Algebra

1. Finding the factors of $x^2 + 3x + 2$ with algebraic tiles (4 marks)

Materials needed: Cardboard or chart‑paper tiles of three kinds —

Activity (students work in groups of 4–5):

Step 1. Give each group exactly 1 big square ($x^2$), 3 long rectangles ($x$) and 2 small squares ($1$) — that is, the tiles representing $x^2 + 3x + 2$.

Step 2. Ask the students to arrange all the tiles so that they form a complete rectangle with no gap and no overlap. Let them try and fail a few times; the teacher only guides by asking questions.

Step 3. The successful arrangement is:

• Put the $x^2$ tile at the top‑left corner.
• Place two $x$ tiles standing upright (each $x$ tall, $1$ wide) to its right — this makes the top part $x$ tall and $(x+2)$ wide.
• Place the third $x$ tile lying flat ( $x$ wide, $1$ tall) just below the $x^2$ tile.
• Fill the remaining $2 \times 1$ space at the bottom‑right with the 2 unit tiles.

The whole figure is now a rectangle whose length is $x + 2$ and breadth is $x + 1$.

Step 4. Since area of rectangle = length × breadth,

$x^2 + 3x + 2 = (x + 1)(x + 2)$

Step 5. Students verify by multiplying back: $(x+1)(x+2) = x^2 + 2x + x + 2 = x^2 + 3x + 2$. ✔

Pedagogical value: the two upright $x$ tiles and the one flat $x$ tile show clearly why $3x$ is broken as $2x + 1x$; the 2 unit tiles show why the constant is 2. The child thus sees the factorisation before writing it in symbols.

Limitation: tiles can be used only when the coefficients are small positive numbers (and for $x^2$ with coefficient 1). So after this concrete stage we must move to a symbolic method.

2. The suitable method to teach after tiles (2 marks)

The middle‑term splitting (breaking) method, also called factorisation by grouping, is the most suitable method. It is suitable because it is the direct symbolic translation of what the tiles showed: the middle term $3x$ is split into two parts whose product equals the product of the first and last terms. It works for all numerical coefficients (positive or negative) and for expressions such as $2x^2 + 7x + 3$, where tiles cannot be used. Thus the teaching sequence follows the concrete → pictorial → abstract order.

3. Explanation of the method with examples (4 marks)

Steps:

1. For $ax^2 + bx + c$, find the product $a \times c$.
2. Find two numbers whose product $= a\times c$ and whose sum $= b$.
3. Split the middle term $bx$ into two terms using these numbers.
4. Group the four terms in pairs and take out the common factor from each pair.
5. Take out the common binomial factor; the two factors are obtained.

Example 1: $x^2 + 3x + 2$ Here $a\times c = 1 \times 2 = 2$, and $b = 3$. The two numbers are $2$ and $1$ (product $2$, sum $3$).

$x^2 + 3x + 2 = x^2 + 2x + x + 2 = x(x+2) + 1(x+2) = (x+2)(x+1)$

Example 2: $x^2 + 7x + 12$ $a\times c = 12$, $b = 7$; the numbers are $4$ and $3$.

$x^2 + 7x + 12 = x^2 + 4x + 3x + 12 = x(x+4) + 3(x+4) = (x+4)(x+3)$

Example 3: $2x^2 + 7x + 3$ $a\times c = 2 \times 3 = 6$, $b = 7$; the numbers are $6$ and $1$.

$2x^2 + 7x + 3 = 2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (x+3)(2x+1)$

Example 4 (with a negative term): $x^2 - 5x + 6$ $a\times c = 6$, $b = -5$; the numbers are $-2$ and $-3$.

$x^2 - 5x + 6 = x^2 - 2x - 3x + 6 = x(x-2) - 3(x-2) = (x-2)(x-3)$

Verification: in each case, multiplying the two binomials gives back the original expression, so the factors are correct. In the classroom, the teacher should first let students do the tile activity, then ask them to write the same result by middle‑term splitting, and finally give practice sums so that the skill becomes permanent.

a) Union of sets and teaching the formula

Union of sets: The union of two sets $A$ and $B$ is the set of all elements that belong to $A$, or to $B$, or to both. It is denoted by $A \cup B$ and read as “$A$ union $B$”.

Symbolically, $A \cup B = \{x : x \in A \text{ or } x \in B\}$

Example: If $A=\{1,2,3\}$ and $B=\{3,4,5\}$, then $A \cup B = \{1,2,3,4,5\}$ Here the common element $3$ is written only once. In a Venn diagram, the union is shown by shading both circles completely.

Teaching $n(A \cup B)=n(A)+n(B)-n(A \cap B)$:

1. Review: Recall set, elements, cardinality $n(A)$, intersection $A \cap B$, and union $A \cup B$.
2. Activity: Write two overlapping sets on the board: $A=\{a,b,c\},\quad B=\{c,d,e\}$ Ask students to list $A \cup B$ and count its elements. They get $A \cup B=\{a,b,c,d,e\}$, so $n(A \cup B)=5$.
3. Compare: Now count $n(A)=3$, $n(B)=3$, and $n(A \cap B)=1$ because $c$ is common. Ask: “Why is $3+3=6$ but the union has only $5$ elements?” Students see that $c$ is counted twice.
4. Derive: To remove the double counting, subtract the common elements once: $n(A \cup B)=n(A)+n(B)-n(A \cap B)$ $5=3+3-1$
5. Venn diagram: Show two overlapping circles. The overlap is counted in both $A$ and $B$; hence it must be subtracted once.
6. Verify with disjoint sets: If $A=\{1,2\}$ and $B=\{3,4\}$, then $n(A \cap B)=0$, so $n(A \cup B)=2+2-0=4$
7. Practice: Give similar problems and ask students to find union, intersection, and verify the formula. Use inductive method and group work.

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b) Differences between median and mode with examples

Median: The median is the middle value of data arranged in ascending or descending order. It divides the data into two equal parts. It is a positional average.

• For odd number of observations: $\text{Median} = \left(\frac{n+1}{2}\right)^{th}$ term.
• For even number of observations: $\text{Median} = \frac{\left(\frac{n}{2}\right)^{th} \text{ term} + \left(\frac{n}{2}+1\right)^{th} \text{ term}}{2}$.

Example: Data: $2,3,4,5,6$. Here $n=5$, so median $=3^{rd}$ term $=4$. Data: $2,4,6,8$. Here $n=4$, so median $=\frac{4+6}{2}=5$.

Mode: The mode is the value that occurs most frequently in a data set. It is the most repeated observation.

Example: Data: $2,3,3,4,5$. Mode $=3$. Data: $1,2,2,3,3,4$. Modes $=2$ and $3$ (bimodal). Data: $1,2,3,4,5$. No mode.

Differences:

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Teaching $\bar{x}=\frac{\sum fm}{N}$ with examples

Symbols:

• $\bar{x}$ = arithmetic mean
• $f$ = frequency of each class
• $m$ = mid-value of the class $=\frac{\text{lower limit} + \text{upper limit}}{2}$
• $N = \sum f$ = total frequency
• $\sum fm$ = sum of the products of frequency and mid-value

Teaching steps:

1. First teach mean for ungrouped data: $\bar{x}=\frac{\text{sum of observations}}{\text{number of observations}}$ Example: Marks $10,20,30,40,50$; mean $=\frac{150}{5}=30$.

2. Explain that in grouped data individual values are not known, so each class is represented by its mid-value $m$.

3. Show that $fm$ gives the total of all observations in that class approximately. Then $\sum fm$ gives the total of all observations.

4. Divide $\sum fm$ by total frequency $N$ to get the mean.

Example table:

$\bar{x}=\frac{\sum fm}{N}=\frac{180}{10}=18$

5. Give another example for practice: Classes $5–10, 10–15$ with $f=4,6$; mid-values $7.5,12.5$; $fm=30,75$; $N=10$; $\sum fm=105$; mean $=10.5$.

6. Ask students to prepare the table, calculate $m$, $fm$, $\sum fm$, and then the mean. Emphasize that the mean must lie between the smallest and largest class limits.

Determinant

A determinant is a scalar value associated with a square matrix, computed by a fixed rule. For a $2\times2$ matrix, $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc,$ and for a $3\times3$ matrix (expanding along the first row), $\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg).$ It is written $|A|$ or $\det A$.

Properties

1. $|A^T|=|A|$.
2. Interchanging two rows (or columns) changes the sign of the determinant.
3. If two rows (or columns) are identical or proportional, the determinant is zero.
4. Multiplying every element of one row (or column) by $k$ multiplies the determinant by $k$.
5. Adding a multiple of one row (or column) to another row (or column) does not change the determinant.
6. If a row (or column) consists wholly of zeros, the determinant is zero.
7. $|AB|=|A||B|$, and $|kA|=k^n|A|$ for an $n\times n$ matrix.
8. For a triangular matrix, the determinant is the product of the diagonal elements.
9. A matrix is invertible if and only if its determinant is non-zero.

Solution by the determinant (Cramer's) method

$x-2y-7z=-6,\qquad 2x+3y+z=-8,\qquad 4x-y+z=-7$

$D=\begin{vmatrix}1&-2&-7\\2&3&1\\4&-1&1\end{vmatrix}=1(3+1)+2(2-4)-7(-2-12)=4-4+98=98$

$D_x=\begin{vmatrix}-6&-2&-7\\-8&3&1\\-7&-1&1\end{vmatrix}=-6(4)+2(-8+7)-7(8+21)=-24-2-203=-229$

$D_y=\begin{vmatrix}1&-6&-7\\2&-8&1\\4&-7&1\end{vmatrix}=1(-8+7)+6(2-4)-7(-14+32)=-1-12-126=-139$

$D_z=\begin{vmatrix}1&-2&-6\\2&3&-8\\4&-1&-7\end{vmatrix}=1(-21-8)+2(-14+32)-6(-2-12)=-29+36+84=91$

Since $D\ne0$, the system has a unique solution: $x=\frac{D_x}{D}=-\frac{229}{98},\qquad y=\frac{D_y}{D}=-\frac{139}{98},\qquad z=\frac{D_z}{D}=\frac{91}{98}=\frac{13}{14}$

Check (equation 2): $2\left(-\tfrac{229}{98}\right)+3\left(-\tfrac{139}{98}\right)+\tfrac{91}{98}=\dfrac{-458-417+91}{98}=-\dfrac{784}{98}=-8$ ✓

1. Concept of Quadratic Equation and Nature of Roots (2 marks)

Quadratic equation: An equation of the second degree in one variable, reducible to the standard form

$ax^2 + bx + c = 0,\quad a \neq 0$

where $a, b, c$ are real numbers and $x$ is the variable. Examples: $x^2 - 5x + 6 = 0$, $2x^2 + 3x - 2 = 0$.

Nature of roots: Solving by the quadratic formula,

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

the expression $D = b^2 - 4ac$ is called the discriminant. It decides the nature of the roots:

Teaching in the classroom: Begin with real-life situations (area of a rectangular field, projectile path) to motivate the topic; derive the formula by the completing-the-square method; then let pupils compute $D$ for several equations and discover the four cases themselves before the table is formalised. Use charts, GeoGebra/graph plotting, and group work; give graded practice (find roots, then predict nature without solving).

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2. Procedure for Teaching Factors of $px^2 + qx + c$ (4 marks)

Objective: To factorise a quadratic trinomial of the form $px^2 + qx + c$ by splitting the middle term.

Step 1 — Recall: Revise multiplication of binomials, e.g. $(x+2)(x+3) = x^2 + 5x + 6$, and the identity $(x+a)(x+b) = x^2 + (a+b)x + ab$.

Step 2 — State the rule: For $px^2 + qx + c$, find two numbers whose product $= p \times c$ and whose sum $= q$.

Step 3 — Demonstrate on the board (worked example):

Factorise $6x^2 + 11x + 4$.

• Here $p = 6$, $q = 11$, $c = 4$; so product $= 6 \times 4 = 24$, sum $= 11$.
• Pairs of factors of 24: $(1,24), (2,12), (3,8), (4,6)$. The pair with sum 11 is $(3, 8)$.
• Split the middle term: $6x^2 + 3x + 8x + 4$
• Group: $3x(2x + 1) + 4(2x + 1)$
• Take the common bracket: $(2x + 1)(3x + 4)$

Step 4 — Verification: Multiply back: $(2x+1)(3x+4) = 6x^2 + 8x + 3x + 4 = 6x^2 + 11x + 4$. ✓

Step 5 — Second example (sign variation): Factorise $2x^2 - 7x + 3$: product $= 6$, sum $= -7$ → pair $(-1, -6)$; $2x^2 - x - 6x + 3 = x(2x-1) -3(2x-1) = (2x-1)(x-3)$.

Step 6 — Pupils' practice: Give exercises such as $3x^2 + 10x + 8$, $5x^2 - 13x + 6$, $4x^2 + 4x - 3$; move among pupils, correct errors, and ask them to explain each step aloud.

Step 7 — Application: Show use in solving quadratic equations by the factor method, e.g. $6x^2 + 11x + 4 = 0 \Rightarrow x = -\tfrac12, -\tfrac43$.

Teaching aids: Chart of the rule, flash cards of number pairs, algebra tiles for visual grouping.

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3. Strategies for Teaching H.C.F. and L.C.M. (4 marks)

Concept clarity: H.C.F. (Highest Common Factor) is the product of the lowest powers of the common factors; L.C.M. (Lowest Common Multiple) is the product of the highest powers of all factors (common and non-common).

Strategies:

1. Concrete-to-abstract approach: Start with numbers (H.C.F. of 12 and 18; L.C.M. of 4 and 6) using factor trees and Venn diagrams, then move to algebraic expressions.
2. Venn diagram method: Write the prime factors of each expression in overlapping circles; the overlap gives H.C.F., the whole diagram gives L.C.M. This makes the "common" idea visible.
3. Factorisation first: Insist that pupils factorise each expression completely (by splitting the middle term, taking common factors, using identities) before finding H.C.F./L.C.M.
4. Worked model + guided practice: Solve one example fully on the board, then give a parallel problem for pupils to attempt in pairs (I do – we do – you do).
5. Comparison table: Have pupils record results in a two-column table (H.C.F. – lowest common powers; L.C.M. – highest powers) to contrast the two ideas.
6. Verification strategy: Check the relation $\text{H.C.F.} \times \text{L.C.M.} = \text{product of the two expressions}$ for two expressions, and check by substitution of a value of $x$.
7. Remedial and enrichment: Give slow learners single-variable monomials first; give fast learners three expressions at a time and problems involving fractions.
8. Use of ICT and games: GeoGebra, factor-matching card games, and board races to sustain interest.

Example (board work): Find H.C.F. and L.C.M. of $x^2 - 4$ and $x^2 + 5x + 6$. $x^2 - 4 = (x-2)(x+2)$; $x^2 + 5x + 6 = (x+2)(x+3)$. H.C.F. $= (x+2)$; L.C.M. $= (x-2)(x+2)(x+3)$.

1. Definitions

Matrix:
A matrix is a rectangular arrangement of numbers, symbols or expressions in rows and columns, enclosed within brackets.
Example:
$A=\begin{bmatrix}1&2&3\\2&1&1\\3&1&2\end{bmatrix}$

Determinant:
A determinant is a single scalar value associated with a square matrix. It is obtained by expanding the elements of the square matrix in a fixed order.
Example:
$|A|=\begin{vmatrix}1&2&3\\2&1&1\\3&1&2\end{vmatrix}$

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2. Basic differences between matrix and determinant

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3. Solve by matrix inverse method

Given equations: $x+2y+3z=7$ $2x+y+z=4$ $3x+y+2z=13$

In matrix form: $AX=B$

where

$$A=\begin{bmatrix}1&2&3\\2&1&1\\3&1&2\end{bmatrix},\quad X=\begin{bmatrix}x\\y\\z\end{bmatrix},\quad B=\begin{bmatrix}7\\4\\13\end{bmatrix}$$

Now, $|A|=\begin{vmatrix}1&2&3\\2&1&1\\3&1&2\end{vmatrix}$

Expanding along the first row: $=1(1\cdot2-1\cdot1)-2(2\cdot2-1\cdot3)+3(2\cdot1-1\cdot3)$ $=1(2-1)-2(4-3)+3(2-3)$ $=1-2-3=-4$

Since (|A|\neq0), inverse exists.

Cofactors: $C_{11}=1,\quad C_{12}=-1,\quad C_{13}=-1$ $C_{21}=-1,\quad C_{22}=-7,\quad C_{23}=5$ $C_{31}=-1,\quad C_{32}=5,\quad C_{33}=-3$

So cofactor matrix: $C=\begin{bmatrix}1&-1&-1\\-1&-7&5\\-1&5&-3\end{bmatrix}$

Adjoint matrix: $\operatorname{adj}A=C^T=\begin{bmatrix}1&-1&-1\\-1&-7&5\\-1&5&-3\end{bmatrix}$

Thus, $A^{-1}=\frac{\operatorname{adj}A}{|A|}$ $A^{-1}=-\frac14\begin{bmatrix}1&-1&-1\\-1&-7&5\\-1&5&-3\end{bmatrix}$ $A^{-1}=\begin{bmatrix}-\frac14&\frac14&\frac14\\[2mm]\frac14&\frac74&-\frac54\\[2mm]\frac14&-\frac54&\frac34\end{bmatrix}$

Now, $X=A^{-1}B$

$$\begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}-\frac14&\frac14&\frac14\\[2mm]\frac14&\frac74&-\frac54\\[2mm]\frac14&-\frac54&\frac34\end{bmatrix} \begin{bmatrix}7\\4\\13\end{bmatrix}$$

So, $x=-\frac14(7)+\frac14(4)+\frac14(13)=\frac{10}{4}=\frac52$

$y=\frac14(7)+\frac74(4)-\frac54(13)=\frac{-30}{4}=-\frac{15}{2}$

$z=\frac14(7)-\frac54(4)+\frac34(13)=\frac{26}{4}=\frac{13}{2}$

Therefore, $\boxed{x=\frac52,\quad y=-\frac{15}{2},\quad z=\frac{13}{2}}$