Mathematics for SGT and TET Paper 1A (Classes III-VIII) · Chapter 11

Factorisation, Polynomial Division and Linear Graphs

What to remember

To factorise is to write an expression as a product of factors. Methods: common factor, grouping, identities, and splitting the middle term.
Dividend = Divisor × Quotient + Remainder. By the remainder theorem, the remainder of p(x) ÷ (x − a) is p(a). If p(a) = 0, then (x − a) is a factor.
The graph of a linear equation ax + by + c = 0 is a straight line. In y = mx + c, m is the slope and c is the y-intercept.

Factorisation

A factor divides an expression exactly. Factorisation is the reverse of multiplication: 3x(x + 2) = 3x² + 6x, so factorising 3x² + 6x gives 3x(x + 2). Always check by multiplying back.

Method 1: Common factor. Take out the highest common factor (HCF) of all terms.

6x² + 9x = 3x(2x + 3).
12a²b − 18ab² = 6ab(2a − 3b).

Method 2: Grouping. Group terms that have a common factor.

ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y).
2xy + 3x − 2y − 3 = x(2y + 3) − 1(2y + 3) = (x − 1)(2y + 3).

Method 3: Identities.

Form	Factors
a² + 2ab + b²	(a + b)²
a² − 2ab + b²	(a − b)²
a² − b²	(a + b)(a − b)
a³ + b³	(a + b)(a² − ab + b²)
a³ − b³	(a − b)(a² + ab + b²)

x² + 10x + 25 = (x + 5)².
49x² − 36 = (7x + 6)(7x − 6).
4x² − 12x + 9 = (2x − 3)².
x³ + 8 = (x + 2)(x² − 2x + 4).

Method 4: Splitting the middle term for x² + bx + c. Find two numbers whose sum is b and whose product is c.

x² + 7x + 12: numbers 3 and 4 (sum 7, product 12), so (x + 3)(x + 4).
x² − 5x + 6: numbers −2 and −3, so (x − 2)(x − 3).
x² + x − 12: numbers 4 and −3, so (x + 4)(x − 3).
x² − 2x − 15: numbers −5 and 3, so (x − 5)(x + 3).

For ax² + bx + c (a ≠ 1), find two numbers whose sum is b and product is a × c. For 6x² + 17x + 5: a × c = 30, numbers 15 and 2. So 6x² + 15x + 2x + 5 = 3x(2x + 5) + 1(2x + 5) = (3x + 1)(2x + 5).

Factorise completely: continue until no factor can be factorised further. 2x² − 8 = 2(x² − 4) = 2(x + 2)(x − 2).

Division of polynomials

Monomial ÷ monomial: divide the numbers and subtract the powers: 24x⁵ ÷ 6x² = 4x³.

Polynomial ÷ monomial: divide each term. (6x³ + 9x² − 3x) ÷ 3x = 2x² + 3x − 1.

Polynomial ÷ polynomial by factorising: (x² − 9) ÷ (x + 3) = (x + 3)(x − 3) ÷ (x + 3) = x − 3.

Long division method (divide the leading term each time):

Divide x² + 5x + 6 by x + 2.

1. x² ÷ x = x. Multiply: x(x + 2) = x² + 2x. Subtract: 3x + 6.
2. 3x ÷ x = 3. Multiply: 3(x + 2) = 3x + 6. Subtract: 0.

Quotient = x + 3, remainder = 0.

Divide 2x² + 3x + 5 by x + 1.

1. 2x² ÷ x = 2x. Subtract 2x² + 2x: left x + 5.
2. x ÷ x = 1. Subtract x + 1: left 4.

Quotient = 2x + 1, remainder = 4. Check: (x + 1)(2x + 1) + 4 = 2x² + 3x + 1 + 4 = 2x² + 3x + 5.

Division rule: Dividend = Divisor × Quotient + Remainder. The degree of the remainder is less than the degree of the divisor. Write the terms in descending powers and put a zero coefficient for a missing power.

Remainder theorem: The remainder when p(x) is divided by (x − a) is p(a).

p(x) = x² + 3x + 5 divided by (x − 1): remainder = 1 + 3 + 5 = 9.
p(x) divided by (x + 2): put x = −2.

Factor theorem: (x − a) is a factor of p(x) exactly when p(a) = 0.

x³ − 6x² + 11x − 6: p(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor.
If (x − 2) is a factor of x² + kx − 10, then 4 + 2k − 10 = 0, so k = 3.

Linear graphs

Cartesian plane: Two perpendicular number lines meet at the origin O(0, 0). The horizontal line is the x-axis and the vertical line is the y-axis. The plane has four quadrants.

Quadrant	Sign of (x, y)
I	(+, +)
II	(−, +)
III	(−, −)
IV	(+, −)

A point is written (x, y). x is the abscissa (distance from the y-axis) and y is the ordinate (distance from the x-axis). A point on the x-axis has y = 0, such as (4, 0). A point on the y-axis has x = 0, such as (0, −3). The point (3, 5) is different from (5, 3), so the order matters.

Graph of a linear equation: An equation ax + by + c = 0 (a and b not both zero) has a graph that is a straight line. Every point on the line satisfies the equation.

Steps to draw:

1. Make a table of at least 2 points (take 3 so you can check).
2. Plot the points on graph paper with a suitable scale.
3. Join them with a ruler and extend both ends.

Example: y = 2x + 1. For x = 0, y = 1; x = 1, y = 3; x = 2, y = 5. Points (0, 1), (1, 3), (2, 5) lie on one straight line.

Special lines:

Equation	Graph
y = 0	The x-axis
x = 0	The y-axis
x = a	A vertical line parallel to the y-axis at distance a
y = b	A horizontal line parallel to the x-axis at distance b
y = mx	A line through the origin

Slope-intercept form: y = mx + c. m = slope (steepness) = rise ÷ run = (y₂ − y₁) ÷ (x₂ − x₁). c = the y-intercept, where the line cuts the y-axis. For y = 3x − 2, the slope is 3 and the y-intercept is −2.

To find the x-intercept, put y = 0. For 2x + 3y = 6: x = 3 and y-intercept (put x = 0) is y = 2.
Parallel lines have equal slopes.
A horizontal line has slope 0.

Distance from two points (optional extension): the distance between (x₁, y₁) and (x₂, y₂) is √[(x₂ − x₁)² + (y₂ − y₁)²]. The distance from the origin to (3, 4) is 5.

Line graphs in daily life: A distance-time graph with constant speed is a straight line, and the slope is the speed. A graph of the cost of n pens at a fixed price is also a straight line through the origin.

Classroom angle

Let children plot their own seating positions as points on a grid drawn on the floor to understand coordinates. Use a "treasure map" game with (x, y) pairs. For factorisation, use area models: rectangles with sides x + 3 and x + 4 show why x² + 7x + 12 factorises. Ask children to multiply back to check each factorisation. Link graphs with real examples such as the cost of notebooks against the number bought.

Exam traps

Taking out a common factor leaves 1, not 0, when a term equals the common factor: 4x + 4 = 4(x + 1).
x² + 9 cannot be factorised with real numbers, as the sum of two squares has no real factors.
In x² − x − 6, the numbers must multiply to −6 and add to −1, which are −3 and 2.
The factor theorem uses p(a) for (x − a), so for (x + 3) use p(−3).
The remainder must have a lower degree than the divisor.
Point (x, y) has x first. (3, 5) is not (5, 3).
Point (0, 5) lies on the y-axis, not the x-axis.
In the line y = 2x + 1, the y-intercept is 1; the slope is 2.

One-liners

Factorisation is the reverse of multiplication.
a² − b² = (a + b)(a − b).
x² + (a + b)x + ab = (x + a)(x + b).
Dividend = Divisor × Quotient + Remainder.
Remainder theorem: remainder of p(x) ÷ (x − a) is p(a).
Factor theorem: p(a) = 0 means (x − a) is a factor.
The origin is (0, 0).
Abscissa is the x-coordinate; ordinate is the y-coordinate.
The graph of a linear equation is a straight line.
y = mx + c: m is slope and c is the y-intercept.
x = a is a vertical line; y = b is a horizontal line.
Point (−2, 3) lies in quadrant II.

Practice questions

Factorising 6x² + 9x gives
1. 6x(x + 9)
2. 3x(2x + 3)
3. x(6x + 9)
4. 3(2x² + 3x)
Answer
B. 3x(2x + 3)
The HCF of 6x² and 9x is 3x.
x² − 49 factorises as
1. (x + 7)²
2. (x − 7)²
3. (x − 49)(x + 1)
4. (x + 7)(x − 7)
Answer
D. (x + 7)(x − 7)
a² − b² = (a + b)(a − b).
x² + 10x + 25 equals
1. (x − 5)²
2. (x + 5)²
3. (x + 5)(x − 5)
4. (x + 25)²
Answer
B. (x + 5)²
a² + 2ab + b² = (a + b)² with a = x, b = 5.
x² + 7x + 12 factorises as
1. (x + 3)(x + 4)
2. (x + 2)(x + 6)
3. (x + 1)(x + 12)
4. (x − 3)(x − 4)
Answer
A. (x + 3)(x + 4)
3 + 4 = 7 and 3 × 4 = 12.
x² − 5x + 6 factorises as
1. (x + 2)(x + 3)
2. (x − 1)(x − 6)
3. (x − 2)(x − 3)
4. (x + 2)(x − 3)
Answer
C. (x − 2)(x − 3)
−2 − 3 = −5 and (−2)(−3) = 6.
x² + x − 12 factorises as
1. (x − 4)(x + 3)
2. (x + 12)(x − 1)
3. (x + 6)(x − 2)
4. (x + 4)(x − 3)
Answer
D. (x + 4)(x − 3)
4 + (−3) = 1 and 4 × (−3) = −12.
x² − 2x − 15 factorises as
1. (x − 5)(x − 3)
2. (x − 15)(x + 1)
3. (x − 5)(x + 3)
4. (x + 5)(x − 3)
Answer
C. (x − 5)(x + 3)
−5 + 3 = −2 and (−5)(3) = −15.
6x² + 17x + 5 factorises as
1. (3x + 1)(2x + 5)
2. (3x + 5)(2x + 1)
3. (6x + 1)(x + 5)
4. (2x + 1)(3x − 5)
Answer
A. (3x + 1)(2x + 5)
Split 17x as 15x + 2x: 3x(2x + 5) + 1(2x + 5).
The factors of 2x² − 8 are
1. 2(x² − 4)
2. 2(x + 2)(x − 2)
3. 2(x − 2)²
4. (2x + 4)(x − 2)
Answer
B. 2(x + 2)(x − 2)
2(x² − 4) = 2(x + 2)(x − 2); it is the complete factorisation.
4x² − 12x + 9 equals
1. (4x − 3)²
2. (2x + 3)²
3. (2x − 3)(2x + 3)
4. (2x − 3)²
Answer
D. (2x − 3)²
(2x)² − 2(2x)(3) + 3² = (2x − 3)².
The factors of x³ + 8 are
1. (x + 2)³
2. (x + 2)(x² + 2x + 4)
3. (x − 2)(x² + 2x + 4)
4. (x + 2)(x² − 2x + 4)
Answer
D. (x + 2)(x² − 2x + 4)
a³ + b³ = (a + b)(a² − ab + b²) with a = x, b = 2.
ax + ay + bx + by factorises as
1. (a + b)(x − y)
2. (a + b)(x + y)
3. ab(x + y)
4. (a − b)(x + y)
Answer
B. (a + b)(x + y)
Group: a(x + y) + b(x + y).
(6x³ + 9x² − 3x) ÷ 3x equals
1. 2x² + 3x − 1
2. 2x³ + 3x² − 1
3. 2x² + 3x
4. 2x² + 3 − x
Answer
A. 2x² + 3x − 1
Divide each term by 3x.
The quotient when x² + 5x + 6 is divided by x + 2 is
1. x + 2
2. x − 3
3. x + 3
4. x + 6
Answer
C. x + 3
x² + 5x + 6 = (x + 2)(x + 3).
When 2x² + 3x + 5 is divided by x + 1, the remainder is
1. 0
2. 2
3. 5
4. 4
Answer
D. 4
Quotient 2x + 1 and remainder 4; or p(−1) = 2 − 3 + 5 = 4.
The division rule for polynomials is
1. Dividend = Quotient ÷ Divisor + Remainder
2. Dividend = Divisor × Quotient + Remainder
3. Divisor = Dividend × Quotient + Remainder
4. Dividend = Divisor + Quotient × Remainder
Answer
B. Dividend = Divisor × Quotient + Remainder
This is the standard division algorithm.
The remainder when x² + 3x + 5 is divided by (x − 1) is
1. 9
2. 10
3. 5
4. 8
Answer
A. 9
p(1) = 1 + 3 + 5 = 9.
If (x − 2) is a factor of x² + kx − 10, then k equals
1. −3
2. 2
3. 3
4. 5
Answer
C. 3
p(2) = 4 + 2k − 10 = 0, so k = 3.
Which is a factor of x³ − 6x² + 11x − 6?
1. x − 4
2. x + 1
3. x + 6
4. x − 1
Answer
D. x − 1
p(1) = 1 − 6 + 11 − 6 = 0.
The point where the x-axis and y-axis meet is
1. (0, 0)
2. (1, 0)
3. (1, 1)
4. (0, 1)
Answer
A. (0, 0)
The origin has both coordinates zero.
The point (−2, 3) lies in which quadrant?
1. I
2. II
3. IV
4. III
Answer
B. II
Negative x and positive y is quadrant II.
The x-coordinate of a point is called its
1. quadrant
2. origin
3. abscissa
4. ordinate
Answer
C. abscissa
Abscissa is the x-coordinate; ordinate is the y-coordinate.
The graph of a linear equation in two variables is
1. a straight line
2. a circle
3. a parabola
4. a triangle
Answer
A. a straight line
Every linear equation ax + by + c = 0 gives a straight line.
The graph of y = 0 is the
1. y-axis
2. x-axis
3. line through (1, 1)
4. origin only
Answer
B. x-axis
All points with ordinate 0 form the x-axis.
The graph of x = 3 is a
1. vertical line parallel to the y-axis
2. horizontal line parallel to the x-axis
3. line through the origin
4. line cutting both axes
Answer
A. vertical line parallel to the y-axis
All points have x = 3 whatever y is.
In y = 3x − 2, the slope is
1. −2
2. 2
3. 3
4. 1
Answer
C. 3
In y = mx + c, m is the slope.
In y = 3x − 2, the y-intercept is
1. 3
2. −2
3. 2
4. −3
Answer
B. −2
c = −2 is where the line cuts the y-axis.
The line 2x + 3y = 6 cuts the y-axis at
1. (3, 0)
2. (0, 6)
3. (0, 3)
4. (0, 2)
Answer
D. (0, 2)
Put x = 0: 3y = 6, so y = 2.
The line 2x + 3y = 6 cuts the x-axis at
1. (6, 0)
2. (0, 3)
3. (3, 0)
4. (2, 0)
Answer
C. (3, 0)
Put y = 0: 2x = 6, so x = 3.
Which point lies on the line y = 2x + 1?
1. (3, 6)
2. (1, 2)
3. (2, 4)
4. (2, 5)
Answer
D. (2, 5)
For x = 2, y = 5.
The slope of the line through (1, 3) and (3, 7) is
1. 1/2
2. 4
3. 3
4. 2
Answer
D. 2
(7 − 3) ÷ (3 − 1) = 2.
In a distance-time graph that is a straight line through the origin, the slope shows
1. the constant speed
2. the acceleration
3. the total distance
4. the total time
Answer
A. the constant speed
Distance ÷ time gives speed, which is the slope.
Which activity is useful for teaching coordinates in Class VIII?
1. Reading rules aloud
2. Copying the graphs from the board
3. Plotting positions of students on a grid
4. Memorising the quadrants
Answer
C. Plotting positions of students on a grid
Real positions on a grid make coordinates meaningful.
Statements: 1. x² + 9 can be written as (x + 3)². 2. x² − 9 = (x + 3)(x − 3). Which is/are correct?
1. Neither 1 nor 2
2. 1 only
3. 2 only
4. Both 1 and 2
Answer
C. 2 only
(x + 3)² = x² + 6x + 9, so statement 1 is wrong.
Statements: 1. If p(a) = 0, then (x − a) is a factor of p(x). 2. The remainder when p(x) is divided by (x − a) is p(a). Which is/are correct?
1. Both 1 and 2
2. Neither 1 nor 2
3. 1 only
4. 2 only
Answer
A. Both 1 and 2
These are the factor and remainder theorems.
Statements: 1. The point (3, 5) is the same as (5, 3). 2. The point (0, 5) lies on the y-axis. Which is/are correct?
1. 2 only
2. Both 1 and 2
3. Neither 1 nor 2
4. 1 only
Answer
A. 2 only
Order matters in coordinates, so (3, 5) and (5, 3) are different.
Statements: 1. The remainder in polynomial division has a degree less than that of the divisor. 2. A missing power in the dividend should be written with a zero coefficient. Which is/are correct?
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are rules of long division.
Statements: 1. Parallel lines have equal slopes. 2. A horizontal line has slope 0. Which is/are correct?
1. Neither 1 nor 2
2. 1 only
3. 2 only
4. Both 1 and 2
Answer
D. Both 1 and 2
Both are true for lines in the plane.
Statements: 1. 4x + 4 = 4x. 2. 12a²b − 18ab² = 6ab(2a − 3b). Which is/are correct?
1. Both 1 and 2
2. Neither 1 nor 2
3. 1 only
4. 2 only
Answer
D. 2 only
4x + 4 = 4(x + 1); 4x is not equal to it.
Match the quadrant with the signs of (x, y): P. I Q. II R. III S. IV 1. (−, +) 2. (+, +) 3. (+, −) 4. (−, −)
1. P-1, Q-2, R-4, S-3
2. P-2, Q-1, R-4, S-3
3. P-2, Q-1, R-3, S-4
4. P-2, Q-4, R-1, S-3
Answer
B. P-2, Q-1, R-4, S-3
Quadrant I (+, +), II (−, +), III (−, −), IV (+, −).
Match the equation with its graph: P. y = 0 Q. x = 0 R. y = 4 S. x = 4 1. y-axis 2. x-axis 3. Horizontal line at height 4 4. Vertical line at distance 4
1. P-1, Q-2, R-3, S-4
2. P-2, Q-1, R-3, S-4
3. P-2, Q-3, R-1, S-4
4. P-2, Q-1, R-4, S-3
Answer
B. P-2, Q-1, R-3, S-4
y = constant is horizontal; x = constant is vertical.
Match the expression with its factors: P. x² − 16 Q. x² + 8x + 16 R. x² − 8x + 16 S. x² + 5x + 6 1. (x − 4)² 2. (x + 2)(x + 3) 3. (x + 4)(x − 4) 4. (x + 4)²
1. P-3, Q-4, R-2, S-1
2. P-3, Q-4, R-1, S-2
3. P-4, Q-3, R-1, S-2
4. P-3, Q-1, R-4, S-2
Answer
B. P-3, Q-4, R-1, S-2
Apply the identities and sum-product rule.
Match each point with its position: P. (0, −3) Q. (4, 0) R. (−1, −1) S. (2, 6) 1. x-axis 2. y-axis 3. Quadrant III 4. Quadrant I
1. P-2, Q-1, R-3, S-4
2. P-2, Q-1, R-4, S-3
3. P-2, Q-3, R-1, S-4
4. P-1, Q-2, R-3, S-4
Answer
A. P-2, Q-1, R-3, S-4
Zero x means y-axis; zero y means x-axis.
The distance of the point (3, 4) from the origin is
1. 7
2. 1
3. 25
4. 5
Answer
D. 5
√(3² + 4²) = √25 = 5.
Which of these can NOT be factorised using real numbers?
1. x² − 9
2. x² − 6x + 9
3. x² + 9
4. x² + 6x + 9
Answer
C. x² + 9
The sum of two squares has no real linear factors.

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1.Rs.900 is divided among A, B, C; the division is such that 1/2 of A's money = 1/3 of B's money = 1/4 of C's money. Find the amount (in Rs.) received by A, B, C.
2.Without actual division, find the remainder when 379843 is divided by 3.
3.In a division sum, the divisor is 13 times the quotient and 6 times the remainder. If the remainder is 39, then the dividend is:
4.Without actual division, find the remainder when 379843 is divided by 3. Te IN Download Test RanKING AP P YOU
5.In a division sum, the divisor is 9 times the quotient and 3 times the remainder. If the remainder is 42, what is the dividend?
6.In a division sum, the divisor is 5 times the quotient and 4 times the remainder. If the remainder is 15, what is the dividend?
7.In a division sum, the divisor is 2 times the quotient and 3 times the remainder. If the remainder is 10, what is the dividend?
8.In a division sum, the divisor is 14 times the quotient and 5 times the remainder. If the remainder is 56, what is the dividend?
9.In finding the HCF of two numbers by division method, the quotients are 1, 5, and 8, respectively, and the last divisor is 57. What is the LCM of the two numbers?
10.In finding the HCF of two numbers by division method, the quotients are 1, 4, and 6, respectively, and the last divisor is 29. What is the LCM of the two numbers?

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