2. Polynomials
medium

Expand each of the following, using suitable identities : $(x+2 y+4 z)^{2}$

Option A
Option B
Option C
Option D

Solution

$( x +2 y +4 z )^{2}$

We have $( x + y + z )^{2}= x ^{2}+ y ^{2}+ z ^{2}+2 xy +2 yz +2 zx$

$\therefore \quad( x +2 y +4 z )^{2}=( x )^{2}+(2 y )^{2}+(4 z )^{2}+2( x )(2 y )+2(2 y )(4 z )+2(4 z )( x )$

$= x ^{2}+4 y ^{2}+16 z ^{2}+4 xy +16 yz +8 zx$

Standard 9
Mathematics

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