Question

Factorise 9x^2 + 4y^2 + 16z^2 + 12xy - 16yz -24xz

Answer 1

Answer:

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

Step-by-step explanation:

Given an algebraic expressionsuch that,

$9 {x}^{2} + 4 {y}^{2} + 16 {z}^{2} + 12xy - 16yz - 24xz$

To factorise it.

Further simplifying, we can write it as,

$= {(3x)}^{2} + {(2y)}^{2} + {( - 4z)}^{2} + 2(3x)(2y) + 2(2y)( - 4z) + 2(3x)( - 4z)$

But, we know that,

${a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ac = {(a + b + c)}^{2}$

Therefore, we will get,

$= {((3x) +( 2y )+ ( - 4z))}^{2} \\ \\ = {(3x + 2y - 4z)}^{2}$

Hence, the required value is (3x+2y-4z)(3x+2y-4z).

Some more identities :-

• ${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$
• ${(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}$
• $(x + y)(x - y) = {x}^{2} - {y}^{2}$

Answer 2

Step-by-step explanation:

${9x}^{2} + 4 {y}^{2} + 16 {z}^{2} + 12xy - 16yz - 24xz \\ = > ( {3x})^{2} + ( {2y})^{2} + ( { - 4z})^{2} + 2.3x.2y + 2.2y.( - 4z) + 2.( - 4z).3x \\ = > ( {3x + 2y - 4z)}^{2} \\ = > (3x + 2y - 4z)(3x + 2y - 4z)$

Using formula:-

(a+b-c)^2=a^2+b^2+c^2+2ab-2bc-2ca