Expand each of the following, using suitable identities:
(i) (x + 2y + 4z)?
(ii) (2x - y +z)
Answers
Step-by-step explanation:
\huge\mathbb{A~N~S~W~E~R}A N S W E R
[1]
\begin{lgathered}= {(x + 2y + 4z)}^{2} \\ \\ = {(a + b + c)}^{2} \\ \\ = {x}^{2} + {(2y)}^{2} + {(4z)}^{2} + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x) \\ \\ = {x}^{2} + 4 {y}^{2} + 16 {z}^{2} + 4xy + 16yz + 8zx\end{lgathered}
=(x+2y+4z)
2
=(a+b+c)
2
=x
2
+(2y)
2
+(4z)
2
+2(x)(2y)+2(2y)(4z)+2(4z)(x)
=x
2
+4y
2
+16z
2
+4xy+16yz+8zx
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[2]
\begin{lgathered}= {(2x - y + z)}^{2} \\ \\ = {4x}^{2} + {( - y)}^{2} + {z}^{2} + 2(2x)( - y) + 2( - y)(z) + 2(z)(2x) \\ \\ = 4 {x}^{2} - {y}^{2} + {z}^{2} - 4xy - 2yz + 4xz\end{lgathered}
=(2x−y+z)
2
=4x
2
+(−y)
2
+z
2
+2(2x)(−y)+2(−y)(z)+2(z)(2x)
=4x
2
−y
2
+z
2
−4xy−2yz+4xz
Answer:
(x+2y+4z) (x+2y+4z)
(x)2+(4y)2 +(4z)2+2(x) (2y)+2(2y)(4z)+2(4z) (x) Using identity 5 xsquare +4y square +16z square +4xy +16yz+8zx and you see there is written square that is your 2 okay and 2. I had done 1. one so you can solve Q.2 you please see Q.1 and solve Q.2 also or you see in Google