Factorise:
(i) $ 4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z $
(ii) $ 2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z $

AcademicMathematicsNCERTClass 9

To do:

We have to factorise the given expressions.

Solution:

We know that,

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

Therefore,

(i) $4x^2 + 9y^2 + 16z^2 +12xy-24yz-16xz = (2x)^2 + (3y)^2 + (-4z)^2 + 2 \times (2x) \times (3y) + 2 \times (3y) \times (-4z) + 2 \times (- 4z) \times (2x)$

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

Hence $4x^2 + 9y^2 + 16z^2 +12xy-24yz-16xz =(2x + 3y - 4z)^2$

(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z = (- \sqrt2x)^2 + (y)^2 + (2 \sqrt2z)2 + 2 \times (- \sqrt2x) \times (y)+ 2 \times (y) \times (2\sqrt2z) + 2 \times (2\sqrt2z) \times (-\sqrt2x)$

$= (-\sqrt2x + y + 2\sqrt2z)^2$

Hence $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z =(-\sqrt2x + y + 2\sqrt2z)^2$

raja
Updated on 10-Oct-2022 13:39:07

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