Question

factorise 4x^2+9y^2+z^2-12xy+6yz-4xz​

Answer 1

Step-by-step explanation:

Given:

$4 {x}^{2} + 9 {y}^{2} + {z}^{2} - 12xy + 6yz - 4xz$

To find: Factorisation

Solution:

To find the factorisation,convert this expression to the form which resembles

$\bold{( {a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca }$

Thus,

$4 {x}^{2} + 9 {y}^{2} + {z}^{2} - 12xy + 6yz - 4xz \\ \\ = ( { - 2x)}^{2} + ( { 3y)}^{2} + {z}^{2} + 2( - 2x)( 3y) + 2(3y)(z) + 2( - 2x)(z)$

On comparison with standard identity

$a = - 2x \\ \\ b = 3y \\ \\ c = z$

So,

Factorisation of

$4 {x}^{2} + 9 {y}^{2} + {z}^{2} - 12xy + 6yz - 4xz = ( { - 2x + 3y + z)}^{2} \\ \\\bold{\blue{4 {x}^{2} + 9 {y}^{2} + {z}^{2} - 12xy + 6yz - 4xz}} \\\\ =\bold{\blue{ ( - 2x + 3y + z)( - 2x + 3y + z)}}$

Hops it helps you.

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