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Ex.2.5 Q4 Polynomials Solution - NCERT Maths Class 9

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Question

Expand each of the following, using suitable identities:

(i) \(\begin{align}(x+2 y+4 z)^{2}\end{align}\)

(ii) \(\begin{align}(2 x-y+z)^{2}\end{align}\)

(iii)\(\begin{align}(-2 x+3 y+2 z)^{2}\end{align}\)

(iv) \(\begin{align}(3 a-7 b-c)^{2}\end{align}\)

(v)\(\begin{align}(-2 x+5 y-3 z)^{2}\end{align}\)

(vi) \(\begin{align}\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}\end{align}\)

 Video Solution
Polynomials
Ex 2.5 | Question 4

Text Solution

Reasoning:

Identity:\(\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}\)

Steps:

\(\begin{align}(\mathrm{i})\;\;(x+2 y+4 z)^{2}\end{align}\)

Identity:\(\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}\)

Taking \(\begin{align}a=x, b=2 y, c=4 z\end{align}\)

\[\begin{align}&{(x + 2y + 4z)^2}&\\&=\left[ \begin{array}{l}{x^2} + {(2y)^2} + {(4z)^2} + 2(x)(2y) + \\2(2y)(4z) + 2(4z)(x)\end{array} \right]\\ &= {x^2} + 4{y^2} + 16{z^2} + 4xy + 16yz + 8zx\end{align}\]

\(\begin{align}\text {(ii)}\left(2 x-y+z)^{2}\right.\end{align}\)

Identity: \(\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}\)

Taking \(\begin{align}a=2 x, b=-y, c=z\end{align}\)

\[\begin{align}&{(2x - y + z)^2}\\ &= \left[ {\begin{array}{*{20}{l}}\begin{array}{l}{(2x)^2} + {( - y)^2} + {(z)^2}\\ + 2(x)( - y) + \end{array}\\{2( - y)(z) + 2(z)(2x)}\end{array}} \right]\\&= 4{x^2} + {y^2} + {z^2} + 4xy + 2yz + 4zx\end{align}\]

\(\begin{align}\text { (iii) }\;\;(-2 x+3 y+2 z)^{2}\end{align}\)

Identity: \(\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+\\c^{2}+2 a b+2 b c+2 c a\end{array} \right]\end{align}\)

Taking \(\begin{align}a=2 x, b=3 y, c=2 z\end{align}\)

\[\begin{align}&{( - 2x + 3y + 2z)^2}\\ &= \left[ {\begin{array}{*{20}{l}}\begin{array}{l}{( - 2x)^2} + {(3y)^2} + {(2z)^2}\\ + 2( - 2x)(3y) + \end{array}\\{2(3y)(2z) + 2(2z)(2x)}\end{array}} \right]\\ &= 4{x^2} + 9{y^2} + 4{z^2} - 12xy + 12yz - 8zx\end{align}\]

\(\begin{align}(\text{iv})(3 a-7 b-c)^{2}\end{align}\)

Identity: \(\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}\)

Taking \(\begin{align}a=3 a, b=-7 b, c=-c\end{align}\)

\[\begin{align}&{(3a - 7b - c)^2}\\ &= \left[ \begin{array}{l}{(3a)^2} + {( - 7b)^2} + {( - c)^2}\\ + 2(3a)( - 7b)\\ + 2( - 7b)( - c) + 2( - c)(3a)\end{array} \right]\\ &= 9{a^2} + 49{b^2} + {c^2} - 42ab + 14bc - 6ca\end{align}\]

\(\begin{align}(\mathrm{v})\;\;(-2 x+5 y-3 z)^{2}\end{align}\)

Identity: \(\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}\)

Taking \(\begin{align}a=-2 x, b=5 y, c=-3 z\end{align}\)

\[\begin{align}&{( - 2x + 5y - 3z)^2}\\ &= \left[ \begin{array}{l}{( - 2x)^2} + {(5y)^2} + {( - 3z)^2} +\\ 2( - 2x)(5y)\\ + 2(5y)( - 3z) + 2( - 3z)( - 2x)\end{array} \right]\\& = \!4{x^2}\!\!+\! 25{y^2} + 9{z^2} \!\!-\! 20xy - 30yz \!+\!\! 12zx\end{align}\]

\(\begin{align}(\mathrm{vi})\;\;\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2} \end{align}\)

Identity: \(\begin{align}\left[ \begin{array}{l}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+\\2 a b+2 b c+2 c a\end{array} \right]\end{align}\)

Taking \(\begin{align}a=\frac{1}{4} a, b=\frac{-1}{2} b, c=1 \end{align}\)

\[\begin{align}&{\left( {\frac{1}{4}a - \frac{1}{2}b + 1} \right)^2}\\&= \left[ \begin{array}{l}{\left( {\frac{1}{4}a} \right)^2} + {\left( {\frac{{ - 1}}{2}b} \right)^2} + {(1)^2} + \\2\left( {\frac{1}{4}a} \right)\left( {\frac{{ - 1}}{2}b} \right) + \\2\left( {\frac{{ - 1}}{2}b} \right)(1) + 2(1)\left( {\frac{1}{4}a} \right)\end{array} \right]\\&= \frac{1}{{16}}{a^2} + \frac{1}{4}{b^2} + 1 - \frac{1}{4}ab - b + \frac{1}{2}a\end{align}\]

 Video Solution
Polynomials
Ex 2.5 | Question 4