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CLASS 9 MATHS CHAPTER 2 POLYNOMIALS
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Answer:
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Step-by-step explanation:
Solutions:-
4)
I) Given that : (x+2y+4z)^2
It is in the form of (a+b+c)^2
Where a = x ,b = 2y and c = 4z
We know that
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca
(x+2y+4z)^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
(x+2y+4z)^2 = x^2+4y^2+16z^2+4xy+16yz+8zx
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ii)Given that : (2x-y+z)^2
=> [2x+(-y)+z]^2
It is in the form of (a+b+c)^2
Where a = 2x ,b = -y and c = z
We know that
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca
(2x-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+4zx
(2x-y+z)^2 = 4x^2+y^2+z^2-4xy-2yz+4zx
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Given that : (-2x+3y+2z)^2
It is in the form of (a+b+c)^2
Where a = -2x ,b = 3y and c = 2z
We know that
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca
(-2x+3y+2z)^2
=> (-2x)^2+(3y)^2+(2z)^2+2(-2x)(3y)+2(3y)(2z)+2(2z)(-2x)
=>4x^2+9y^2+4z^2-12xy+12yz-8zx
(-2x+3y+2z)^2=4x^2+9y^2+4z^2-12xy+12yz-8zx
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Given that : (3a-7b-c)^2
=> [(3a+(-7b)+(-c)]^2
It is in the form of (a+b+c)^2
Where a = 3a ,b = -7b and c = -c
We know that
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca
[(3a+(-7b)+(-c)]^2
=> (3a)^2+(-7b)^2+(-c)^2+2(3a)(-7b)+2(-7b)(-c)+2(-c)(3a)
=>9a^2+49b^2+c^2-42ab+14bc-6ac
(3a-7b-c)^2 = 9a^2+49b^2+c^2-42ab+14bc-6ac
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Given that : ( -2x+5y-3z)^2
=> (-2x+5y+(-3z))^2
It is in the form of (a+b+c)^2
Where a = -2x ,b = 5y and c = -3z
We know that
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca
(-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+12zx
(-2x+5y-3z)^2=4x^2+25y^2+9z^2-20xy-30yz
+12zx
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Given that : (1/4 a - 1/2b + 1)^2
It is in the form of (a+b+c)^2
Where a = 1/4 a ,b = 1/2 b and c = 1
We know that
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca
(1/4 a - 1/2b + 1)^2
=> (1/4 a)^2+(-1/2 b)^2+(1)^2+2(1/4 a)(-1/2 b).+ 2(-1/2 b)(1)+2(1)(1/4 a)
=> 1/16 a^2 +1/4 b^2+ 1 -1/4 ab -b+ 1/2 a
(1/4 a - 1/2b + 1)^2 = 1/16 a^2 +1/4 b^2+ 1 -1/4 ab -b+ 1/2 a
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5.
1)Given expression is
4x^2+9y^2+16z^2+12xy-24yz-16xz
It can be written as
(2x)^2+(3y)^2+(-4z)^2+2(2x)(3y)+2(3y)(-4z)+2(-4z)(2x)
It is in the form of a^2+b^2+c^2+2ab+2bc+2c
Where a = 2x ,b= 3y,c = -4z
We know that
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca
=> (2x+3y-4z)^2
=> (2x+3y-4z)(2x+3y-4z)
4x^2+9y^2+16z^2+12xy-24yz-16xz = (2x+3y-4z)(2x+3y-4z)
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ii)Given expression is
2x^2+y^2+8z^2-2√2xy+4√2yz-8xz
It can be written as
(-√2x)^2+(y)^2+(2√2z)^2+2(-√2x)(y)+2(y)(2√2z)+2(2√2z)(-√2x)
It is in the form of a^2+b^2+c^2+2ab+2bc+2c
Where a = -√2x ,b= y,c = 2√2z
We know that
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca
=> (-√2x+y-2√2z)^2
=>(-√2x+y-2√2z)(-√2x+y-2√2z)
2x^2+y^2+8z^2-2√2xy+4√2yz-8xz= (-√2x+y-2√2z)(-√2x+y-2√2z)
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Used formulae:-
(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca