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Factorize the following using appropriate identities :
i)\((x + 2y + 4z)^2\)
ii)\((2x - y + z)^2\)
iii)\((-2x + 3y + 2z)^2\)
iv)\((3a - 7b - c)^2\)
v)\((-2x + 5y - 3z)^2\)
vi)\([\frac{1}{4} a - \frac{1}{2} b + 1]^2\)


Answer :

i)\((x + 2y + 4z)^2 \)
\( = (x)^2 + (2y)^2 +(4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(x)(4z)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= x^2 + 4y^2 + 16z^2 + 4xy + 16yz + 8xz\)


ii)\((2x - y + z)^2 \)
\( = (2x)^2 + (-y)^2 + (z)^2 - 2(2x)(y) - 2(y)(z) + 2(2x)(z)\)
\([ \because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= 4x^2 + y^2 + z^2 - 4xy - 2yz + 4xz\)


iii)\((-2x + 3y + 2z)^2 \)
\( = (-2x)^2 + (3y)^2 +(2z)^2 - 2(2x)(3y) + 2(3y)(2z) - 2(2x)(2z)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= 4x^2 + 9y^2 + 4z^2 - 12xy + 12yz - 8xz\)


iv)\((3a - 7b - c)^2 \)
\( = (3a)^2 + (-7b)^2 + (-c)^2 - 2(3a)(7b) + 2(-7b)(-c) - 2(3a)(c)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= 9a^2 + 49b^2 + c^2 - 42ab + 14bc - 6ac\)


v)\((-2x + 5y - 3z)^2 \)
\( = (-2x)^2 + (5y)^2 +(-3z)^2 - 2(2x)(5y) - 2(5y)(3z) + 2(2x)(3z)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= 4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12xz\)


vi)\([\frac{1}{4} a - \frac{1}{2} b + 1]^2 \)
\( = (\frac{1}{4} a )^2 + (\frac{-1}{2} b )^2 + (1)^2 - 2(\frac{1}{4} a )(\frac{1}{2} b ) - 2(\frac{-1}{2} b )(1) + 2(\frac{1}{4} a )(1)\)
\([\because (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac]\)
\(= \frac{a^2}{16} + \frac{b^2}{4} + 1 - \frac{ab}{4} - b + \frac{a}{2} \)

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