Question

urgent i've tried numerous times but i've been unable to factor it

Answer 1

Answer:

$(2c+a+3b)(2c-a-3b)$

Step-by-step explanation:

Given expression:  $4c^2-a^2-6ab-9b^2$

Factor $-a^2-6ab-9b^2$

$\implies -(a^2+6ab+9b^2)$

$\implies -(a+3b)(a+3b)$

$\implies -(a+3b)^2$

Factor $4c^2$

$\implies(2c)(2c)$

$\implies (2c)^2$

Therefore,

$(2c)^2-(a+3b)^2$

Difference of Two Squares: $x^2-y^2=(x+y)(x-y)$

$x^2=(2c)^2\implies x=2c$

$y^2=(a+3b)^2 \implies y=a+3b$

Therefore,

$\implies (2c + a + 3b)(2c - (a + 3b))$

$\implies (2c+a+3b)(2c-a-3b)$

Answer 2

Answer:

(2c - a + 3b)(2c + a + 3b)

Step-by-step explanation:

4c² - a² - 6ab - 9b²

= 4c² - (a² + 6ab + 9b²) ← the terms in the parenthesiis are a perfect square )

= 4c² - (a + 3b)² ← this is a difference of squares which factors in general as

a² - b² = (a - b)(a + b) , then

= (2c)² - (a + 3b)²

= (2c - (a + 3b) )(2c + (a + 3b) )

= (2c - a - 3b)(2c + a + 3b)