Factorize the following.
(i) 21x²y³ + 27x³y²
(ii) a³ - 4a² + 12 - 3a
(iii) 4x² - 20x + 25
(iv) y²/9 - 9
(v) x⁴ - 256

Solution:

(i) 21x²y³ + 27x³y²

Factorizing,

21x²y³ + 27x³y² = [3 × 7 × x × x × y × y × y] + [ 3 × 3 × 3 × x × x × x × y × y]

Taking common factors, using ab + ac = a (b + c)

= 3 × x² × y² [7y + 9x] = 3x²y² [7y + 9x]

(ii) a³ - 4a² + 12 - 3a

Rearranging,

a³ - 4a² - 3a + 12

= a²(a - 4) - 3(a - 4)

= (a² - 3) (a - 4)

=(a + √3)(a - √3)(a - 4)

(iii) 4x² - 20x + 25

= (2x)² - (2 × 5 × 2x) + (5)²

It represents the identity : a² - 2ab + b² = (a - b)²

= (2x - 5)²

(iv) y²/9 - 9

= (y/3)² - (3)²

It represents the identity: a² - b² = (a + b) (a - b)

= (y/3 + 3) (y/3 - 3)

(v) x⁴ - 256

= (x²)² - (16)²

It represents the identity a² - b² = (a + b) (a - b)

= (x² + 16)(x² - 16) , further we may notice that (x² - 16) = x² - 4² = (x - 4)(x + 4)

= (x² + 16)(x - 4) (x + 4)

✦ Try This: Factorize the following (i) 24x⁵ y⁶ + 27x⁶ y⁵, (ii) x³ - 5x² + 15 - 3x, (iii) 9x² + 36x + 4

(i) 24x⁵ y⁶ + 27x⁶ y⁵ = (2 × 2 × 2 × 3 × x × x × x × x × x × y × y × y × y × y × y ) + (3 × 3 × 3 × x × x × x × x × x × x × y × y × y × y × y )

Taking common factors, using ab + ac = a (b + c)

= 3x⁵ y⁵(8y + 9x)

(ii) x³ - 5x² + 15 - 3x

Rearranging,

x³ - 5x² - 3x + 15

= x²( x - 5) - 3 ( x - 5)

= (x² - 3) ( x - 5)

= (x - √3)(x + √3) ( x - 5)

(iii) 9x² + 30x + 25

=(3x)² + (2 × 3x × 5) + (5)²

It represents the identity : a² + 2ab + b² = (a + b)²

=(3x + 5)²

☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9

NCERT Exemplar Class 8 Maths Chapter 7 Sample Problem 12

Factorize the following. (i) 21x²y³ + 27x³y² (ii) a³ - 4a² + 12 - 3a (iii) 4x² - 20x + 25 (iv) y²/9 - 9 (v) x⁴ - 256

Summary:

On factorising the following (i) 21x²y³ + 27x³y² (ii) a³ - 4a² + 12 - 3a (iii) 4x² - 20x + 25 (iv) y²/9 - 9 (v) x⁴ - 256, we get, 3x²y² [7y + 9x], (a² - 3) (a - 4), (2x - 5)², (y/3 + 3) (y/3 - 3), and (x² + 16)(x - 4) (x + 4) respectively

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