Recall that an algebraic luently 13 a su
ables occurring in it. You have studied the follo
Identity 1: (x + y)= x2 + 2xy + y2
Identity II : (x - y)2 = x2 - 2xy + y2
Answers
Answer:
We know,
(a+b+c)
2
=a
2
+b
2
+c
2
+2ab+2bc+2ac
i) (x+2y+4z)
2
=x
2
+(2y)
2
+(4z)
2
+2(x)(2y)+2(2y)(4z)+2(x)(4z)
=x
2
+4y
2
+16z
2
+4xy+16yz+8xz
ii) (2x−y+z)
2
=(2x)
2
+(−y)
2
+(z)
2
+2(2x)(−y)+2(−y)(z)+2(2x)(z)
=4x
2
+y
2
+z
2
−4xy−2yz+4xz
iii) (−2x+3y+2z)
2
=(−2x)
2
+(3y)
2
+(2z)
2
+2(−2x)(3y)+2(3y)(2z)+2(−2x)(2z)
=4x
2
+9y
2
+4z
2
−12xy+12yz−8xz
iv) (3a−7b−c)
2
=(3a)
2
+(−7b)
2
+(−c)
2
+2(3a)(−7b)+2(−7b)(−c)+2(3a)(−c)
=9a
2
+49b
2
+c
2
−42ab+14bc−6ac
v) (−2x+5y−3z)
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
iv) (
4
1
a−
2
1
b+1)
2
=(
4
1
a)
2
+(−
2
1
b)
2
+(1)
2
+2(
4
1
a×−
2
1
b)+2(−
2
1
b×1)+2(
4
1
a×1)
=
16
1
a
2
+
4
1
b
2
+1−
4
1
ab−b+
2
1
a
Answer:
1.(x+y)²
(x+y)(x+y)
x(x+y)+y(x+y)
x²+xy+xy+y²
x²+2xy+y²
2.(3+2)²=3²+2(3)(2)+2²
(5)²+9+12+4
25=25
Step-by-step explanation: