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Question:

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(i) $(2 a-5 b-7 c)^{2}$

(ii) $(-3 a+4 b-5 c)^{2}$

(iii) $\left(\frac{1}{2} a-\frac{1}{4} a+2\right)^{2}$

Solution:

(i) $(2 a-5 b-7 c)^{2}=[(2 a)+(-5 b)+(-7 c)]^{2}$

$=(2 a)^{2}+(-5 b)^{2}+(-7 c)^{2}+2(2 a)(-5 b)+2(-5 b)(-7 c)+2(2 a)(-7 c)$

$=4 a^{2}+25 b^{2}+49 c^{2}-20 a b+70 b c-28 a c$

(ii) $(-3 a+4 b-5 c)^{2}=[(-3 a)+(4 b)+(-5 c)]^{2}$

$=(-3 a)^{2}+(4 b)^{2}+(-5 c)^{2}+2(-3 a)(4 b)+2(4 b)(-5 c)+2(-3 a)(-5 c)$

$=9 a^{2}+16 b^{2}+25 c^{2}-24 a b-40 b c+30 a c$

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

$=\left(\frac{a}{2}\right)^{2}+\left(-\frac{b}{4}\right)^{2}+(2)^{2}+2\left(\frac{a}{2}\right)\left(\frac{-b}{4}\right)+2\left(-\frac{b}{4}\right)(2)+2\left(\frac{a}{2}\right)(2)$

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

 

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