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# Factorise: (i) 4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

(ii) 2x^{2} + y^{2} + 8z^{2} – 2√2 xy + 4√2 yz – 8xz

**Solution:**

We will be using the algebraic identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca to solve the following.

i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz

This can be re-written as:

(2x)² +(3y)² +(-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)

This is of the form a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²

Here a = 2x, b = 3y, c = -4z

Hence, 4x² + 9y² + 16z² +12xy - 24yz - 16xz = (2x + 3y - 4z)²

ii) 2x² + y² + 8z² - 2√2xy + 4√2yz - 8xz

We know that, 2 = (√2)^{2}, √8 = 2√2

Thus, the given expression can be re-written as:

(-√2x)² + (y)² + (2√2z )² + 2(-√2x)(y) + 2(y)(2√2z) + 2(2√2z)(-√2x)

This is of the form a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²

Here a = -√2x, b = y, c = 2√2z

Hence 2x² + y² + 8z² - 2√2xy + 4√2yz - 8xz = (-√2x + y + 2√2z)²

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 2

**Video Solution:**

## Factorise: (i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz (ii) 2x² + y² + 8z² - 2√2xy + 4√2yz - 8xz

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.5 Question 5

**Summary:**

The factorized form of 4x² + 9y² + 16z² + 12xy - 24yz - 16xz and 2x² + y² + 8z² - 2√2xy + 4√2yz - 8xz are (2x + 3y − 4z)² and (−√2x +y +2√2z)² respectively.

**☛ Related Questions:**

- Use suitable identities to find the following products: (i) (x + 4) (x + 10) (ii) (x + 8) (x - 10) (iii) (3x + 4) (3x - 5) (iv) (y^2 + 3/2)(y^2 - 3/2) (v) (3 - 2x) (3 + 2x)
- Evaluate the following products without multiplying directly:(i) 103 × 107(ii) 95 × 96(iii) 104 × 96
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