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CDS Mathematics

Algebraic Expressions & Polynomials

📐 Algebra · MC07 CDS Elementary Mathematics 🎯 High Priority

Algebraic Expressions and Polynomials is the foundation for all CDS Algebra. Master the identities and the remainder/factor theorems here, and linear/quadratic equations (MC08) become fast and mechanical. CDS tests these at Class 9–10 NCERT level — no advanced algebra — but questions use identities and theorems in clever, non-obvious ways.

📌 CDS exam focus: (1) Applying identities to simplify — especially (a±b)², a³±b³, and "if a+b+c=0 then a³+b³+c³=3abc"; (2) Finding value of expressions like x²+1/x² when x+1/x is given; (3) Remainder theorem — find remainder without division; (4) Factor theorem — determine unknown coefficients; (5) Laws of indices — same-base rules and negative/fractional exponents; (6) Set theory Venn diagrams — counting elements.

Topics at a Glance

① Identities
(a±b)², a²−b², (a±b)³, a³±b³
② Factorisation
Common factor, grouping, splitting middle term
③ Remainder Theorem
P(a) = remainder when divided by (x−a)
④ Factor Theorem
P(a)=0 ↔ (x−a) is a factor
⑤ Laws of Indices
aᵐ·aⁿ = aᵐ⁺ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁻ⁿ = 1/aⁿ
⑥ Set Theory
n(A∪B) = n(A)+n(B)−n(A∩B)

1. Algebraic Identities — The Complete Set

1.1
Memorise All — Recognise the Pattern Before Calculating
Identities cut calculation time by 70% in CDS paper
⚡ Standard Identities
SQUARE IDENTITIES: (a+b)² = a²+2ab+b² (a−b)² = a²−2ab+b² a²−b² = (a+b)(a−b) ← difference of squares DERIVED: (a+b)²+(a−b)² = 2(a²+b²) (a+b)²−(a−b)² = 4ab a²+b² = (a+b)²−2ab a²+b² = (a−b)²+2ab CUBE IDENTITIES: (a+b)³ = a³+3a²b+3ab²+b³ = a³+b³+3ab(a+b) (a−b)³ = a³−3a²b+3ab²−b³ = a³−b³−3ab(a−b) a³+b³ = (a+b)(a²−ab+b²) a³−b³ = (a−b)(a²+ab+b²) ★ KEY RESULT (CDS favourite): If a+b+c = 0 → a³+b³+c³ = 3abc (no calculation needed!)
When you see a+b+c = 0, immediately apply 3abc. This appears in at least one CDS paper every year.
Worked Examples — Identity Application

If x + 1/x = 5, find x² + 1/x² and x³ + 1/x³.
x²+1/x² = (x+1/x)²−2 = 25−2 = 23.
x³+1/x³ = (x+1/x)(x²−1+1/x²) = 5×(23−1) = 5×22 = 110.

If a+b+c=0 and a=3, b=−5, find a³+b³+c³.
c = −(a+b) = −(3−5) = 2. a³+b³+c³ = 3abc = 3×3×(−5)×2 = −90.

2. Factorisation

2.1
Methods in Order: Common Factor → Grouping → Identity → Splitting
Always try the simplest method first

Common Factor

  • Find GCF of all terms, factor out
  • 3x²+6x = 3x(x+2)
  • Try first — always

Grouping

  • Group pairs; common binomial emerges
  • ax+bx+ay+by = (x+y)(a+b)
  • Rearrange if first grouping fails

Splitting Middle Term

  • ax²+bx+c: find p,q: p×q=ac, p+q=b
  • x²+5x+6: p=2,q=3 → (x+2)(x+3)
  • Most common for quadratics

3. Remainder Theorem & Factor Theorem

3.1
Find Remainders and Factors by Substitution — No Long Division
Directly tested in CDS — 1–2 questions per paper

Remainder Theorem

  • Dividing P(x) by (x−a): remainder = P(a)
  • Dividing by (x+a): substitute x = −a
  • Dividing by (ax−b): substitute x = b/a
  • Example: P(x)=x³−6x²+11x−6 ÷ (x−2): P(2)=8−24+22−6=0

Factor Theorem

  • (x−a) is a factor ⟺ P(a) = 0
  • Use to find unknown coefficients: P(a)=0, solve for k
  • Example: (x−1) factor of x²+kx−6: 1+k−6=0 → k=5
  • Converse: remainder=0 means it divides exactly

4. Laws of Indices

4.1
All Index Laws with CDS-Type Examples
Index questions thread through Number System, Log, and Algebra sections
⚡ Complete Laws of Indices
aᵐ×aⁿ = aᵐ⁺ⁿ aᵐ÷aⁿ = aᵐ⁻ⁿ (aᵐ)ⁿ = aᵐⁿ (ab)ⁿ = aⁿbⁿ a⁰ = 1 a⁻ⁿ = 1/aⁿ a^(1/n) = ⁿ√a a^(m/n) = ⁿ√(aᵐ) Same base: aˣ = aʸ → x = y (if a≠0,1,−1) Equal powers: aⁿ = bⁿ → a = b (for positive bases)

5. Set Theory

5.1
Venn Diagram Counting — Inclusion-Exclusion Formula
Word problems with surveys — find "both", "only one", "neither"
Universal set U A only A∩B (both) B only Neither Key Formulas n(A∪B) = n(A)+n(B)−n(A∩B) Only A = n(A) − n(A∩B) Only B = n(B) − n(A∩B) Neither = U − n(A∪B) 3 sets: add all, subtract pairs, add back triple overlap
Worked Example — Venn Diagram

In a class of 50, 30 play cricket, 25 play football, 10 play both. Find who play neither.
n(C∪F)=30+25−10=45. Neither=50−45=5.

📋 TOPIC-WISE PYQ
Algebraic Expressions — CDS Questions
Q1. If a+b=5 and a²+b²=17, find ab.
  • (a) 3   (b) 4   (c) 6   (d) 8
Answer: (b) 4
(a+b)²=25=17+2ab → ab=4.
Q2. If x+y+z=0, what is x³+y³+z³?
  • (a) 0   (b) 3xyz   (c) xyz   (d) 3(x+y+z)
Answer: (b) 3xyz
Identity: a+b+c=0 → a³+b³+c³=3abc=3xyz.
Q3. Find the remainder when x³−6x²+11x−6 is divided by (x−2).
  • (a) 0   (b) 2   (c) 4   (d) −2
Answer: (a) 0
P(2)=8−24+22−6=0. So (x−2) is a factor.
Q4. If x−1/x=4, find x²+1/x².
  • (a) 14   (b) 16   (c) 18   (d) 20
Answer: (c) 18
(x−1/x)²=x²−2+1/x²=16 → x²+1/x²=18.
Q5. The value of (256)^0.16 × (256)^0.09 is:
  • (a) 4   (b) 8   (c) 16   (d) 2
Answer: (a) 4
256^(0.16+0.09)=256^0.25=256^(1/4)=⁴√256=4.
Q6. In a survey of 100 people, 72 like tea, 52 like coffee. Minimum who like both?
  • (a) 20   (b) 24   (c) 28   (d) 32
Answer: (b) 24
Min n(A∩B)=n(A)+n(B)−n(U)=72+52−100=24.
🔥 TRICKY QUESTIONS
Algebra — Classic CDS Traps
🧩 T1. If x−1/x=3, find x³−1/x³.
Solution: 36.
x²+1/x²=(x−1/x)²+2=11. x³−1/x³=(x−1/x)(x²+1+1/x²)=3×12=36.
Key: middle factor is x²+1+1/x², not x²+1/x².
🧩 T2. Factorise: a³+b³+c³−3abc when a+b+c≠0.
Solution: (a+b+c)(a²+b²+c²−ab−bc−ca).
This is the standard factorisation. Also note: a²+b²+c²−ab−bc−ca = ½[(a−b)²+(b−c)²+(c−a)²] ≥ 0.
When a+b+c=0: the expression = 3abc (use that shortcut instead).
🧩 T3. If (x−2) is a factor of x³+kx²+2x−8, find k and the other factors.
Solution: k=−1; factors: (x−2)(x²+x+4).
P(2)=8+4k+4−8=4k+4=0 → k=−1. Polynomial: x³−x²+2x−8.
Divide by (x−2): x³−x²+2x−8 = (x−2)(x²+x+4). Discriminant of x²+x+4 = 1−16 < 0 → no real roots.

📐 Formula Sheet — MC07

Square Identities
  • (a+b)²=a²+2ab+b²
  • (a−b)²=a²−2ab+b²
  • a²−b²=(a+b)(a−b)
  • a²+b²=(a+b)²−2ab
Cube Identities
  • (a+b)³=a³+b³+3ab(a+b)
  • a³+b³=(a+b)(a²−ab+b²)
  • a³−b³=(a−b)(a²+ab+b²)
  • a+b+c=0 → a³+b³+c³=3abc
x+1/x Shortcuts
  • x²+1/x²=(x+1/x)²−2
  • x²+1/x²=(x−1/x)²+2
  • x³+1/x³=(x+1/x)(x²−1+1/x²)
  • x³−1/x³=(x−1/x)(x²+1+1/x²)
Theorems
  • Remainder=P(a) when dividing by (x−a)
  • Factor: P(a)=0 ↔ (x−a) is factor
  • Divisor (x+a): sub x=−a
Index Laws
  • aᵐ·aⁿ=aᵐ⁺ⁿ; aᵐ÷aⁿ=aᵐ⁻ⁿ
  • (aᵐ)ⁿ=aᵐⁿ; a⁰=1; a⁻ⁿ=1/aⁿ
  • a^(m/n)=ⁿ√(aᵐ)
Set Theory
  • n(A∪B)=n(A)+n(B)−n(A∩B)
  • Only A=n(A)−n(A∩B)
  • Neither=U−n(A∪B)
  • Min overlap=n(A)+n(B)−U

⚡ Quick Revision Booster — MC07

Identity Shortcuts
  • a+b+c=0 → a³+b³+c³=3abc
  • x²+1/x²=(x±1/x)²∓2
  • a²+b²=(a+b)²−2ab
Remainder / Factor
  • Sub root of divisor into P(x)
  • P(a)=0 → (x−a) is a factor
  • Find k: use factor condition
Index Rules
  • Same base: add/subtract exponents
  • Power of power: multiply
  • a^(1/4)=⁴√a; (256)^0.25=4
Set Formula
  • n(A∪B)=n(A)+n(B)−n(A∩B)
  • Min overlap=n(A)+n(B)−U
Factorisation Order
  • 1. Common factor
  • 2. Grouping
  • 3. Identity
  • 4. Splitting middle term
🚨 Traps
  • (a+b)²≠a²+b²
  • Middle factor of x³±1/x³ has ±1
  • a+b+c=0: use 3abc instantly
✏️ PRACTICE EXERCISE
Test Yourself — MC07
E1. If x+y=10 and xy=21, find x³+y³.
  • (a) 370   (b) 430   (c) 1000−63×10   (d) 370
💡 x³+y³=(x+y)³−3xy(x+y)=1000−3×21×10.
E2. Remainder when 2x³+3x²−x+5 is divided by (x+2)?
  • (a) 3   (b) 5   (c) 7   (d) 11
💡 P(−2)=2(−8)+3(4)−(−2)+5=−16+12+2+5.
E3. Simplify: (a+b+c)²−(a²+b²+c²).
  • (a) 2(ab+bc+ca)   (b) ab+bc+ca   (c) 2abc   (d) 0
💡 Expand (a+b+c)²=a²+b²+c²+2(ab+bc+ca), then subtract.
E4. If (x−2) is factor of x³+ax²+bx−12 and a=1, find b.
  • (a) −1   (b) 0   (c) −2   (d) 2
💡 P(2)=0: 8+4(1)+2b−12=0. Solve for b.
E5. In a group of 70 students, 45 study Maths, 30 study Science, and all study at least one. How many study both?
  • (a) 5   (b) 10   (c) 15   (d) 20
💡 n(M∪S)=70=n(M)+n(S)−n(M∩S)=45+30−n(M∩S).
E6. Factorise: x²−5x+6.
  • (a) (x−2)(x−3)   (b) (x+2)(x+3)   (c) (x−2)(x+3)   (d) (x+2)(x−3)
💡 Find p,q: p×q=6, p+q=−5. Both negative.
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