Question

factorise :(4a-b+2c)^2​

Answer 1

Answer:

$(4a - b + 2c) {}^{2} \\ = (4a) {}^{2} + ( - b) {}^{2} + (2c) {}^{2} \\ + 2 \times (4a)( - b) + 2( - b)(2c) + 2(2c)(4a) \\ = 16a {}^{2} + b {}^{2} + 4c {}^{2} - 8ab - 4bc + 16ac$

Using the identity

$(x + y + z) {}^{2} \\ x {}^{2} + y { }^{2} + z {}^{2} + 2xy + 2yz + 2zx$

where

x = 4a , (-b) = y and 2c = z

Answer 2

$\huge\underline\bold\red{AnSwEr:-}$

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ᴜꜱɪɴɢ ᴛʜᴇ 4ᴛʜ ɪᴅᴇɴᴛɪᴛʏ

${a+b+c}^{2}$

=>(4ᴀ)^2+(-ʙ)^2+(2ᴄ^2)+2(4ᴀ×-ʙ)+2(-ʙ×2ᴄ)+2(2ᴄ×4ᴀ)

=>16ᴀ^2+ʙ^2+4ᴄ^2-8ᴀʙ-4ʙᴄ+16ᴄᴀ.

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Hope this helps u dear mate ✌

keep smiling $\huge\pink{\ddot\smile}$