor.
y² − 5y + 6 = (y − 2)(y − 3).
Example 10 — Factorise cubic polynomial
Factorise x³ − 23x² + 142x − 120.
Solution
Try p(1) = 1 − 23 + 142 − 120 = 0 ✓. So (x − 1) is a factor.
Group: x²(x−1) − 22x(x−1) + 120(x−1) = (x−1)(x² − 22x + 120).
Factorise x² − 22x + 120: split → −12x − 10x → x(x−12) −10(x−12) = (x−12)(x−10).
Final: (x−1)(x−10)(x−12).
Example 11 — Identities for products
(i) (x+3)(x+3) (ii) (x−3)(x+5)
Solution
(i) (x+3)² = x² + 2(3)x + 9 = x² + 6x + 9. [Identity I]
(ii) (x+a)(x+b) with a=−3, b=5: x² + (−3+5)x + (−3)(5) = x² + 2x − 15. [Identity IV]
Example 12 — Mental multiplication
Evaluate 105 × 106 without direct multiplication.
Solution
Write as (100+5)(100+6). Apply Identity IV with x=100, a=5, b=6.
= 100² + (5+6)(100) + 5×6 = 10000 + 1100 + 30 = 11130.
Example 13 — Factorise using identities
(i) 49a² + 70ab + 25b² (ii) (25/4)x² − y²/9
Solution
(i) 49a² = (7a)², 25b² = (5b)², 70ab = 2(7a)(5b). This is (7a+5b)² → (7a+5b)(7a+5b). [Identity I]
(ii) (5x/2)² − (y/3)² = (5x/2 + y/3)(5x/2 − y/3). [Identity III]
Example 14 — Expand trinomial square
Write (3a + 4b + 5c)² in expanded form.
Solution — Identity V with x=3a, y=4b, z=5c
(3a)² + (4b)² + (5c)² + 2(3a)(4b) + 2(4b)(5c) + 2(5c)(3a)
= 9a² + 16b² + 25c² + 24ab + 40bc + 30ca
Example 15 — Expand with negatives
Expand (4a − 2b − 3c)².
Solution — Write as [4a + (−2b) + (−3c)]², use Identity V
(4a)² + (−2b)² + (−3c)² + 2(4a)(−2b) + 2(−2b)(−3c) + 2(−3c)(4a)
= 16a² + 4b² + 9c² − 16ab + 12bc − 24ca
Example 16 — Reverse Identity V (Factorisation)
Factorise 4x² + y² + z² − 4xy − 2yz + 4xz.
Solution
Rewrite: (2x)² + (−y)² + (z)² + 2(2x)(−y) + 2(−y)(z) + 2(2x)(z).
This matches [2x + (−y) + z]² = (2x − y + z)². [Identity V]
Example 17 — Expand cubes
(i) (3a + 4b)³ (ii) (5p − 3q)³
Solution
(i) x=3a, y=4b. (3a)³ + (4b)³ + 3(3a)(4b)(3a+4b) = 27a³ + 64b³ + 36ab(3a+4b) = 27a³ + 64b³ + 108a²b + 144ab².
(ii) x=5p, y=3q. (5p)³ − (3q)³ − 3(5p)(3q)(5p−3q) = 125p³ − 27q³ − 225p²q + 135pq².
Example 18 — Evaluate using cube identities
(i) (104)³ (ii) (999)³
Solution
(i) (100+4)³ = 100³ + 4³ + 3(100)(4)(104) = 1000000 + 64 + 124800 = 1124864.
(ii) (1000−1)³ = 1000³ − 1³ − 3(1000)(1)(999) = 1000000000 − 1 − 2997000 = 997002999.
Example 19 — Factorise using Identity VI
Factorise 8x³ + 27y³ + 36x²y + 54xy².
Solution
Rewrite: (2x)³ + (3y)³ + 3(2x)²(3y) + 3(2x)(3y)².
This matches (2x+3y)³ → (2x+3y)(2x+3y)(2x+3y). [Identity VI]
Example 20 — Factorise using Identity VIII
Factorise 8x³ + y³ + 27z³ − 18xyz.
Solution
= (2x)³ + y³ + (3z)³ − 3(2x)(y)(3z). Compare with Identity VIII.
= (2x+y+3z)[(2x)²+y²+(3z)²−(2x)(y)−(y)(3z)−(2x)(3z)]
= (2x+y+3z)(4x²+y²+9z²−2xy−3yz−6xz).
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Exercises — Fully Solved
Exercise 2.1
Q1 — Which are polynomials in one variable?
(i) 4x²−3x+7 (ii) y²+√2 (iii) 3√t + t√2 (iv) y+2/y (v) x¹⁰+y³+t⁵⁰
Solution
(i) 4x²−3x+7 → exponents 2,1,0 (all whole numbers) → YES, polynomial in x.
(ii) y²+√2 → exponents 2, 0 (whole numbers) → YES, polynomial in y.
(iii) 3√t + t√2 = 3t^(1/2) + t^1·√2 → exponent 1/2 is not a whole number → NO.
(iv) y + 2/y = y + 2y⁻¹ → exponent −1 is not a whole number → NO.
(v) x¹⁰+y³+t⁵⁰ → three different variables → NOT a polynomial in one variable.
Q2 — Coefficients of x²
(i) 2+x²+x (ii) 2−x²+x³ (iii) (π/2)x²+x (iv) √2·x−1
Solution
(i) Coefficient of x² = 1
(ii) Coefficient of x² = −1
(iii) Coefficient of x² = π/2
(iv) No x² term → coefficient = 0
Q4 — Degree of polynomials
(i) 5x³+4x²+7x (ii) 4−y² (iii) 5t−√7 (iv) 3
Solution
(i) Highest power = x³ → degree 3
(ii) Highest power = y² → degree 2
(iii) Highest power = t¹ → degree 1
(iv) 3 = 3·x⁰ → degree 0
Q5 — Classify as linear, quadratic or cubic
(i) x²+x (ii) x−x³ (iii) y+y²+4 (iv) 1+x (v) 3t (vi) r² (vii) 7x³
Solution
(i) x²+x → degree 2 → Quadratic
(ii) x−x³ → degree 3 → Cubic
(iii) y+y²+4 → degree 2 → Quadratic
(iv) 1+x → degree 1 → Linear
(v) 3t → degree 1 → Linear
(vi) r² → degree 2 → Quadratic
(vii) 7x³ → degree 3 → Cubic
Exercise 2.2
Q1 — Value of 5x − 4x² + 3
Find value at (i) x=0 (ii) x=−1 (iii) x=2
Solution
(i) 5(0)−4(0)+3 = 3
(ii) 5(−1)−4(1)+3 = −5−4+3 = −6
(iii) 5(2)−4(4)+3 = 10−16+3 = −3
Q3 — Verify zeroes
Selected: (i) p(x)=3x+1, x=−1/3 (iii) p(x)=x²−1, x=1,−1 (vii) p(x)=3x²−1, x=−1/√3, 2/√3
Solution
(i) p(−1/3)=3(−1/3)+1=−1+1=0 ✓ → Yes, zero.
(iii) p(1)=1−1=0 ✓; p(−1)=1−1=0 ✓ → Both are zeroes.
(vii) p(−1/√3)=3(1/3)−1=1−1=0 ✓; p(2/√3)=3(4/3)−1=4−1=3≠0 → Only −1/√3 is a zero.
Q4 — Find zeroes
(i) p(x)=x+5 (ii) p(x)=x−5 (iii) p(x)=2x+5 (iv) p(x)=3x−2 (v) p(x)=3x (vi) p(x)=ax, a≠0
Solution
(i) x+5=0 → x=−5
(ii) x−5=0 → x=5
(iii) 2x+5=0 → x=−5/2
(iv) 3x−2=0 → x=2/3
(v) 3x=0 → x=0
(vi) ax=0 → x=0
Exercise 2.3
Q1 — Does (x+1) divide?
(i) x³+x²+x+1 (ii) x⁴+x³+x²+x+1 (iii) x⁴+3x³+3x²+x+1 (iv) x³−x²−(2+√2)x+√2
Solution — test p(−1) for each
(i) p(−1)=−1+1−1+1=0 → (x+1) IS a factor.
(ii) p(−1)=1−1+1−1+1=1≠0 → (x+1) is NOT a factor.
(iii) p(−1)=1−3+3−1+1=1≠0 → (x+1) is NOT a factor.
(iv) p(−1)=−1−1+(2+√2)−√2=−1−1+2+√2−√2=0 → (x+1) IS a factor.
Q3 — Find k when (x−1) is a factor
(i) p(x)=x²+x+k (ii) p(x)=2x²+kx+√2 (iii) p(x)=kx²−√2x+1 (iv) p(x)=kx²−3x+k
Solution — set p(1)=0
(i) 1+1+k=0 → k=−2
(ii) 2+k+√2=0 → k=−(2+√2)
(iii) k−√2+1=0 → k=√2−1
(iv) k−3+k=0 → 2k=3 → k=3/2
Q4 — Factorise quadratics
(i) 12x²−7x+1 (ii) 2x²+7x+3 (iii) 6x²+5x−6 (iv) 3x²−x−4
Solution
(i) p·q=12, p+q=−7 → −4,−3. 12x²−4x−3x+1=4x(3x−1)−1(3x−1)=(4x−1)(3x−1)
(ii) p·q=6, p+q=7 → 6,1. 2x²+6x+x+3=2x(x+3)+(x+3)=(2x+1)(x+3)
(iii) p·q=−36, p+q=5 → 9,−4. 6x²+9x−4x−6=3x(2x+3)−2(2x+3)=(3x−2)(2x+3)
(iv) p·q=−12, p+q=−1 → −4,3. 3x²−4x+3x−4=x(3x−4)+(3x−4)=(x+1)(3x−4)
Q5 — Factorise cubics
(i) x³−2x²−x+2 (ii) x³−3x²−9x−5 (iii) x³+13x²+32x+20 (iv) 2y³+y²−2y−1
Solution
(i) p(1)=1−2−1+2=0; p(−1)=−1−2+1+2=0; p(2)=8−8−2+2=0. Factors: (x−1)(x+1)(x−2) → (x−1)(x+1)(x−2)
(ii) p(−1)=−1−3+9−5=0 → (x+1) factor. Divide: x³−3x²−9x−5=(x+1)(x²−4x−5)=(x+1)(x−5)(x+1)=(x+1)²(x−5)
(iii) p(−1)=−1+13−32+20=0 → (x+1). Divide → (x+1)(x²+12x+20)=(x+1)(x+2)(x+10)=(x+1)(x+2)(x+10)
(iv) p(1)=2+1−2−1=0 → (y−1). Divide → (y−1)(2y²+3y+1)=(y−1)(2y+1)(y+1)=(y−1)(y+1)(2y+1)
Exercise 2.4 (Selected)
Q7 — Evaluate cubes using identities
(i) (99)³ (ii) (102)³ (iii) (998)³
Solution
(i) (100−1)³=100³−1³−3(100)(1)(99)=1000000−1−29700=970299
(ii) (100+2)³=100³+2³+3(100)(2)(102)=1000000+8+61200=1061208
(iii) (1000−2)³=1000³−2³−3(1000)(2)(998)=1000000000−8−5988000=994011992
Q14 — Without calculating cubes
(i) (−12)³+(7)³+(5)³ (ii) (28)³+(−15)³+(−13)³
Solution — If x+y+z=0, then x³+y³+z³=3xyz
(i) −12+7+5=0 → (−12)³+7³+5³ = 3(−12)(7)(5) = −1260.
(ii) 28+(−15)+(−13)=0 → 28³+(−15)³+(−13)³ = 3(28)(−15)(−13) = 16380.
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Tips & Tricks
Remainder = p(a)
When dividing p(x) by (x − a), never do long division! Just substitute x = a and compute p(a). That's your remainder.
Factor Test Shortcut
To test if (x − a) is a factor: just check if p(a) = 0. No division needed to confirm — division only for finding the other factor.
Rational Root Trick
For cubic p(x) = x³ + bx² + cx + d, always try ±1, ±2, ... (factors of d) first. Start with ±1 as it's easiest.
Middle Term Split
For ax² + bx + c: find two numbers with SUM = b and PRODUCT = ac. Common mistake: product = c (wrong — it should be ac).
x+y+z=0 Magic
If three numbers sum to zero, their cubes sum to 3 times their product. Saves enormous calculation in exam questions!
Mental Maths with Identity IV
To multiply numbers close to a base: 97×103 = (100−3)(100+3) = 10000−9 = 9991. Identity III!
Identity V Pattern
(x+y+z)² has 3 square terms + 3 cross terms (each doubled). Count 6 terms total. If sign is negative, treat as (x+(−y)+(−z))².
Cube Identity Memory Aid
(x+y)³ → all +; (x−y)³ → alternating signs: +, −, −, +. Memorise: "plus-minus-minus-plus" for (x−y)³ = x³ − 3x²y + 3xy² − y³.
Finding k Problems
If (x−a) is given as a factor, then p(a) = 0. Substitute x = a, set equal to 0, solve for k. Never forget to actually equate to ZERO.
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Formula & Fact Sheet
Zero of Linear Polynomial
ax + b = 0 → x = −b/a
One and only one zero for linear poly
Remainder Theorem
p(x) ÷ (x−a) → remainder = p(a)
No long division needed
Factor Theorem
(x−a) | p(x) ⟺ p(a) = 0
Both directions are true
Identity I
(x+y)² = x²+2xy+y²
Perfect square (sum)
Identity II
(x−y)² = x²−2xy+y²
Perfect square (diff)
Identity III
x²−y² = (x+y)(x−y)
Difference of squares
Identity IV
(x+a)(x+b) = x²+(a+b)x+ab
Mental multiplication
Identity V
(x+y+z)² = x²+y²+z²+2xy+2yz+2zx
3 squares + 3 cross terms
Identity VI
(x+y)³ = x³+y³+3xy(x+y)
Sum cube
Identity VII
(x−y)³ = x³−y³−3xy(x−y)
Difference cube
Identity VIII
x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²−xy−yz−zx)
Sum of three cubes
Special Case VIII
If x+y+z=0 → x³+y³+z³ = 3xyz
Powerful shortcut!
Derived Identities
x³ + y³ = (x+y)(x²−xy+y²)
x³ − y³ = (x−y)(x²+xy+y²)
Expanded forms to memorise:
(x+y)³ = x³ + 3x²y + 3xy² + y³
(x−y)³ = x³ − 3x²y + 3xy² − y³
Degree rules: Degree of product = sum of degrees | Degree of zero polynomial = undefined.
(x+y)³ = x³ + 3x²y + 3xy² + y³
(x−y)³ = x³ − 3x²y + 3xy² − y³
Degree rules: Degree of product = sum of degrees | Degree of zero polynomial = undefined.
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50 Practice Questions
Covering all CBSE types: MCQ · 1-mark · 2-mark · 3-mark · 5-mark · Proof. Click Show Answer to reveal.
MCQ (1 mark each)
Which of the following is NOT a polynomial?
- A) x² + 3x + 1
- B) 5t − √7
- C) x + 1/x
- D) 3
Show Answer
x + 1/x = x + x⁻¹. The exponent −1 is not a whole number, so it is NOT a polynomial. (C)
Degree of the polynomial 5y⁶ − 4y² − 6 is:
- A) 2
- B) 6
- C) 5
- D) 0
Show Answer
Highest power = y⁶. Degree = 6. (B)
The zero of the polynomial p(x) = 3x − 5 is:
- A) 3/5
- B) 5/3
- C) −5/3
- D) 0
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3x − 5 = 0 → x = 5/3. (B)
When p(x) = x³ − ax² + x + 6 is divided by (x − 2), the remainder is 0. What is a?
- A) 1
- B) 2
- C) 4
- D) 6
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p(2) = 0 → 8 − 4a + 2 + 6 = 0 → 16 − 4a = 0 → a = 4. (C)
Which identity gives (x + y)² = x² + 2xy + y²?
- A) Identity I
- B) Identity II
- C) Identity III
- D) Identity IV
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Identity I. (A)
The number of zeroes of a linear polynomial is:
- A) 0
- B) 1
- C) 2
- D) infinite
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A linear polynomial has exactly 1 zero. (B)
If (x − 1) is a factor of p(x) = x² + x + k, then k =
- A) 0
- B) 1
- C) −2
- D) 2
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p(1) = 1+1+k = 0 → k = −2. (C)
Value of (x+y+z)² using Identity V contains how many terms?
- A) 3
- B) 5
- C) 6
- D) 9
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3 square terms + 3 cross-product terms = 6 terms. (C)
If x + y + z = 0, then x³ + y³ + z³ =
- A) 0
- B) 3xyz
- C) (x+y+z)³
- D) 3x²y²z²
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Special case of Identity VIII: x³+y³+z³ = 3xyz. (B)
The coefficient of x² in p(x) = (√2)x − 1 is:
- A) √2
- B) 1
- C) 0
- D) −1
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There is no x² term. Coefficient = 0. (C)
1-Mark Questions
State the Remainder Theorem in one line.
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If polynomial p(x) is divided by (x−a), the remainder is p(a).
Is √x + 3 a polynomial? Give reason.
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No. √x = x^(1/2) has exponent 1/2, which is not a whole number.
Write the degree of the zero polynomial.
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The degree of the zero polynomial is not defined.
What is the zero of the polynomial p(x) = ax, where a ≠ 0?
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p(x) = ax = 0 → x = 0. Zero is 0.
Give an example of a binomial of degree 35.
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x³⁵ + 1 (two terms, highest power 35).
In Identity V, how many terms does the expansion of (x+y+z)² have?
Show Answer
6 terms: x², y², z², 2xy, 2yz, 2zx.
Find p(−1) for p(x) = x³ − x².
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p(−1) = (−1)³ − (−1)² = −1 − 1 = −2.
A cubic polynomial in one variable can have at most how many terms?
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4 terms (degree + 1 = 3 + 1 = 4).
2-Mark Questions
Find the remainder when p(x) = x³ − 6x² + 2x − 4 is divided by (x − 3).
Show Answer
By Remainder Theorem, remainder = p(3).
p(3) = 27 − 54 + 6 − 4 = −25.
p(3) = 27 − 54 + 6 − 4 = −25.
Evaluate 99² using an algebraic identity.
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99² = (100−1)² = 100² − 2(100)(1) + 1² = 10000 − 200 + 1 = 9801. [Identity II]
Verify whether (x+3) is a factor of 2x² + 5x − 3.
Show Answer
Zero of (x+3) is −3. p(−3) = 2(9) + 5(−3) − 3 = 18 − 15 − 3 = 0. Since p(−3) = 0, (x+3) IS a factor.
Factorise: 9x² − 4y².
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= (3x)² − (2y)² = (3x+2y)(3x−2y). [Identity III]
Expand (2x + 3y)².
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(2x+3y)² = (2x)²+2(2x)(3y)+(3y)² = 4x²+12xy+9y². [Identity I]
If p(x) = x² − 4x + 3, find p(1) and p(3). Are 1 and 3 zeroes?
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p(1) = 1−4+3 = 0 ✓; p(3) = 9−12+3 = 0 ✓. Yes, both 1 and 3 are zeroes.
Evaluate 103 × 97 using an identity.
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= (100+3)(100−3) = 100² − 3² = 10000 − 9 = 9991. [Identity III]
Factorise a² − 2a − 15.
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Find p+q = −2, pq = −15: p=−5, q=3. Split: a²−5a+3a−15 = a(a−5)+3(a−5) = (a+3)(a−5).
Write the expanded form of (x − 2y − z)².
Show Answer
= [x+(−2y)+(−z)]²
= x²+4y²+z²+2(x)(−2y)+2(−2y)(−z)+2(x)(−z)
= x²+4y²+z²−4xy+4yz−2xz
= x²+4y²+z²+2(x)(−2y)+2(−2y)(−z)+2(x)(−z)
= x²+4y²+z²−4xy+4yz−2xz
If (2x + 1) is a factor of 6x² + 5x + k, find k.
Show Answer
Zero of (2x+1) is −1/2. p(−1/2) = 6(1/4)+5(−1/2)+k = 3/2−5/2+k = −1+k = 0 → k = 1.
3-Mark Questions
Factorise: x³ + 3x² − 9x − 5.
Show Answer
Try p(1) = 1+3−9−5 = −10 ≠ 0. Try p(−1) = −1+3+9−5 = 6 ≠ 0. Try p(5) = 125+75−45−5 = 150 ≠ 0. Try p(−5) = −125+75+45−5 = −10 ≠ 0. Try p(−1): already tried. Wait: p(1) = −10, let's re-check p(−5) = −125+75+45−5 = −10. Try p(5) again: 125+75−45−5=150. Try: Actually for x³−3x²−9x−5: p(−1)=−1−3+9−5=0! ✓
For THIS problem x³+3x²−9x−5: p(−5)=−125+75+45−5=−10. p(1)=1+3−9−5=−10. p(5)=125+75−45−5=150. p(−1)=−1+3+9−5=6.
Note: (x+5): p(−5)=−10 ≠0. Try factoring directly: x³+3x²−9x−5. Group: x²(x+5)−2x(x+5)−(x+5) wait... x²(x+5)=(x³+5x²)≠x³+3x².
Correct: p(1)=−10, p(−1)=6, p(5)=150. Note: actually we find (x+5) by computing differently.
Actually: x³+3x²−9x−5. Let's try p = x+5: need x=−5; p(−5)=−125+75+45−5=−10. Not zero.
Try
Actually the textbook example uses x³−3x²−9x−5, not +3x². For x³−3x²−9x−5: p(−1)=−1−3+9−5=0. Factor (x+1). Divide: x³−3x²−9x−5÷(x+1)=x²−4x−5=(x+1)(x−5). Final: (x+1)²(x−5).
For THIS problem x³+3x²−9x−5: p(−5)=−125+75+45−5=−10. p(1)=1+3−9−5=−10. p(5)=125+75−45−5=150. p(−1)=−1+3+9−5=6.
Note: (x+5): p(−5)=−10 ≠0. Try factoring directly: x³+3x²−9x−5. Group: x²(x+5)−2x(x+5)−(x+5) wait... x²(x+5)=(x³+5x²)≠x³+3x².
Correct: p(1)=−10, p(−1)=6, p(5)=150. Note: actually we find (x+5) by computing differently.
Actually: x³+3x²−9x−5. Let's try p = x+5: need x=−5; p(−5)=−125+75+45−5=−10. Not zero.
Try
(3x²−9x) + (x³−5) → not helpful. Use splitting: x³+3x²−9x−5 = x²(x−1)+4x(x−1)+5(x−1)? Check: x²(x−1)=x³−x², 4x(x−1)=4x²−4x, 5(x−1)=5x−5. Sum=x³+3x²+x−5 ≠ original.Actually the textbook example uses x³−3x²−9x−5, not +3x². For x³−3x²−9x−5: p(−1)=−1−3+9−5=0. Factor (x+1). Divide: x³−3x²−9x−5÷(x+1)=x²−4x−5=(x+1)(x−5). Final: (x+1)²(x−5).
If x + y + z = 9 and xy + yz + zx = 26, find x² + y² + z².
Show Answer
(x+y+z)² = x²+y²+z² + 2(xy+yz+zx)
9² = x²+y²+z² + 2(26)
81 = x²+y²+z² + 52
x²+y²+z² = 29.
9² = x²+y²+z² + 2(26)
81 = x²+y²+z² + 52
x²+y²+z² = 29.
Without actual division, show that x³ − 3x² − 13x + 15 has (x+2) as a factor. Then factorise completely.
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p(−2) = −8−12+26+15 = 21 ≠ 0. So (x+2) is NOT a factor. But for (x−5): p(5)=125−75−65+15=0 ✓. Divide: (x−5)(x²+2x−3) = (x−5)(x+3)(x−1). Answer: (x−5)(x−1)(x+3).
Factorise: 2y³ + y² − 2y − 1.
Show Answer
p(1) = 2+1−2−1 = 0 → (y−1) is a factor.
Divide: 2y³+y²−2y−1 = (y−1)(2y²+3y+1)
Factorise 2y²+3y+1: split → 2y²+2y+y+1 = 2y(y+1)+(y+1) = (2y+1)(y+1).
Final: (y−1)(y+1)(2y+1).
Divide: 2y³+y²−2y−1 = (y−1)(2y²+3y+1)
Factorise 2y²+3y+1: split → 2y²+2y+y+1 = 2y(y+1)+(y+1) = (2y+1)(y+1).
Final: (y−1)(y+1)(2y+1).
Expand (2a − 3b)³ and verify the pattern.
Show Answer
(x−y)³ = x³−y³−3xy(x−y) with x=2a, y=3b:
= (2a)³−(3b)³−3(2a)(3b)(2a−3b)
= 8a³−27b³−18ab(2a−3b)
= 8a³−27b³−36a²b+54ab².
= (2a)³−(3b)³−3(2a)(3b)(2a−3b)
= 8a³−27b³−18ab(2a−3b)
= 8a³−27b³−36a²b+54ab².
If x = 2 and x = 0 are zeroes of p(x) = 2x³ − 5x² + ax + b, find a and b.
Show Answer
p(0) = 0 → b = 0.
p(2) = 16−20+2a+0 = 0 → −4+2a = 0 → a = 2. So a=2, b=0.
p(2) = 16−20+2a+0 = 0 → −4+2a = 0 → a = 2. So a=2, b=0.
Evaluate (98)³ using a suitable identity without a calculator.
Show Answer
(98)³ = (100−2)³ = 100³ − 2³ − 3(100)(2)(100−2)
= 1000000 − 8 − 600(98)
= 1000000 − 8 − 58800
= 941192.
= 1000000 − 8 − 600(98)
= 1000000 − 8 − 58800
= 941192.
Factorise 27x³ + y³ + z³ − 9xyz.
Show Answer
= (3x)³ + y³ + z³ − 3(3x)(y)(z)
[Identity VIII with a=3x, b=y, c=z]
= (3x+y+z)[(3x)²+y²+z²−(3x)(y)−(y)(z)−(3x)(z)]
= (3x+y+z)(9x²+y²+z²−3xy−yz−3xz).
[Identity VIII with a=3x, b=y, c=z]
= (3x+y+z)[(3x)²+y²+z²−(3x)(y)−(y)(z)−(3x)(z)]
= (3x+y+z)(9x²+y²+z²−3xy−yz−3xz).
5-Mark & Proof Questions
Prove the Factor Theorem: (x − a) is a factor of p(x) if and only if p(a) = 0.
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Proof:
By the Remainder Theorem, when p(x) is divided by (x − a):
p(x) = (x − a)·q(x) + p(a) ...(1)
Part (i): Prove: if p(a) = 0, then (x − a) is a factor.
Substituting p(a) = 0 in (1): p(x) = (x − a)·q(x).
This shows (x − a) divides p(x) exactly, i.e., (x − a) is a factor of p(x). ✓
Part (ii): Prove: if (x − a) is a factor of p(x), then p(a) = 0.
Since (x − a) is a factor: p(x) = (x − a)·g(x) for some polynomial g(x).
Putting x = a: p(a) = (a − a)·g(a) = 0·g(a) = 0. ✓
Hence (x − a) is a factor of p(x) ⟺ p(a) = 0. Proved.
By the Remainder Theorem, when p(x) is divided by (x − a):
p(x) = (x − a)·q(x) + p(a) ...(1)
Part (i): Prove: if p(a) = 0, then (x − a) is a factor.
Substituting p(a) = 0 in (1): p(x) = (x − a)·q(x).
This shows (x − a) divides p(x) exactly, i.e., (x − a) is a factor of p(x). ✓
Part (ii): Prove: if (x − a) is a factor of p(x), then p(a) = 0.
Since (x − a) is a factor: p(x) = (x − a)·g(x) for some polynomial g(x).
Putting x = a: p(a) = (a − a)·g(a) = 0·g(a) = 0. ✓
Hence (x − a) is a factor of p(x) ⟺ p(a) = 0. Proved.
Factorise x³ + 13x² + 32x + 20 completely, showing all steps.
Show Answer
Let p(x) = x³+13x²+32x+20.
Step 1: Try p(−1) = −1+13−32+20 = 0 ✓ → (x+1) is a factor.
Step 2: Divide: x³+13x²+32x+20 = (x+1)(x²+12x+20).
Verification: (x+1)(x²+12x+20) = x³+12x²+20x+x²+12x+20 = x³+13x²+32x+20 ✓
Step 3: Factorise x²+12x+20: need p+q=12, pq=20 → 10, 2.
x²+10x+2x+20 = x(x+10)+2(x+10) = (x+2)(x+10).
Final: (x+1)(x+2)(x+10).
Step 1: Try p(−1) = −1+13−32+20 = 0 ✓ → (x+1) is a factor.
Step 2: Divide: x³+13x²+32x+20 = (x+1)(x²+12x+20).
Verification: (x+1)(x²+12x+20) = x³+12x²+20x+x²+12x+20 = x³+13x²+32x+20 ✓
Step 3: Factorise x²+12x+20: need p+q=12, pq=20 → 10, 2.
x²+10x+2x+20 = x(x+10)+2(x+10) = (x+2)(x+10).
Final: (x+1)(x+2)(x+10).
Prove that x³ + y³ = (x + y)(x² − xy + y²). Hence factorise 8a³ + 27b³.
Show Answer
Proof:
RHS = (x+y)(x²−xy+y²)
= x(x²−xy+y²) + y(x²−xy+y²)
= x³−x²y+xy² + x²y−xy²+y³
= x³ + y³ (all middle terms cancel) = LHS ✓
Factorisation of 8a³ + 27b³:
= (2a)³ + (3b)³
Using x³+y³ = (x+y)(x²−xy+y²) with x=2a, y=3b:
= (2a+3b)[(2a)²−(2a)(3b)+(3b)²]
= (2a+3b)(4a²−6ab+9b²).
RHS = (x+y)(x²−xy+y²)
= x(x²−xy+y²) + y(x²−xy+y²)
= x³−x²y+xy² + x²y−xy²+y³
= x³ + y³ (all middle terms cancel) = LHS ✓
Factorisation of 8a³ + 27b³:
= (2a)³ + (3b)³
Using x³+y³ = (x+y)(x²−xy+y²) with x=2a, y=3b:
= (2a+3b)[(2a)²−(2a)(3b)+(3b)²]
= (2a+3b)(4a²−6ab+9b²).
If x = 2 − ∛4 + ∛2, find the value of x³ − 6x² + 12x − 8.
Show Answer
Notice that x − 2 = −∛4 + ∛2.
x³−6x²+12x−8 = (x−2)³ [using Identity VII with a=x, b=2]
= (−∛4 + ∛2)³
Let u = ∛2. Then −∛4 = −u². So (x−2) = −u²+u = u(1−u).
(x−2)³ = [u(1−u)]³ = u³(1−u)³ = 2(1−∛2)³.
Now (1−∛2)³ = 1−3∛2+3∛4−2 = −1−3∛2+3∛4.
So answer = 2(−1−3∛2+3∛4) = −2−6∛2+6∛4.
Alternate shortcut: The expression (x−2)³ = (−∛4+∛2)³. Note ∛2 ≈ 1.26, so −∛4+∛2 ≈ −1.587+1.26 ≈ −0.327. Answer ≈ (−0.327)³ ≈ −0.035. (Exact: 2−6∛2+6∛4 form).
x³−6x²+12x−8 = (x−2)³ [using Identity VII with a=x, b=2]
= (−∛4 + ∛2)³
Let u = ∛2. Then −∛4 = −u². So (x−2) = −u²+u = u(1−u).
(x−2)³ = [u(1−u)]³ = u³(1−u)³ = 2(1−∛2)³.
Now (1−∛2)³ = 1−3∛2+3∛4−2 = −1−3∛2+3∛4.
So answer = 2(−1−3∛2+3∛4) = −2−6∛2+6∛4.
Alternate shortcut: The expression (x−2)³ = (−∛4+∛2)³. Note ∛2 ≈ 1.26, so −∛4+∛2 ≈ −1.587+1.26 ≈ −0.327. Answer ≈ (−0.327)³ ≈ −0.035. (Exact: 2−6∛2+6∛4 form).
Verify the identity: x³+y³+z³−3xyz = ½(x+y+z)[(x−y)²+(y−z)²+(z−x)²].
Show Answer
LHS: x³+y³+z³−3xyz = (x+y+z)(x²+y²+z²−xy−yz−zx) ... (1)
RHS: Expand ½(x+y+z)[(x−y)²+(y−z)²+(z−x)²]
= ½(x+y+z)[x²−2xy+y² + y²−2yz+z² + z²−2zx+x²]
= ½(x+y+z)[2x²+2y²+2z²−2xy−2yz−2zx]
= ½(x+y+z)·2(x²+y²+z²−xy−yz−zx)
= (x+y+z)(x²+y²+z²−xy−yz−zx) ... (2)
From (1) and (2): LHS = RHS. Proved ✓
RHS: Expand ½(x+y+z)[(x−y)²+(y−z)²+(z−x)²]
= ½(x+y+z)[x²−2xy+y² + y²−2yz+z² + z²−2zx+x²]
= ½(x+y+z)[2x²+2y²+2z²−2xy−2yz−2zx]
= ½(x+y+z)·2(x²+y²+z²−xy−yz−zx)
= (x+y+z)(x²+y²+z²−xy−yz−zx) ... (2)
From (1) and (2): LHS = RHS. Proved ✓
Mixed Additional Practice
Use Identity IV: evaluate 101 × 103.
Show Answer
= (100+1)(100+3) = 100²+(1+3)(100)+1×3 = 10000+400+3 = 10403.
Factorise 27 − 125a³ − 135a + 225a².
Show Answer
= 3³ − (5a)³ − 3·9·(5a) + 3·3·(5a)² ... wait: = (3−5a)³ = (3)³−(5a)³−3(3)²(5a)+3(3)(5a)² → 27−125a³−135a+225a². Yes! (3−5a)³.
If x² + y² = 29 and xy = 10, find (x+y)² and (x−y)².
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(x+y)² = x²+2xy+y² = 29+20 = 49. So x+y = 7.
(x−y)² = x²−2xy+y² = 29−20 = 9. So x−y = 3.
(x−y)² = x²−2xy+y² = 29−20 = 9. So x−y = 3.
Factorise 64m³ − 343n³.
Show Answer
= (4m)³ − (7n)³
Using x³−y³ = (x−y)(x²+xy+y²):
= (4m−7n)[(4m)²+(4m)(7n)+(7n)²]
= (4m−7n)(16m²+28mn+49n²).
Using x³−y³ = (x−y)(x²+xy+y²):
= (4m−7n)[(4m)²+(4m)(7n)+(7n)²]
= (4m−7n)(16m²+28mn+49n²).
Possible expressions for length and breadth of rectangle with area 25a² − 35a + 12.
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Factorise: 25a²−35a+12. p×q=300, p+q=−35 → −20, −15. 25a²−20a−15a+12 = 5a(5a−4)−3(5a−4) = (5a−3)(5a−4). Length = (5a−3), Breadth = (5a−4).
If a + b + c = 0, show that a³ + b³ + c³ = 3abc. (Proof)
Show Answer
Proof: Using Identity VIII:
a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)
Given a+b+c = 0, so:
a³+b³+c³−3abc = 0·(a²+b²+c²−ab−bc−ca) = 0
∴ a³+b³+c³ = 3abc. Proved ✓
a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)
Given a+b+c = 0, so:
a³+b³+c³−3abc = 0·(a²+b²+c²−ab−bc−ca) = 0
∴ a³+b³+c³ = 3abc. Proved ✓
Find the value of (28)³ + (−15)³ + (−13)³ without calculating individual cubes.
Show Answer
Check sum: 28+(−15)+(−13) = 0. So by the identity: x³+y³+z³ = 3xyz when sum = 0.
= 3(28)(−15)(−13) = 3(28)(195) = 3×5460 = 16380.
= 3(28)(−15)(−13) = 3(28)(195) = 3×5460 = 16380.
Factorise completely: 8a³ − b³ − 12a²b + 6ab². State the identity used.
Show Answer
8a³−b³−12a²b+6ab²
= (2a)³ − b³ − 3(2a)²(b) + 3(2a)(b)²
= (2a)³ − 3(2a)²b + 3(2a)b² − b³
This matches (x−y)³ = x³−3x²y+3xy²−y³ with x=2a, y=b.
= (2a−b)³ = (2a−b)(2a−b)(2a−b). [Identity VII]
= (2a)³ − b³ − 3(2a)²(b) + 3(2a)(b)²
= (2a)³ − 3(2a)²b + 3(2a)b² − b³
This matches (x−y)³ = x³−3x²y+3xy²−y³ with x=2a, y=b.
= (2a−b)³ = (2a−b)(2a−b)(2a−b). [Identity VII]
Prove that x³ − y³ = (x − y)(x² + xy + y²). Using it, factorise 27y³ + 125z³.
Show Answer
Proof of x³ − y³:
RHS = (x−y)(x²+xy+y²)
= x(x²+xy+y²) − y(x²+xy+y²)
= x³+x²y+xy² − x²y−xy²−y³
= x³−y³ = LHS ✓
Factorise 27y³ + 125z³:
Note: this is a SUM of cubes, so use x³+y³ = (x+y)(x²−xy+y²).
= (3y)³ + (5z)³
= (3y+5z)[(3y)²−(3y)(5z)+(5z)²]
= (3y+5z)(9y²−15yz+25z²).
RHS = (x−y)(x²+xy+y²)
= x(x²+xy+y²) − y(x²+xy+y²)
= x³+x²y+xy² − x²y−xy²−y³
= x³−y³ = LHS ✓
Factorise 27y³ + 125z³:
Note: this is a SUM of cubes, so use x³+y³ = (x+y)(x²−xy+y²).
= (3y)³ + (5z)³
= (3y+5z)[(3y)²−(3y)(5z)+(5z)²]
= (3y+5z)(9y²−15yz+25z²).