Expand each of the following, using suitable identities:
(i) $ (x+2 y+4 z)^{2} $
(ii) $ (2 x-y+z)^{2} $
(iii) $ (-2 x+3 y+2 z)^{2} $
(iv) $ (3 a-7 b-c)^{2} $
(v) $ (-2 x+5 y-3 z)^{2} $
(vi) $ \left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2} $

To do:

We have to expand each of the given expressions using suitable identities.

Solution:

We know that,

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

Therefore,

(i) \( (x+2 y+4 z)^{2} \)

Here, $a=x, b=2y$ and $c=4z$

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

$=x^2+4y^2+16z^2+4xy+16yz+8xz$

(ii) \( (2 x-y+z)^{2} \)

Here, $a=2x, b=-y$ and $c=z$

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

$=4x^2+y^2+z^2-4xy-2yz+4xz$

(iii) \( (-2 x+3 y+2 z)^{2} \)

Here, $a=-2x, b=3y$ and $c=2z$

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

$=4x^2+9y^2+4z^2-12xy+12yz-8xz$

(iv) \( (3 a-7 b-c)^{2} \)

Here, $a=3a, b=-7b$ and $c=-c$

$(3 a-7b-c)^{2}=(3a)^2+(-7b)^2+(-c)^2+2(3a)(-7b)+2(-7b)(-c)+2(-c)(3a)$

$=9a^2+49y^2+c^2-42ab+14bc-6ac$

(v) \( (-2 x+5 y-3 z)^{2} \)

Here, $a=-2x, b=5y$ and $c=-3z$

$(-2 x+5 y-3 z)^{2}=(-2x)^2+(5y)^2+(-3z)^2+2(-2x)(5y)+2(5y)(-3z)+2(-3z)(-2x)$

$=4x^2+25y^2+9z^2-20xy-30yz+12xz$

(vi) \( \left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2} \)

Here, $a=\frac{1}{4} a, b=-\frac{1}{2} b$ and $c=1$

$[\frac{1}{4} a-\frac{1}{2} b+1]^{2}=(\frac{1}{4} \mathrm{a})^{2}+(-\frac{1}{2} b)^{2}+(1)^{2}+2 \times \frac{1}{4}a \times(-\frac{1}{2} b)+2 \times(-\frac{1}{2} b) \times 1+2 \times 1 \times \frac{1}{4} a$

$=\frac{1}{16} a^{2}+\frac{1}{4}b^{2}+1^{2}-\frac{2}{8}ab-\frac{2}{2} b+\frac{2}{4}a$

$=\frac{1}{16}a^{2}+\frac{1}{4}b^{2}+1-\frac{1}{4} ab-b+\frac{1}{2} a$

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Updated on: 10-Oct-2022

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