Question

Factorise : 9x2 + y2 - 16z2 + 6xy.​

Answer 1

1. Step-by-step explanation:

(3x)2+(y)2+2×3x×y-(4z)2

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

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

Answer 2

The factorisation of 9$x^2$ + $y^2$ - 16$z^2$ + 6xy is (3x + y + 4z)(3x + y - 4z).

Step-by-step explanation:

We have,

9$x^2$ + $y^2$ - 16$z^2$ + 6xy

To find, the factorisation of 9$x^2$ + $y^2$ - 16$z^2$ + 6xy = ?

∴ $9x^2+y^2-16z^2+6xy$

= $(9x^2+y^2+6xy)-16z^2$

= $[(3x)^2+y^2+2(3x)(y)]-16z^2$

Using the algebraic identity,

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

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

Using the algebraic identity,

$a^2-b^2=(a+b)(a-b)$

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

∴ The factorisation of 9$x^2$ + $y^2$ - 16$z^2$ + 6xy = (3x + y + 4z)(3x + y - 4z)

Thus, the factorisation of 9$x^2$ + $y^2$ - 16$z^2$ + 6xy is (3x + y + 4z)(3x + y - 4z).