Question

Factorise



4x² - 12ax - y² - z² - 2yz + 9a²​

Answer 1

Answer:

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

Step-by-step explanation:

(2x)² - 2.2x.3a + (3a)² - (y² + 2.y.z + z²)

(2x - 3a)² - (y + z)²

so applying a²-b² = (a+b)(a-b)

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

Answer 2

ATQ, We have to factorize 4x² - 12ax - y² - z² - 2yz + 9a²​

.

When we are asked to factorize something, we are basically asked to find it's factors, which when multiplied will give you the polynomial you started with.

$\sf \Longrightarrow 4x^2 - 12ax - y^2 - z^2 - 2yz + 9a^2$

| Re-arranging the terms we get:

$\sf \Longrightarrow 4x^2 - 12ax + 9a^2 - y^2 - z^2 - 2yz$

$\sf \Longrightarrow 4x^2 - 12ax + 9a^2 - \Big(y^2 + z^2 + 2yz\Big)$

| Let's try to express 4x² - 12ax + 9a²​ in the form of (a - b)² = a² - 2ab + b²

$\sf \Longrightarrow \Big(2x\Big)^2 - 2\Big(2x\Big) \Big(3a\Big) + \Big(3a\Big)^2 - \Big(y^2 + z^2 + 2yz\Big)$

| If a = 2x, b = 3a,

| We can write it as (2x - 3a)² since we have (2x)² - 2(2x)(3a) + (3a)²

$\sf \Longrightarrow \Big(2x - 3a\Big)^2 - \Big(y^2 + z^2 + 2yz\Big)$

| Similarly, let a = y & b = z.

| ∴ y² + z² + 2yz can be written as (y + z)²

| Using the identity (a + b)² = a² + 2ab + b² we get:

$\sf \Longrightarrow \Big(2x - 3a\Big)^2 - \Big(y + z\Big)^{2}$

| We know that a² - b² = (a + b)(a - b)

| Here, let a = 2x - 3a and let b = y + z

Therefore:

$\sf \Longrightarrow \Big(2x - 3a + y + z \Big) \Big(2x - 3a - (y + z) \Big)$

$\sf \Longrightarrow \Big(2x - 3a + y + z \Big) \Big(2x - 3a - y - z \Big)$

∴ Answer: (2x - 3a + y + z) (2x - 3a - y - z)