Class 8 Maths Chapter 12 Practice Question Answers - Factorisation

Q1: Solve for (4x² – 100) ÷ 6(x + 5)
Sol:

= ⅔ (x – 5)

Q2: 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

Q3: Find the common factors of the following:
(i) 6 xyz, 24 xy² and 12 x²y
(ii) 3x² y³, 10x³ y² 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, 10x³ y² and 6x² y² z are x², y² and, (x² × y²) = x² y²

Q4: 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
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

Q5: 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 + 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)

Q6: 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

Q7: 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)

Q8: Factorize x² + 6x – 16
Sol:
To factorize, it should be checked that the sum of factors of 16 should be equal to 6.
Here, 16 = -2 × 8 and 8 + (-2) = 6
So,
x² + 6x – 16 = x² – 2x + 8x – 16
= x(x – 2) + 8(x – 2)
= (x + 8) (x – 2)
Hence, x² + 6x – 16 = (x + 8) (x – 2)