SEBA Class 10 Maths Revision 4 Solutions – Complete Exam Preparation Guide

Q1. Find common factors of the following:

(i) 14pq, 28p²q
Answer: Given: 14pq, 28p²q
We will find common numerical and variable factors.
Common number = 14
Common variables = p, q
So, common factor = \(14pq\)

(ii) 16x³ − 4x², 32x
Answer: Given: 16x³ − 4x², 32x
Factor first expression:
\(16x^3 - 4x^2 = 4x^2(4x - 1)\)
Common factor with 32x is \(4x\)
Final Answer: \(4x\)

(iii) 20pq, 30qr, 40rp
Answer: Given: 20pq, 30qr, 40rp
Common number = 10
No common variable in all three terms
Final Answer: 10

(iv) 3x²y³, 10x³y², 6x²y²z
Answer: Given: 3x²y³, 10x³y², 6x²y²z
Common number = 1
Common variables = \(x^2, y^2\)
Final Answer: \(x^2y^2\)

Q2. Factorise:

(i) 4a² + 8a³
Answer: Given: \(4a^2 + 8a^3\)
Take common factor \(4a^2\):
\[ 4a^2(1 + 2a) \]

(ii) 7x²y − 21xy²
Answer: Given: \(7x^2y - 21xy^2\)
Common factor = \(7xy\)
\[ 7xy(x - 3y) \]

(iii) a²bc + abc² + abc²
Answer: Given: \(a^2bc + abc^2 + abc^2\)
Common factor = \(abc\)
\[ abc(a + c + c) \] \[ = abc(a + 2c) \]

(iv) a³ − a²b
Answer: Given: \(a^3 - a^2b\)
Common factor = \(a^2\)
\[ a^2(a - b) \]

Q3. Factorise:

(i) x² + xy + 6x + 6y
Solution:
Given: \(x^2 + xy + 6x + 6y\)

Group terms:
\[ x^2 + xy + 6x + 6y = (x^2 + xy) + (6x + 6y) \] Take common factors:
\[ = x(x + y) + 6(x + y) \] \[ = (x + y)(x + 6) \]

(ii) xy + x + y + 1
Solution:
Given: \(xy + x + y + 1\)

Group terms:
\[ xy + x + y + 1 = (xy + x) + (y + 1) \] Take common factors:
\[ = x(y+1) + 1(y+1) \] \[ = (x+1)(y+1) \]

(iii) 24x²y + 12x² − 12xy − 6x
Solution:
Given: \(24x^2y + 12x^2 - 12xy - 6x\)

Group terms:
\[ = (24x^2y + 12x^2) - (12xy + 6x) \] Take common factors:
\[ = 12x^2(2y+1) - 6x(2y+1) \] Take common binomial factor:
\[ = (12x^2 - 6x)(2y+1) \] Factor further:
\[ = 6x(2x -1)(2y+1) \]

(iv) z − 7 + 7xy − xyz
Solution:
Given: \(z - 7 + 7xy - xyz\)

Rearrange terms:
\[ = (z - xyz) + (7xy - 7) \] Take common factors:
\[ = z(1 - xy) + 7(xy -1) \] \[ = z(1 - xy) - 7(1 - xy) \] Take common factor:
\[ = (z - 7)(1 - xy) \]

Q4. Express in factors:

(i) 4x² + 12x + 9
Solution:
Given: \(4x^2 + 12x + 9\)

We compare with identity:
\[ a^2 + 2ab + b^2 = (a+b)^2 \] Here,
\[ (2x)^2 = 4x^2,\; 2(2x)(3) = 12x,\; 3^2 = 9 \] So,
\[ 4x^2 + 12x + 9 = (2x + 3)^2 \]

(ii) 25m² + 30m + 9
Solution:
Given: \(25m^2 + 30m + 9\)

Compare with identity:
\[ a^2 + 2ab + b^2 \] \[ (5m)^2 = 25m^2,\; 2(5m)(3) = 30m,\; 3^2 = 9 \] So,
\[ 25m^2 + 30m + 9 = (5m + 3)^2 \]

(iii) x² − 10x + 25
Solution:
Given: \(x^2 - 10x + 25\)

Compare with identity:
\[ a^2 - 2ab + b^2 = (a-b)^2 \] \[ x^2 - 10x + 25 = x^2 - 2(x)(5) + 5^2 \] So,
\[ = (x - 5)^2 \]

(iv) 121b² − 88bc + 16c²
Solution:
Given: \(121b^2 - 88bc + 16c^2\)

Compare with identity:
\[ a^2 - 2ab + b^2 \] \[ (11b)^2 = 121b^2,\; 2(11b)(4c) = 88bc,\; (4c)^2 = 16c^2 \] So,
\[ = (11b - 4c)^2 \]

(v) 9p² − 16q²
Solution:
Given: \(9p^2 - 16q^2\)

This is a difference of squares:
\[ a^2 - b^2 = (a-b)(a+b) \] \[ 9p^2 = (3p)^2,\; 16q^2 = (4q)^2 \] So,
\[ = (3p - 4q)(3p + 4q) \]

(vi) (l + m)² − (l − m)²
Solution:
Given: \((l + m)^2 - (l - m)^2\)

Use identity:
\[ a^2 - b^2 = (a-b)(a+b) \] \[ = [(l+m) - (l-m)][(l+m) + (l-m)] \] Simplify:
\[ = (l + m - l + m)(l + m + l - m) \] \[ = (2m)(2l) \] \[ = 4lm \]

(vii) x² − 13x − 30
Solution:
Given: \(x^2 - 13x - 30\)

We find two numbers whose:
Sum = −13 and Product = −30

Numbers are −15 and 2

Split middle term:
\[ x^2 - 15x + 2x - 30 \] Group terms:
\[ = (x^2 - 15x) + (2x - 30) \] \[ = x(x - 15) + 2(x - 15) \] Take common factor:
\[ = (x - 15)(x + 2) \]

(viii) y² − 5y − 36
Solution:
Given: \(y^2 - 5y - 36\)

Find two numbers whose:
Sum = −5 and Product = −36

Numbers are −9 and 4

Split middle term:
\[ y^2 - 9y + 4y - 36 \] Group terms:
\[ = (y^2 - 9y) + (4y - 36) \] \[ = y(y - 9) + 4(y - 9) \] Take common factor:
\[ = (y - 9)(y + 4) \]

(ix) 4y² + 25y − 21
Solution:
Given: \(4y^2 + 25y - 21\)

Multiply 4 × (−21) = −84
Find two numbers whose sum = 25 and product = −84

Numbers are 28 and −3

Split middle term:
\[ 4y^2 + 28y - 3y - 21 \] Group terms:
\[ = (4y^2 + 28y) - (3y + 21) \] \[ = 4y(y + 7) - 3(y + 7) \] Take common factor:
\[ = (4y - 3)(y + 7) \]

(x) 3x⁶ − 6x²y − 45x²y²
Solution:
Given: \(3x^6 - 6x^2y - 45x^2y^2\)

Take common factor:
\[ = 3x^2(x^4 - 2y - 15y^2) \] Now factorise inside:
\[ x^4 - 2y - 15y^2 \] Find two numbers whose product = −15y² and sum = −2y
Numbers are −5y and 3y

Split middle term:
\[ x^4 - 5y + 3y - 15y^2 \] (Grouping may not simplify further neatly, so final form is:)
\[ = 3x^2(x^4 - 2y - 15y^2) \]