Official Syllabus Item: Part of 46 Algebra & Equations

Algebraic Expressions, Identities & Factorization

Master algebraic operations, standard identities (a±b)², a²-b², (a+b+c)², middle-term splitting, polynomial long division, and multi-step factorization for RIMC Entrance Exam.

Overview & Fundamental Concept

Algebraic Expressions, Identities, and Factorization constitute one of the highest-scoring and most heavily weighted segments of the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Item 46 of Chapter 8 in the official RIMC brochure, questions in this module appear regularly in Part A (5 Marks) and Part B (10 Marks). The syllabus tests mastery across four primary competencies: performing polynomial operations (addition, subtraction, horizontal/column multiplication, and polynomial long division), expanding and evaluating algebraic identities including $(a \pm b)^2$, $a^2 - b^2$, and $(a + b + c)^2$, factoring quadratic trinomials by rigorously splitting the middle term ($p \cdot q = ac$, $p + q = b$), and factoring multi-variable expressions via strategic regrouping and nested differences of squares. In the descriptive exam booklet, students must declare given identities before substitution, show detailed middle-term splits with explicit product-sum side columns, and present final algebraic factors in irreducible form.

Official Syllabus Coverage Breakdown
Syllabus Item Prescribed Scope Exam Relevance & Pedagogical Objective
Part of Item 46 Addition, Subtraction, Multiplication, Division, Simple Factors based on A²-B², split the Middle Term, formulas (a-b)², (a+b+c)² Explicitly specified in Chapter 8 of RIMC brochure. Forms the core of descriptive algebra testing manipulation, factor theorem, and polynomial division.
1. Terminology, Polynomial Operations & Degree

An algebraic expression is formed by combining variables and constants using the fundamental operations of addition, subtraction, multiplication, and division.

• Monomial: An expression containing exactly one non-zero term (e.g., $7x^2$, $-3xy$).
• Binomial: An expression containing exactly two unlike terms (e.g., $3x - 5y$, $a^2 + b^2$).
• Trinomial: An expression containing three unlike terms (e.g., $2x^2 + 5x - 7$).
• Polynomial: An algebraic expression with one or more terms with non-negative integer exponents on variables.

Like Terms vs. Unlike Terms: Terms having the exact same algebraic factors (same variables raised to the same exponents) are called like terms. Only like terms can be combined through addition or subtraction. For instance, $4x^2y$ and $-7x^2y$ are like terms ($4x^2y - 7x^2y = -3x^2y$), but $4x^2y$ and $4xy^2$ are unlike terms and cannot be combined.

Horizontal vs. Column Multiplication of Polynomials

When multiplying a binomial by a trinomial, such as $(2x + 3)(x^2 - 4x + 5)$:

  • Distributive Method: Multiply each term of the first polynomial by every term of the second:
    $$(2x + 3)(x^2 - 4x + 5) = 2x(x^2 - 4x + 5) + 3(x^2 - 4x + 5)$$
  • $$= (2x^3 - 8x^2 + 10x) + (3x^2 - 12x + 15)$$
  • Combine Like Terms: $2x^3 + (-8x^2 + 3x^2) + (10x - 12x) + 15 = \mathbf{2x^3 - 5x^2 - 2x + 15}$.
2. The Prescribed Standard Algebraic Identities

An identity is an equality that remains true for all possible numerical values of the variables involved. The official RIMC brochure prescribes six foundational identities:

Corollaries & Transformed Identities for High-Speed Evaluation

  • $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$
  • $(a + b)^2 - (a - b)^2 = 4ab$
  • $x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = \left(x - \frac{1}{x}\right)^2 + 2$
  • $(a - b)^2 = (a + b)^2 - 4ab$ and $(a + b)^2 = (a - b)^2 + 4ab$.
No.Standard Algebraic IdentityName / Pedagogical Focus
I$(a + b)^2 = a^2 + 2ab + b^2$Square of a Binomial Sum
II$(a - b)^2 = a^2 - 2ab + b^2$Square of a Binomial Difference
III$a^2 - b^2 = (a - b)(a + b)$Difference of Two Squares ($A^2 - B^2$)
IV$(x + a)(x + b) = x^2 + (a + b)x + ab$Product of Two Binomials with Common Term
V$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$Square of a Trinomial
VI$\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$Symmetric Reciprocal Identity
3. Complete Factorization Techniques for RIMC

Factorization is the reverse process of expansion: it decomposes an algebraic polynomial into a product of simpler irreducible factors. The four essential techniques are:

Method 1: Extraction of Common Monomial & Binomial Factors

Always inspect the expression for an overall greatest common divisor (HCF) before applying advanced methods:

  • Example: $12a^2b + 15ab^2 = 3ab(4a + 5b)$. Here $3ab$ is the common monomial factor.
  • Binomial Common: $5(x - 2y) - (x - 2y)^2 = (x - 2y)[5 - (x - 2y)] = (x - 2y)(5 - x + 2y)$.

Method 2: Factoring by Regrouping Terms (4-Term Expressions)

When an expression has four terms without an overall common factor, group them in pairs $(2 + 2)$ or as $(3 - 1)$:

  • Pairwise Grouping: $ax + bx - ay - by = x(a + b) - y(a + b) = (a + b)(x - y)$.
  • Trinomial Square minus Monomial: $x^2 + 2xy + y^2 - z^2 = (x + y)^2 - z^2 = (x + y - z)(x + y + z)$.

Method 3: Difference of Two Squares $(A^2 - B^2)$

Express each term as a complete perfect square: $A^2 - B^2 = (A - B)(A + B)$:

  • Nested Example: $x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$.
  • Never stop at $(x^2 - 9)$! Examiners deduct marks for incomplete factorization.

Method 4: Factoring Quadratic Trinomials by Splitting the Middle Term

To factorize the general quadratic trinomial $ax^2 + bx + c$:

  • Step 1: Compute the target product $P = a \times c$ and target sum $S = b$.
  • Step 2: Find two integers $p$ and $q$ such that $p \times q = P$ and $p + q = S$.
  • Step 3: Rewrite the middle term $bx$ as $px + qx$: $ax^2 + px + qx + c$.
  • Step 4: Group in pairs and factor out the common binomial factor: $(mx + n)(kx + l)$.
4. Polynomial Long Division Algorithm

In RIMC Part B (10-mark problems), long division of a polynomial $P(x)$ by a divisor $D(x)$ is frequently tested.

Division Algorithm:

$$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$$

$$P(x) = D(x) \cdot Q(x) + R(x) \quad \text{where } R(x) = 0 \text{ or } \deg R(x) < \deg D(x)$$

Rules for Polynomial Long Division:

1. Arrange both the dividend and divisor in descending powers of the variable (standard index form).
2. If any power is missing, insert it with coefficient $0$ (e.g., write $x^3 - 1$ as $x^3 + 0x^2 + 0x - 1$).
3. Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
4. Multiply the entire divisor by this quotient term and subtract from the dividend.
5. Repeat until the degree of the remainder is strictly less than the degree of the divisor.

Illustrated Conceptual Diagrams & Visual Schematics
Visual Geometric Proof of Standard Algebraic Identities
Geometric Proof: (a + b)² = a² + 2ab + b² a² ab ab b² a b a b Total Area = a² + ab + ab + b² = a² + 2ab + b² The 4 Core RIMC Algebraic Identities Identity 1: Square of Sum (a + b)² = a² + 2ab + b² Identity 2: Square of Difference (a - b)² = a² - 2ab + b² Identity 3: Difference of Squares a² - b² = (a - b)(a + b) Identity 4: Product of Binomials (x + a)(x + b) = x² + (a+b)x + ab
Geometric Proof of $(a+b)^2$: A large square of side $(a+b)$ is divided into 4 regions: square of area $a^2$, two rectangles of area $ab$, and small square of area $b^2$. Total Area = $a^2 + 2ab + b^2$. Similarly, $a^2 - b^2 = (a-b)(a+b)$.

Topper Strategy & Descriptive Step Breakdown

Mastering RIMC Mathematics Standard

In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.

Official Descriptive Marking Scheme & Step Allocation

In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:

Step 1: Problem Setup & Variable / Identity Declaration 1 - 2 Marks

State the given algebraic expression, arrange in descending order of powers, and declare the identity to be used.

Booklet Layout:
Given quadratic trinomial: $6x^2 + 17x + 5$\nHere, $a = 6$, $b = 17$, $c = 5$.\nProduct $ac = 6 \times 5 = 30$; Sum $b = 17$.
Step 2: Splitting & Factor Pair Selection 2 - 3 Marks

Show the two split factors with explicit multiplication and addition checks in the side column.

Booklet Layout:
Finding two numbers whose product is $30$ and sum is $17$:\n$15 \times 2 = 30$ and $15 + 2 = 17$.\nSplitting the middle term $17x$ as $15x + 2x$:\n$= 6x^2 + 15x + 2x + 5$
Step 3: Pairwise Grouping & Common Binomial Extraction 2 - 3 Marks

Group the four terms in pairs, factor out the HCF of each pair, and extract the identical binomial factor.

Booklet Layout:
$$= 3x(2x + 5) + 1(2x + 5)$$
$$= (2x + 5)(3x + 1)$$
Step 4: Boxed Final Factors & Irreducibility Check 1 - 2 Marks

Confirm that neither factor can be factorized further, and present the final answer in a clear box.

Booklet Layout:
$$\mathbf{\text{Answer: } (2x + 5)(3x + 1)}$$
Examiner Traps & Common Blunders (Marks Penalties)
Forgetting the Middle 2ab Term in Binomial Squares -2 to -3 Marks
Common Mistake: Writing $(a + b)^2 = a^2 + b^2$ or $(a - b)^2 = a^2 - b^2$.
Cadet Antidote: Always write out all three terms: $(a \pm b)^2 = a^2 \pm 2ab + b^2$. The cross term $2ab$ is never zero for non-zero variables.
Incomplete Factorization of Differences of Squares -2 Marks
Common Mistake: Leaving $x^4 - 16 = (x^2 - 4)(x^2 + 4)$ as the final answer.
Cadet Antidote: Check every factor! $(x^2 - 4)$ is itself a difference of squares: $(x - 2)(x + 2)$. The complete factorization is $(x - 2)(x + 2)(x^2 + 4)$.
Sign Errors when Subtracting in Polynomial Long Division -3 to -4 Marks
Common Mistake: Forgetting to invert signs of all terms when subtracting the product line in long division.
Cadet Antidote: Always draw circled opposite signs below each subtracted term before adding vertically.
Middle Term Splitting Sign Blunders with Negative Constants -2 Marks
Common Mistake: For $x^2 - 2x - 15$, choosing $-5$ and $-3$ ($(-5) \times (-3) = +15 \neq -15$).
Cadet Antidote: When $c < 0$, the two factors must have opposite signs. Here $-5$ and $+3$: $(-5) \times 3 = -15$ and $(-5) + 3 = -2$.
Cadet Topper Mental Verification Shortcuts
Synthetic Value Test (x = 1 Check)
To verify whether your factors $(2x+5)(3x+1)$ are correct for $6x^2+17x+5$, substitute $x = 1$ in both: Original $= 6(1) + 17(1) + 5 = 28$. Factors $= (2+5)(3+1) = 7 \times 4 = 28$. If both numbers match, your factorization is guaranteed 100% correct!
Original at $x=1$: 28. Factors at $x=1$: 28. Perfect match in 5 seconds!
Direct Reciprocal Doubling for Power Calculations
To find $x^4 + \frac{1}{x^4}$ given $x + \frac{1}{x} = k$: first compute $x^2 + \frac{1}{x^2} = k^2 - 2 = M$. Then compute $x^4 + \frac{1}{x^4} = M^2 - 2$ directly.
If $x + \frac{1}{x} = 5$: $x^2 + \frac{1}{x^2} = 5^2 - 2 = 23$. Then $x^4 + \frac{1}{x^4} = 23^2 - 2 = 529 - 2 = 527$.
Symmetric Rearrangement for Trinomial Squares
When expanding $(a - b - c)^2$, treat it as $[a + (-b) + (-c)]^2$ to systematically eliminate sign confusion in cross terms: $a^2 + b^2 + c^2 - 2ab + 2bc - 2ca$.
The product of two negatives is positive: $2(-b)(-c) = +2bc$.

Key Formulas & Conversion Cheat Sheet

Square of Binomial Sum
$$(a + b)^2 = a^2 + 2ab + b^2$$

Sum of squares plus twice their product.

Square of Binomial Difference
$$(a - b)^2 = a^2 - 2ab + b^2$$

Sum of squares minus twice their product.

Difference of Two Squares
$$a^2 - b^2 = (a - b)(a + b)$$

Factorizes into product of sum and difference.

Product of Binomials with Common Term
$$(x + a)(x + b) = x^2 + (a + b)x + ab$$

Basis of middle term splitting for quadratics.

Square of a Trinomial
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$

Sum of three individual squares plus twice the cyclic pairwise products.

Reciprocal Square Identity
$$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = \left(x - \frac{1}{x}\right)^2 + 2$$

Used to compute higher powers of reciprocal sums without solving for x.

Quadratic Middle Term Split Formula
$$ax^2 + bx + c = ax^2 + px + qx + c \quad (p \cdot q = ac,\; p + q = b)$$

Splits linear coefficient into two factors whose product equals ac.

Polynomial Division Algorithm
$$P(x) = D(x) \cdot Q(x) + R(x) \quad (\deg R < \deg D)$$

Dividend equals Divisor times Quotient plus Remainder.

Solved Model Examples (Descriptive Format with Step Mark Distribution)
Model 1: Evaluating Identities without Direct Multiplication 5 Marks
Evaluate the following using suitable algebraic identities showing all intermediate steps:
(a) $105^2 - 95^2$
(b) $103 \times 97$
Step-by-Step Model Solution:
Step 1: Evaluate Part (a) using $a^2 - b^2$ 2.5 Marks
The given expression is $105^2 - 95^2$.
Using the identity $a^2 - b^2 = (a + b)(a - b)$, where $a = 105$ and $b = 95$:
$$105^2 - 95^2 = (105 + 95)(105 - 95)$$
$$= (200) \times (10) = 2000$$
Step 2: Evaluate Part (b) using $(a + b)(a - b)$ 2.5 Marks
Express both numbers relative to base $100$:
$103 = 100 + 3$ and $97 = 100 - 3$.
Using $(a + b)(a - b) = a^2 - b^2$, where $a = 100$ and $b = 3$:
$$103 \times 97 = (100 + 3)(100 - 3) = 100^2 - 3^2$$
$$= 10000 - 9 = 9991$$
Examiner Scoring Key: Candidates multiplying 105 * 105 and 95 * 95 directly receive zero marks because the question specifies 'using suitable algebraic identities'.
Model 2: Reciprocal Powers Identity 5 Marks
If $x - \frac{1}{x} = 4$, find the exact numerical values of:
(a) $x^2 + \frac{1}{x^2}$
(b) $x^4 + \frac{1}{x^4}$
Step-by-Step Model Solution:
Step 1: Square Both Sides for Part (a) 2.5 Marks
Given: $x - \frac{1}{x} = 4$.
Squaring both sides:
$$\left(x - \frac{1}{x}\right)^2 = 4^2$$
Using identity $(a - b)^2 = a^2 - 2ab + b^2$ where $a = x$ and $b = \frac{1}{x}$:
$$x^2 - 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^2} = 16$$
$$x^2 - 2 + \frac{1}{x^2} = 16 \implies x^2 + \frac{1}{x^2} = 16 + 2 = \mathbf{18}$$
Step 2: Square the Result for Part (b) 2.5 Marks
Now take $x^2 + \frac{1}{x^2} = 18$ and square both sides:
$$\left(x^2 + \frac{1}{x^2}\right)^2 = 18^2$$
Using $(a + b)^2 = a^2 + 2ab + b^2$ where $a = x^2$ and $b = \frac{1}{x^2}$:
$$(x^2)^2 + 2(x^2)\left(\frac{1}{x^2}\right) + \left(\frac{1}{x^2}\right)^2 = 324$$
$$x^4 + 2 + \frac{1}{x^4} = 324 \implies x^4 + \frac{1}{x^4} = 324 - 2 = \mathbf{322}$$
Examiner Scoring Key: Canceling x * (1/x) = 1 in the middle term must be shown clearly. Box both final values.
Model 3: Trinomial Square Expansion & Evaluation 7 Marks
Expand the expression $(2x - 3y + 4z)^2$ using standard algebraic identities. Then, find its numerical value when $x = 1$, $y = 2$, and $z = -1$.
Step-by-Step Model Solution:
Step 1: State Identity & Rewrite in Standard Form 2 Marks
Using the trinomial square identity:
$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$$
Let $a = 2x$, $b = -3y$, and $c = 4z$.
Step 2: Expand Step-by-Step 3 Marks
$$(2x - 3y + 4z)^2 = (2x)^2 + (-3y)^2 + (4z)^2 + 2(2x)(-3y) + 2(-3y)(4z) + 2(4z)(2x)$$
$$= 4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16zx$$
Step 3: Substitute Numerical Values & Evaluate 2 Marks
Substitute $x = 1, y = 2, z = -1$:
Direct check in base: $2(1) - 3(2) + 4(-1) = 2 - 6 - 4 = -8$. Square: $(-8)^2 = 64$.
In expanded form:
$$= 4(1)^2 + 9(2)^2 + 16(-1)^2 - 12(1)(2) - 24(2)(-1) + 16(-1)(1)$$
$$= 4(1) + 9(4) + 16(1) - 24 + 48 - 16$$
$$= 4 + 36 + 16 - 24 + 48 - 16 = 104 - 40 = \mathbf{64}$$
$$\mathbf{\text{Answer: } 4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16zx; \quad \text{Value = } 64}$$
Examiner Scoring Key: Careful sign evaluation: 2(-3y)(4z) = -24yz and 2(4z)(2x) = +16zx. Direct check confirms expansion is accurate.
Model 4: Splitting the Middle Term with Leading Coefficient a > 1 7 Marks
Factorize the following quadratic trinomial completely by splitting the middle term:
$$6x^2 + 17x + 5$$
Step-by-Step Model Solution:
Step 1: Identify Target Product & Sum 2 Marks
The given quadratic expression is in standard form $ax^2 + bx + c$, where:
$a = 6$, $b = 17$, $c = 5$.
Target Product $P = a \times c = 6 \times 5 = 30$.
Target Sum $S = b = 17$.
Step 2: Find the Factor Pair 2 Marks
List pairs of factors of $30$:
$1 \times 30$ (sum = 31)
$2 \times 15$ (sum = 17) -> Correct pair!
$3 \times 10$ (sum = 13)
$5 \times 6$ (sum = 11)
The required factors are $15$ and $2$, since $15 \times 2 = 30$ and $15 + 2 = 17$.
Step 3: Split Middle Term & Factor by Grouping 2 Marks
Rewrite the expression with the split middle term:
$$6x^2 + 17x + 5 = 6x^2 + 15x + 2x + 5$$
Group into pairs:
$$= (6x^2 + 15x) + (2x + 5)$$
Take out common factors from each pair:
$$= 3x(2x + 5) + 1(2x + 5)$$
Step 4: Extract Common Binomial & State Answer 1 Mark
Factor out the common binomial $(2x + 5)$:
$$= (2x + 5)(3x + 1)$$
$$\mathbf{\text{Answer: } (2x + 5)(3x + 1)}$$
Examiner Scoring Key: Showing 1(2x + 5) before factoring out (2x + 5) ensures students don't miss the +1 term in (3x + 1).
Model 5: Advanced Multi-Step Factorization 10 Marks
Factorize the following expressions completely:
(a) $a^4 - 2a^2b^2 + b^4 - 16c^2$
(b) $x^2 - y^2 - 2yz - z^2$
Step-by-Step Model Solution:
Step 1: Factorize Part (a) using Trinomial Square and A² - B² 5 Marks
Observe the first three terms of $a^4 - 2a^2b^2 + b^4 - 16c^2$:
$$(a^4 - 2a^2b^2 + b^4) = [(a^2)^2 - 2(a^2)(b^2) + (b^2)^2] = (a^2 - b^2)^2$$
Now rewrite the entire expression:
$$= (a^2 - b^2)^2 - (4c)^2$$
This is in the form $A^2 - B^2 = (A - B)(A + B)$, where $A = a^2 - b^2$ and $B = 4c$:
$$= (a^2 - b^2 - 4c)(a^2 - b^2 + 4c)$$
$$\mathbf{\text{Answer (a): } (a^2 - b^2 - 4c)(a^2 - b^2 + 4c)}$$
Step 2: Factorize Part (b) by Grouping Negative Trinomial 5 Marks
The given expression is $x^2 - y^2 - 2yz - z^2$.
Group the last three terms by factoring out $-1$:
$$= x^2 - (y^2 + 2yz + z^2)$$
Recognize $y^2 + 2yz + z^2 = (y + z)^2$:
$$= x^2 - (y + z)^2$$
This is now a difference of two squares $A^2 - B^2$ where $A = x$ and $B = (y + z)$:
$$= [x - (y + z)][x + (y + z)]$$
Distribute the negative sign in the first bracket:
$$= (x - y - z)(x + y + z)$$
$$\mathbf{\text{Answer (b): } (x - y - z)(x + y + z)}$$
Examiner Scoring Key: In part (b), distributing the minus sign to obtain (x - y - z) is where most candidates make a sign blunder (writing x - y + z). Deduct 2 marks for incorrect sign.
Model 6: Polynomial Long Division & Verification Algorithm 10 Marks
Divide the polynomial $P(x) = 6x^3 + 11x^2 - 39x - 65$ by the linear divisor $D(x) = 3x - 5$. State the quotient $Q(x)$ and remainder $R$. Verify your result using the Division Algorithm.
Step-by-Step Model Solution:
Step 1: First Division Step 2.5 Marks
Divide first term of dividend by first term of divisor: $\frac{6x^3}{3x} = 2x^2$.
Multiply divisor $(3x - 5)$ by $2x^2$:
$2x^2(3x - 5) = 6x^3 - 10x^2$.
Subtract from dividend:
$(6x^3 + 11x^2) - (6x^3 - 10x^2) = 21x^2$.
Bring down next term: $21x^2 - 39x$.
Step 2: Second Division Step 2.5 Marks
Divide: $\frac{21x^2}{3x} = 7x$.
Multiply divisor $(3x - 5)$ by $7x$:
$7x(3x - 5) = 21x^2 - 35x$.
Subtract:
$(21x^2 - 39x) - (21x^2 - 35x) = -4x$.
Bring down next term: $-4x - 65$.
Step 3: Third Division Step to Obtain Remainder 2.5 Marks
Notice that $\frac{-4x}{3x} = -\frac{4}{3}$.
Wait, let us check coefficients: if dividend is $6x^3 + 11x^2 - 39x - 20$?
Here $\frac{-4x}{3x} = -\frac{4}{3}$, remainder $= -65 - \left(-\frac{4}{3} \times (-5)\right) = -65 - \frac{20}{3} = -\frac{215}{3}$.
Quotient $Q(x) = 2x^2 + 7x - \frac{4}{3}$ and Remainder $R = -\frac{215}{3}$.
Step 4: Division Algorithm Verification 2.5 Marks
Division Algorithm: $\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$
$$(3x - 5)\left(2x^2 + 7x - \frac{4}{3}\right) + \left(-\frac{215}{3}\right)$$
$$= 3x\left(2x^2 + 7x - \frac{4}{3}\right) - 5\left(2x^2 + 7x - \frac{4}{3}\right) - \frac{215}{3}$$
$$= (6x^3 + 21x^2 - 4x) - (10x^2 + 35x - \frac{20}{3}) - \frac{215}{3}$$
$$= 6x^3 + (21 - 10)x^2 + (-4 - 35)x + \left(\frac{20}{3} - \frac{215}{3}\right)$$
$$= 6x^3 + 11x^2 - 39x - \frac{195}{3} = 6x^3 + 11x^2 - 39x - 65 = \text{Dividend}$$
$$\mathbf{\text{Verified Successfully!}}$$
Examiner Scoring Key: Award full 10 marks when the long division layout is accompanied by the algebraic multiplication verification.

RIMC Practice & Drill Exercises

Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.

Part A: Short Descriptive Problems (5 Marks) RIMC Descriptive Pattern
Question 1 5 Marks
Expand the algebraic expression $(3a - 4b)^2$ using a standard identity, and find its value when $a = 2$ and $b = 3$.
Model Descriptive Solution:
Step 1: State Identity & Expand 2.5 Marks
Using $(x - y)^2 = x^2 - 2xy + y^2$ where $x = 3a$ and $y = 4b$:
$$(3a - 4b)^2 = (3a)^2 - 2(3a)(4b) + (4b)^2 = 9a^2 - 24ab + 16b^2$$
Step 2: Substitute Values & Evaluate 2.5 Marks
Substitute $a = 2$ and $b = 3$:
$$3a - 4b = 3(2) - 4(3) = 6 - 12 = -6$$
$$(-6)^2 = 36$$
Check in expansion: $9(2^2) - 24(2)(3) + 16(3^2) = 9(4) - 144 + 16(9) = 36 - 144 + 144 = 36$.
Final Answer: Expansion: $9a^2 - 24ab + 16b^2$; Value: $36$
Question 2 5 Marks
Factorize completely using the difference of two squares identity:
(a) $49x^2 - 64y^2$
(b) $81 - 16x^4$
Model Descriptive Solution:
Step 1: Factorize Part (a) 2.5 Marks
$$49x^2 - 64y^2 = (7x)^2 - (8y)^2 = (7x - 8y)(7x + 8y)$$
Step 2: Factorize Part (b) Fully 2.5 Marks
$$81 - 16x^4 = 9^2 - (4x^2)^2 = (9 - 4x^2)(9 + 4x^2)$$
Now factorize $(9 - 4x^2)$ further:
$$9 - 4x^2 = 3^2 - (2x)^2 = (3 - 2x)(3 + 2x)$$
Thus: $81 - 16x^4 = (3 - 2x)(3 + 2x)(9 + 4x^2)$.
Final Answer: (a) $(7x - 8y)(7x + 8y)$; (b) $(3 - 2x)(3 + 2x)(9 + 4x^2)$
Question 3 5 Marks
If $x + \frac{1}{x} = 6$, compute the values of:
(a) $x^2 + \frac{1}{x^2}$
(b) $x - \frac{1}{x}$
Model Descriptive Solution:
Step 1: Compute Part (a) 2.5 Marks
Squaring both sides of $x + \frac{1}{x} = 6$:
$$\left(x + \frac{1}{x}\right)^2 = 6^2 \implies x^2 + 2 + \frac{1}{x^2} = 36$$
$$x^2 + \frac{1}{x^2} = 36 - 2 = 34$$
Step 2: Compute Part (b) 2.5 Marks
Using $\left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2$:
$$\left(x - \frac{1}{x}\right)^2 = 34 - 2 = 32$$
Taking square root: $x - \frac{1}{x} = \pm \sqrt{32} = \pm 4\sqrt{2}$.
Final Answer: (a) $34$; (b) $\pm 4\sqrt{2}$
Part B: Long Descriptive Problems (7-10 Marks) RIMC Descriptive Pattern
Question 4 7 Marks
Factorize the quadratic polynomial by splitting the middle term:
$$2x^2 + 11x + 12$$
Model Descriptive Solution:
Step 1: Determine Target Product and Sum 2 Marks
$a = 2$, $b = 11$, $c = 12$.
Product $ac = 2 \times 12 = 24$. Sum $b = 11$.
Step 2: Find Factors and Split Middle Term 3 Marks
Factors of $24$ adding up to $11$: $8$ and $3$ ($8 \times 3 = 24$ and $8 + 3 = 11$).
$$2x^2 + 11x + 12 = 2x^2 + 8x + 3x + 12$$
Step 3: Factor in Pairs 2 Marks
$$= 2x(x + 4) + 3(x + 4) = (x + 4)(2x + 3)$$
Final Answer: $(x + 4)(2x + 3)$
Question 5 7 Marks
Factorize the quadratic trinomial with negative constant term:
$$x^2 - 8x - 105$$
Model Descriptive Solution:
Step 1: Identify Product and Sum Parameters 2 Marks
$a = 1$, $b = -8$, $c = -105$.
Product $ac = -105$; Sum $b = -8$.
Step 2: Factor Pairs of -105 3 Marks
Since product is negative, one factor is positive and one is negative.
$105 = 3 \times 5 \times 7 = 15 \times 7$.
Notice $-15 + 7 = -8$ and $(-15) \times 7 = -105$.
Rewrite: $x^2 - 15x + 7x - 105$.
Step 3: Group and Factor 2 Marks
$$= x(x - 15) + 7(x - 15) = (x - 15)(x + 7)$$
Final Answer: $(x - 15)(x + 7)$
Question 6 7 Marks
Factorize the following 4-term expression by strategic expansion and regrouping:
$$ab(x^2 + y^2) - xy(a^2 + b^2)$$
Model Descriptive Solution:
Step 1: Expand All Brackets 2.5 Marks
$$abx^2 + aby^2 - a^2xy - b^2xy$$
Step 2: Regroup Terms in Alternate Pairs 2.5 Marks
Group 1st with 3rd, and 2nd with 4th:
$$= (abx^2 - a^2xy) - (b^2xy - aby^2)$$
$$= ax(bx - ay) - by(bx - ay)$$
Step 3: Extract Common Binomial 2 Marks
$$= (bx - ay)(ax - by)$$
Final Answer: $(bx - ay)(ax - by)$
Question 7 7 Marks
Expand the trinomial square $(x - 2y - 3z)^2$ using the identity $(a+b+c)^2$.
Model Descriptive Solution:
Step 1: State Identity with Negative Signs Included 2 Marks
Identity: $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$.
Let $a = x$, $b = -2y$, $c = -3z$.
Step 2: Substitute and Expand 3 Marks
$$= x^2 + (-2y)^2 + (-3z)^2 + 2(x)(-2y) + 2(-2y)(-3z) + 2(-3z)(x)$$
Step 3: Simplify Signs 2 Marks
$$= x^2 + 4y^2 + 9z^2 - 4xy + 12yz - 6zx$$
Final Answer: $x^2 + 4y^2 + 9z^2 - 4xy + 12yz - 6zx$
Question 8 10 Marks
Factorize each of the following expressions completely:
(a) $x^4 + 4x^2 + 3$
(b) $a^2 + 4b^2 - 4ab - 9c^2$
Model Descriptive Solution:
Step 1: Factorize Part (a) via Quadratic in x² 5 Marks
Let $y = x^2$. Then $x^4 + 4x^2 + 3 = y^2 + 4y + 3$.
Splitting middle term: $3 \times 1 = 3$ and $3 + 1 = 4$.
$$y^2 + 3y + y + 3 = y(y + 3) + 1(y + 3) = (y + 3)(y + 1)$$
Substitute back $y = x^2$:
$$= (x^2 + 3)(x^2 + 1)$$
Step 2: Factorize Part (b) via Perfect Square Trinomial 5 Marks
Observe the first three terms of $a^2 + 4b^2 - 4ab - 9c^2$:
$$(a^2 - 4ab + 4b^2) = (a - 2b)^2$$
Now the expression becomes:
$$(a - 2b)^2 - (3c)^2$$
Using $A^2 - B^2 = (A - B)(A + B)$:
$$= (a - 2b - 3c)(a - 2b + 3c)$$
Final Answer: (a) $(x^2 + 3)(x^2 + 1)$; (b) $(a - 2b - 3c)(a - 2b + 3c)$
Question 9 10 Marks
Divide $2x^4 - 9x^3 + 21x^2 - 26x + 12$ by $(2x - 3)$. State the quotient and remainder, and verify using the Division Algorithm.
Model Descriptive Solution:
Step 1: Long Division First Two Terms 3 Marks
$\frac{2x^4}{2x} = x^3$. Multiply $(2x-3) \times x^3 = 2x^4 - 3x^3$.
Subtract: $(-9x^3) - (-3x^3) = -6x^3$. Bring down $21x^2$: $-6x^3 + 21x^2$.
$\frac{-6x^3}{2x} = -3x^2$. Multiply: $-6x^3 + 9x^2$.
Subtract: $21x^2 - 9x^2 = 12x^2$. Bring down $-26x$: $12x^2 - 26x$.
Step 2: Long Division Final Steps 3 Marks
$\frac{12x^2}{2x} = 6x$. Multiply: $12x^2 - 18x$.
Subtract: $(-26x) - (-18x) = -8x$. Bring down $12$: $-8x + 12$.
$\frac{-8x}{2x} = -4$. Multiply: $-8x + 12$.
Subtract: $(-8x + 12) - (-8x + 12) = 0$.
Step 3: Verification 4 Marks
Quotient $Q(x) = x^3 - 3x^2 + 6x - 4$ and Remainder $R = 0$.
Check: $(2x - 3)(x^3 - 3x^2 + 6x - 4) = 2x^4 - 6x^3 + 12x^2 - 8x - 3x^3 + 9x^2 - 18x + 12 = 2x^4 - 9x^3 + 21x^2 - 26x + 12 = \text{Dividend}$.
Final Answer: Quotient: $x^3 - 3x^2 + 6x - 4$; Remainder: $0$ [Verified]
Question 10 10 Marks
If $a + b + c = 9$ and $ab + bc + ca = 26$, determine the numerical value of $a^2 + b^2 + c^2$. Also find the value of $a^3 + b^3 + c^3 - 3abc$ if $abc = 24$.
Model Descriptive Solution:
Step 1: Compute a² + b² + c² using Trinomial Square 5 Marks
We know that $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$.
Substitute given values $a + b + c = 9$ and $ab + bc + ca = 26$:
$$9^2 = a^2 + b^2 + c^2 + 2(26)$$
$$81 = a^2 + b^2 + c^2 + 52$$
$$a^2 + b^2 + c^2 = 81 - 52 = \mathbf{29}$$
Step 2: Compute a³ + b³ + c³ - 3abc 5 Marks
Using identity: $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$$= (a + b + c)[(a^2 + b^2 + c^2) - (ab + bc + ca)]$$
Substitute $a + b + c = 9$, $a^2 + b^2 + c^2 = 29$, and $ab + bc + ca = 26$:
$$= 9 \times [29 - 26] = 9 \times 3 = \mathbf{27}$$
Final Answer: $a^2 + b^2 + c^2 = 29$; $a^3 + b^3 + c^3 - 3abc = 27$

High-Yield FAQs & Revision Notes

What is the difference between an algebraic identity and an algebraic equation?
An identity is true for all possible values of its variables (e.g. $(a+b)^2 = a^2+2ab+b^2$). An equation is true only for specific values called roots (e.g. $2x + 3 = 7$ is true only for $x = 2$).
Why do we split the middle term in quadratic polynomials?
Splitting the linear term $bx$ into $px + qx$ (where $pq = ac$) transforms a 3-term trinomial into a 4-term expression that can be factored in pairs by taking out common binomial factors.
How do you know what signs to use when splitting the middle term?
If $ac > 0$, both factors must have the same sign (both positive if $b > 0$, both negative if $b < 0$). If $ac < 0$, the factors must have opposite signs, and the larger factor takes the sign of $b$.
What is irreducible factorization?
Factorization where no factor can be decomposed into simpler algebraic factors. For example, $x^4 - 16 = (x - 2)(x + 2)(x^2 + 4)$ is irreducible over real numbers.
What is the degree of a polynomial?
The highest power of the variable present in the polynomial. In multivariable terms, it is the highest sum of the exponents of variables in any single term.
Can (x + 1/x) ever equal a value less than 2 for positive x?
No. For any positive real number $x$, $x + \frac{1}{x} \ge 2$ by the AM-GM inequality, with equality holding when $x = 1$.
How do you check your factorization in the exam hall?
Use the Synthetic Value Test: substitute $x = 1$ in the original expression and in your final factored expression. Both must evaluate to the exact same number.
What is the division algorithm for polynomials?
$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$, where the degree of the remainder is strictly less than the degree of the divisor.
Why must like terms be grouped before performing operations?
Only terms with identical variables and powers have the same algebraic dimensions and can be arithmetically combined through addition or subtraction.
How does this module map to the official RIMC syllabus?
Directly covers the core polynomial operations, identities $(a-b)^2, (a+b+c)^2$, difference of squares $A^2-B^2$, and middle-term splitting from Item 46 of Chapter 8 in the RIMC brochure.
How does this topic relate to the official RIMC syllabus?
This module directly covers Serial No. Part of 46 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.
Last-Minute Cadet High-Yield Revision Matrix
Topic / Concept High-Yield Exam Rules & Memory Anchor
Binomial Squares $(a \pm b)^2 = a^2 \pm 2ab + b^2$. Never forget the middle $2ab$ term!
Difference of Squares $a^2 - b^2 = (a - b)(a + b)$. Always check if resulting factors can be factored again.
Trinomial Square $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$. Pay careful attention to negative signs in pairwise cross terms.
Middle Term Split For $ax^2 + bx + c$, find $p, q$ such that $pq = ac$ and $p + q = b$.
Reciprocal Powers $x^2 + 1/x^2 = (x + 1/x)^2 - 2$. Then $(x^2 + 1/x^2)^2 - 2 = x^4 + 1/x^4$.
Verification Shortcut Substitute $x = 1$ in the question and in your factors to verify equality in 5 seconds.