Question

Factorize the following :

$2 {x}^{2} + {y}^{2} + 8 {z}^{2} - 2 \sqrt{2xy} + 4 \sqrt{2yz} - 8xz$


With proper and clear explanation of steps please!!!

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Answer 1

Correct Question :

Factorize

$\rm{2{x}^{2} + {y}^{2} + 8 {z}^{2} - 2 \sqrt{2}xy+ 4 \sqrt{2}yz- 8xz}$

Solution :

We have to Factorize

$\rm{2{x}^{2} + {y}^{2} + 8 {z}^{2} - 2 \sqrt{2}xy + 4 \sqrt{2}yz - 8xz}$

Compare with Formula

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

Then ,

$\rm{2{x}^{2} + {y}^{2} + 8 {z}^{2} - 2 \sqrt{2}xy+ 4 \sqrt{2}yz - 8xz}$

$=\rm{(-\sqrt{2}x)^2+y^2+(2\sqrt{2}z)^2+2(-\sqrt{2}x)y+2y(2\sqrt{2}z+2(2\sqrt{2}z)(-\sqrt{2}x)}$

$=\rm{(-2\sqrt{2}x+y+2\sqrt{2}z)^2}$

$=\rm{(-2\sqrt{2}x+y+2\sqrt{2}z)</p><p>(-2\sqrt{2}x+y+2\sqrt{2}z)}$

$\rule{200}2$

Algeberaic Indentities :

$1)\sf{(a+b)^2=a^2+b^2+2ab}$

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

$3)\sf{(a^2-b^2)=(a+b)(a-b)}$

$4)\sf{(a+b+c)^2={a}^{2} + {b}^{2} + {c}^{2}+2ab+2bc+2ca}$

$5)\sf{(a+b)^3=a^3+b^3+3ab(a+b)}$

Answer 2

Answer:

Given :

Factorize

$2 {x}^{2} + {y}^{2} + 8 {z}^{2} - 2 \sqrt{2xy} + 4 \sqrt{2yz} - 8xz$

SoluTion

To factorize

$2 {x}^{2} + {y}^{2} + 8 {z}^{2} - 2 \sqrt{2xy} + 4 \sqrt{2yz} - 8xz$

Formula applied

${\boxed {(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac }}$

Afterwards

$2 {x}^{2} + {y}^{2} + 8 {z}^{2} - 2 \sqrt{2xy} + 4 \sqrt{2yz} - 8xz$

= (-√2x)² + y² + 8z² - (2√2z)² + 2( √-2x)y + 2y (2√2z) +2 (2√2z ) (√-2x)

= (-2√2x+y+ 2√2z)²

= (-2√2x + y + 2√2z) (-2√2x +y + 2√2z)

Learn more:-

• Some algebraic identities:-

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

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

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

${\boxed{(a+b)(a-b)={a}^{2 }-{b}^{2}}}$

${\boxed{a{}^{3}-{b}^{3}={a}^{3}-3a{}^{2}b+3ab {}^{2}+{b}^{3}}}$

${\boxed{{(a+b)}^{3}={a }^{3}+{b}^{3}+3ab (a+b)}}$

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

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

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

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

${\boxed{{a}^{3}+{b}^{3}+{c}^{3}-3abc=(a+b+c)({a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca}}$