Question

factorise:9x^2-12ax-y^2-z^2-2yz+4a^2

Answer 1

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Answer 2

Answer:

$9x^{2}-12ax-y^{2}-z^{2}-2yz+4a^{2}\\=(3x-2a+y+z)(3x-2a-y-z)$

Step-by-step explanation:

$9x^{2}-12ax-y^{2}-z^{2}-2yz+4a^{2}$

/* Rearranging the terms , we get

$=9x^{2}-12ax+4a^{2}-y^{2}-z^{2}-2yz$

$=[(3x)^{2}-2\times 3x\times 2a+(2a)^{2}]-(y^{2}+z^{2}+2\times y\times z)$

/* By algebraic identities:

1) m²-2mn+n² = (m-n)²

2) m²+2mn+n² = (m+n)² */

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

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

/* By algebraic identity:

m²-n² = (m+n)(m-n) */

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

Therefore,

$9x^{2}-12ax-y^{2}-z^{2}-2yz+4a^{2}\\=(3x-2a+y+z)(3x-2a-y-z)$

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