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Class 8 Maths Chapter 12 Important Question Answers - Factorisation

Q1: Find the common factors of the following terms.
(a) 25x²y, 30xy²
(b) 63m³n, 54mn⁴
Sol:
(a) 25x²y, 30xy²
25x²y = 5 × 5 × x × x × y
30xy² = 2 × 3 × 5 × x × y × y
Common factors are 5× x × y = 5 xy
(b) 63m³n, 54mn⁴
63m³n = 3 × 3 × 7 × m × m × m × n
54mn⁴ = 2 × 3 × 3 × 3 × m × n × n × n × n
Common factors are 3 × 3 × m × n = 9mn

Q2: Factorise the following expressions.
(a) 54m³n + 81m⁴n²
(b) 15x²y³z + 25x³y²z + 35x²y²z²
Sol:
(a) 54m³n + 81m⁴n²
= 2 × 3 × 3 × 3 × m × m × m × n + 3 × 3 × 3 × 3 × m × m × m × m × n × n
= 3 × 3 × 3 × m × m × m × n × (2 + 3 mn)
= 27m³n (2 + 3mn)

(b) 15x²y³z + 25 x³y²z + 35x²y²z²

= 3 × 5 × x × x × y × y × y × z + 5 × 5 × x × x × x × y × y × z + 5 × 7 × x × x × y × y × z × z

= 5x²y²z ( 3y + 5x + 7z)

Q3: Factorise the following polynomials.
(a) 6p(p – 3) + 1 (p – 3)
(b) 14(3y – 5z)³ + 7(3y – 5z)²
Sol: (a) 6p(p – 3) + 1 (p – 3) = (p – 3) (6p + 1)
(b) 14(3y – 5z)³ + 7(3y – 5z)²
= 7(3y – 5z)² [2(3y – 5z) +1]
= 7(3y – 5z)² (6y – 10z + 1)

Q4: Factorise the following:
(a) p²q – pr² – pq + r²
(b) x² + yz + xy + xz
Sol: (a) p²q – pr² – pq + r²
= (p²q – pq) + (-pr² + r²)
= pq(p – 1) – r²(p – 1)
= (p – 1) (pq – r²)
(b) x² + yz + xy + xz
= x² + xy +xz + yz
= x(x + y) + z(x + y)
= (x + y) (x + z)

Q5: Factorise the following polynomials.
(a) xy(z² + 1) + z(x² + y²)
(b) 2axy² + 10x + 3ay² + 15
Sol:
(a) xy(z² + 1) + z(x² + y²)
= xyz² + xy + zx² + zy²
= (xyz² + zx²) + (xy + zy²)
= zx(yz + x) + y(x + yz)
= zx(x + yz) + y(x + yz)
= (x + yz) (zx + y)
(b) 2axy² + 10x + 3ay² + 15
= (2axy² + 3ay²) + (10x + 15)
= ay²(2x + 3) +5(2x + 3)
= (2x + 3) (ay² + 5)

Q6: Factorise the following expressions.
(а) x² + 4x + 8y + 4xy + 4y²
(b) 4p² + 2q² + p²q² + 8
Sol: (a) x² + 4x + 8y + 4xy + 4y²
= (x² + 4xy + 4y²) + (4x + 8y)
= (x + 2y)² + 4(x + 2y)
= (x + 2y)(x + 2y + 4)
(b) 4p² + 2q² + p²q² + 8
= (4p² + 8) + (p²q² + 2q²)
= 4(p² + 2) + q²(p² + 2)
= (p² + 2)(4 + q²)

Q7: Factorise:
(a) a² + 14a + 48
(b) m² – 10m – 56
Sol:
(a) a² + 14a + 48
= a² + 6a + 8a + 48
[6 + 8 = 14 ; 6 × 8 = 48]
= a(a + 6) + 8(a + 6)
= (a + 6) (a + 8)
(b) m² – 10m – 56
= m² – 14m + 4m – 56
[14 – 4 = 10; 4 × 4 = 56]
= m(m – 14) + 4(m – 14)
= (m – 14) (m + 4)

Q8: Factorise:
(a) x⁴ – (x – y)⁴
(b) 4x² + 9 – 12x – a² – b² + 2ab
Sol: (a) x⁴ – (x – y)⁴
= (x²)² – [(x – y)²]²[ Using Identity a² - b² = (a + b)(a - b) ]
= [x² – (x – y)²] [x² + (x – y)²]
= [x + (x – y] [x – (x – y)] [x² + x² – 2xy + y²] [ Using Identity (a - b)² = (a² - 2ab + b²) ]
= (x + x – y) (x – x + y)[2x² – 2xy + y²]
= (2x – y) (y) (2x² – 2xy + y²)
= y(2x – y) (2x² – 2xy + y²)
(b) 4x² + 9 – 12x – a² – b² + 2ab
= (4x² – 12x + 9) – (a² + b² – 2ab)
= (2x – 3)² – (a – b)²
= [(2x – 3) + (a – b)] [(2x – 3) – (a – b)]
= (2x – 3 + a – b)(2x – 3 – a + b)

Q9: Factorise the following polynomials.
(a) 16x⁴ – 81
(b) (a – b)² + 4ab
Sol: (a) 16x⁴ – 81
= (4x²)² – (9)²_{[Using Identity a² - b² = (a + b)(a - b)]}
= (4x² + 9)(4x² – 9)
= (4x² + 9)[(2x)² – (3)²]
= (4x² + 9)[(2x + 3) (2x – 3)]
(b) (a – b)² + 4ab
= a² – 2ab + b² + 4ab
= a² + 2ab + b²
= (a + b)²

Q10: Factorise:
(а) 14m⁵n⁴p² – 42m⁷n³p⁷ – 70m⁶n⁴p³
(b) 2a²(b² – c²) + b²(2c² – 2a²) + 2c²(a² – b²)
Sol:
(a) 14m⁵n⁴p² – 42m⁷n³p⁷ – 70m⁶n⁴p³
= 14m⁵n³p²(n – 3m²p⁵ – 5mp)
(b) 2a²(b² – c²) + b²(2c² – 2a²) + 2c²(a² – b²)
= 2a²(b² – c²) + 2b²(c² – a²) + 2c²(a² – b²)
= 2[a²(b² – c²) + b²(c² – a²) + c²(a² – b²)]

= 2[a²b² – a²c² + b²c² – b²a²+ c²a² – c²b²)]
= 2 × 0
= 0

Q11: Factorise:
(a) (x + y)² – 4xy – 9z²
(b) 25x² – 4y² + 28yz – 49z²
Sol: (a) (x + y)² – 4xy – 9z²
= x² + 2xy + y² – 4xy – 9z²
= (x² – 2xy + y²) – 9z²
= (x – y)² – (3z)²_{[Using Identity a² - b² = (a + b)(a - b)]}
= (x – y + 3z) (x – y – 3z)

(b) 25x² – 4y² + 28yz – 49z²
= 25x² – (4y² – 28yz + 49z²)
= (5x)² – (2y – 7)²_{[Using Identity a² - b² = (a + b)(a - b)]}
= (5x + 2y – 7) [5x – (2y – 7)]
= (5x + 2y – 7) (5x – 2y + 7)

Q12: If one of the factors of (5x² + 70x – 160) is (x – 2). Find the other factor.
Sol: Let the other factor be m.
(x – 2) × m = 5x² + 70x – 160
To find the other factor, we proceed as follows:
1. Factor out the greatest common factor (GCF):
5(x² + 14x - 32)
2. Factorize the quadratic expression inside the parentheses:
x² + 14x - 32 = (x + 16)(x - 2)
3. Combine the factors:
5(x + 16)(x - 2)
Since (x - 2) is a factor, we can divide the expression by (x - 2) to find the other factor:

5(x + 16)(x - 2)(x - 2) = 5(x + 16)

Therefore, the other factor is:
5(x + 16)

Q13: Express the following as in the form of (a+b)(a-b)
(i) a² – 64
(ii) 20a² – 45b²
(iii) 32x²y² – 8
(iv) x² – 2xy + y² – z²
(v) 49x² – 1
Sol: For representing the expressions in (a+b)(a-b) form, use the following formula
a² – b² = (a+b)(a-b)
(i) a² – 64 = a² – 8²= (a + 8)(a – 8)
(ii) 20a² – 45b² = 5(4a² – 9b²) = 5(2a + 3b)(2a – 3b)
(iii) 32x²y² – 8 = 8( 4x²y² – 1) = 8( (2xy)² – 1²) = 8(2xy + 1)(2xy – 1)
(iv) x² – 2xy + y² – z² = (x – y)² – z²= (x – y – z)(x – y + z)
(v) 49x² – 1 = (7x)² – (1)² = (7x + 1)(7x – 1)

Q14: Verify whether the following equations are correct. Rewrite the incorrect equations correctly.
(i) (a + 6)² = a² + 12a + 36
(ii) (2a)² + 5a = 4a + 5a
Sol:
(i) (a + 6)² = a² + 12a + 36
Here, LHS = (a + 6)² = a² + 12a + 36 [Using Identity (a + b)²= a² + b² + 2ab]
Now, RHS = a² + 12a + 36
Hence, LHS = RHS.
(ii) (2a)² + 5a = 4a + 5a
Here, LHS = (2a)² + 5a = 4a² + 5a
Now, RHS = 4a + 5a
So, LHS ≠ RHS
Correct equation: (2a)² + 5a = 4a² + 5a

Q15: For a = 3, simplify a² + 5a + 4 and a² – 5a
Sol:
Substitute the value of a = 3 in the given equations.
a² + 5a + 4 = 3² + 5(3) + 4 = 9 + 15 + 4 = 28
And,
a² – 5a = 3² – 5(3) = 9 – 15 = -6

Q16: Find the common factors of the following:
(i) 6 xyz, 24 xy² and 12 x²y
(ii) 3x² y³, 10x³y² and 6x² y² z
Sol:
(i) 6 xyz = 2 × 3 × x × y × z
24 xy² = 2 × 2 × 2 × 3 × x × y × y
12 x²y = 2 × 2 × 3 × x × x × y
Thus, the common factors are common factors of 6 xyz, 24 xy² and 12 x²y are 2, 3, x, y and, (2 × 3 × x × y) = 6xy
(ii) 3x² y³ = 3 × x × x × y × y × y
10x³ y² = 2 × 5 × x × x × x × y × y
6 x² y² z = 3 × 2 × x × x × y × y × z
Now, the common factors of 3x² y³, 10x³y² and 6x² y² z are x², y² and, (x² × y²) = x² y²

Q17: Factorize the following expressions:
(i) 54x³y + 81x⁴y²
(ii) 14(3x – 5y)³ + 7(3x – 5y)²
(iii) 15xy + 15 + 9y + 25x
Sol:
(i) 54x³y + 81x⁴y²
= 2 × 3 × 3 × 3 × x × x × x × y + 3 × 3 × 3 × 3 × x × x × x × x × y × y
= 3 × 3 × 3 × x × x × x × y × (2 + 3 xy)
= 27x³y (2 + 3 xy)

(ii) 14(3x – 5y)³ + 7(3x – 5y)²
= 7(3x – 5y)² [2(3x – 5y) +1]
= 7(3x – 5y)² (6x – 10y + 1)

(iii) 15xy + 15 + 9y + 25x
Rearrange the terms as:
15xy + 25x + 9y + 15
= 5x(3y + 5) + 3(3y + 5)
Or, (5x + 3)(3y + 5)

Q18: Factorize (x + y)² – 4xy
Sol:
To solve this expression, expand (x + y)²
Use the formula:
(x + y)² = x² + 2xy + y²
(x + y)² – 4xy = x² + 2xy + y² – 4xy
= x² + y² – 2xy
We know, (x – y)² = x² + y² – 2xy
So, factorization of (x + y)² – 4xy = (x – y)²

Q19: When we factorise an expression, we write it as a ______ of factors.
Sol: When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.

Q20: Find the common factors of 12x, 36
(a) 12
(b) 36
(c) x
(d) 12x
Ans: (a)
Sol: Here,
$12x = 2 \times 2 \times 3 \times x$
$36 = 2 \times 2 \times 3 \times 3$
So, common factor = $2 \times 2 \times 3$
$\Rightarrow 4 \times 3 = 12$
$\Rightarrow 4 \times 3 = 12$
Hence the correct option is (a).

Question for Important Questions: Factorisation

Try yourself:Q21: Factorise 49p²- 36

A.
(7p + 6)(7p + 6)
B.
(7p - 6)(7p - 6)
C.
(7p - 6)(7p + 6)
D.
(6p - 7)(7p - 6)

View Solution

Q22: Factorise using identity $x^{2} +10x+25$
Sol:

x^{2} +10x+25 = x^{2} +2 \times 5 \times x+ 5^{2}

Using the identity,

{(a + b)}^{2} = a^{2} + 2 a b + b^{2}

= x^{2} +2 \times 5 \times x+ 5^{2}

= (x+5)^{2}

Q23: Simplify $[1 0^{2} -18 \times 10+81]$

Sol: $[1 0^{2} -18 \times 10+81] = [10^{2} - 2 \times 9 \times 10 + 9^{2}]$
Using the identity, ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
$= [10^{2} - 2 \times 9 \times 10 + 9^{2}]$
$= {[10 - 9]}^{2}$
$= {[1]}^{2}$
$= 1$

Q24: Simplify 4m² - 169n²2m + 13n

Sol:

4m² - 169n²2m + 13n = (2m)² - (13n)²2m + 13n

Using the identity x² - y² = (x + y)(x - y):

(2m + 13n)(2m - 13n)2m + 13n

= (2m - 13n)

Q25: Find x if x(y - z) = 4y² - 4z²y + z .
Sol: Ans: Taking 4 common from the numerator we get

x(y - z) = 4(y² - z²)y + z

Using the identity, x² − y² = (x + y)(x − y) in the numerator

x(y - z) = 4(y - z)(y + z)y + z

x(y - z) = 4(y - z)

On comparing both side we can see that x = 4.

Q26: Find the remainder in the following (x⁴ - a⁴) ÷ (x² + a²)
Sol: For dividing both equations, first we have to simplify them.
So, (x⁴ − a⁴) can be written as (x²)²−(a²)².
Now by applying the identity a² − b² = (a + b)(a − b)

(x²)²− (a²)²= (x²+ a²) ( x²− a²)

Now, (x⁴ - a⁴)(x² + a²) = (x⁴ - a⁴) (x⁴ + a⁴)(x² + a²)

=(x⁴ - a⁴)(x² + a²) = (x² - a²)

Hence, the remainder is (x² - a²).

Q27: Using identity (a - b)² = a² - 2ab + b² find the value of (98)² .
Sol: (98)² can be further written as (100 - 2)²
By comparing the above equation with the identity, we get
a = 100 and b = 2
(98)2 = (100)² − 2(100)(2) + (2)²
=10000−400+4
=9604

Q28: Simplify a² - 16a² - 25 ÷ a² - 2a - 8a² + 10a + 25
Sol: Identities to be used in the question are;
a²− b² = (a + b)(a − b)

a² - 4²a² - 5² ÷ (a² - 2a - 8)(a² + 2 × 5 × a + 5²)

Factoring and simplifying:

(a + 4)(a - 4)(a + 5)(a - 5) ÷ (a - 4)(a + 2)(a + 5)(a + 5)

Rewriting the division as multiplication:

(a + 4)(a - 4)(a + 5)(a - 5) × (a + 5)(a + 5)(a - 4)(a + 2)

Canceling common factors:

(a + 5)(a + 4)(a - 5)(a + 2)

Q29: The area of a rectangle is $6 a^{2} + 36 a$ and its width is 36a. Find its length.
Sol: Let the length of the rectangle is x
Breath = 36a
$Area of rectangle = Length \times Breath$
$6 a^{2} +36a = x \times 36a$

6a² + 36a36a = x

x = 6a(a + 6)36a

x = a + 66

Hence the length of rectangle is a + 66

Q30: The combined area of two squares is $20c m^{2}$ . Each side of one square is twice as long as a side of the other square. Find the length of the sides of each square.

Sol: Let the side of the smaller square be S and that of the bigger square be 2S.
Combined area of the two squares = $20c m^{2}$
$S^{2} + {(2 S)}^{2} = 20$
$S^{2} + 4 S^{2} = 20$
$5 S^{2} = 20$
$S^{2} = 4$
$S = 2 c m$
Hence the side of the smaller square is 2cm and that of the bigger square is 4cm.

Q31: Find the factors of $25 x^{2} -4 y^{2} +28yz-49 z^{2}$
Sol:
$25 x^{2} -4 y^{2} +28yz-49 z^{2} = 25 x^{2} -(4 y^{2} -28yz+49 z^{2})$
$= 25 x^{2} - [(2y)^{2} -2 \times 2 y \times 7z+(7z)^{2}]$
Using the identity , ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$
${(5 x)}^{2} - {(2 y - 7 z)}^{2}$
Now using the identity , $a^{2} - b^{2} = (a + b) (a - b)$
$(5 x + 2 y - 7 z) (5 x - 2 y + 7 y)$

Q32: Factorise 6xy – 4y + 6 – 9x
Which of the following are factors of the given equation?
(a) (3x – 2)
(b) (3x – 2)(2y – 3)
(c) (2y – 3)
(d) (2x – 3)(3y – 2)
Ans: (b)
$6 x y - 4 y + 6 - 9 x = 2 \times 3 \times x \times y - 2 \times 2 \times y + 2 \times 3 - 3 \times 3 \times x$
$= 2 y (3 x - 2) - 3 (3 x - 2)$
$= (3 x - 2) (2 y - 3)$
Hence the correct option is (b).

Q33: Simplify p² + 11p + 28p + 4
Sol:
$(p^{2} +11p+28) = (p^{2} +7p+4p+28)$
$= p (p + 7) + 4 (p + 7)$
$= (p + 7) (p + 4)$
$= (p + 7) (p + 4)$

(p² + 11p + 28)p + 4 = (p + 7)(p + 4)p + 4
= (p + 7)

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FAQs on Class 8 Maths Chapter 12 Important Question Answers - Factorisation

1. What is factorisation in mathematics?

Ans.Factorisation is the process of breaking down an expression into a product of its factors. For example, the factorisation of the quadratic expression $x^2 - 5x + 6$ is $(x - 2)(x - 3)$.

2. Why is factorisation important in solving equations?

Ans.Factorisation is crucial for solving equations because it allows us to transform complex expressions into simpler forms. By setting each factor to zero, we can find the solutions to the equation more easily.

3. What are some common methods of factorisation?

Ans.Common methods of factorisation include taking out the common factor, using the difference of squares, grouping, and applying the quadratic formula for quadratic expressions. Each method suits different types of expressions.

4. How can I factorise a polynomial expression?

Ans.To factorise a polynomial expression, first look for any common factors among the terms. Next, apply methods like grouping or using special products (like square of a binomial) to break down the expression further.

5. Can factorisation be used in real-life applications?

Ans.Yes, factorisation can be used in various real-life applications, such as in areas of physics for solving motion equations, in economics for optimizing functions, and in computer science for algorithm design and analysis.

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