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: \(\begin{align}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\end{align}\)

Steps:

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

Identity: \(\begin{align}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\end{align}\)

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

\[\begin{align}(x+2 y+4 z)^{2} &=x^{2}+(2 y)^{2}+(4 z)^{2}+2(x)(2 y)+2(2 y)(4 z)+2(4 z)(x) \\ &=x^{2}+4 y^{2}+16 z^{2}+4 x y+16 y z+8 z x \end{align}\]

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

Identity: \(\begin{align}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\end{align}\)

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

\[\begin{align}(2 x-y+z)^{2} &=(2 x)^{2}+(-y)^{2}+(z)^{2}+2(x)(-y)+2(-y)(z)+2(z)(2 x) \\ &=4 x^{2}+y^{2}+z^{2}+4 x y+2 y z+4 z x \end{align}\]

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

Identity: \(\begin{align}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\end{align}\)

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

\[\begin{align}(-2 x+3 y+2 z)^{2} &=(-2 x)^{2}+(3 y)^{2}+(2 z)^{2}+2(-2 x)(3 y)+2(3 y)(2 z)+2(2 z)(2 x) \\ &=4 x^{2}+9 y^{2}+4 z^{2}-12 x y+12 y z-8 z x \end{align}\]

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

Identity: \(\begin{align}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\end{align}\)

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

\[\begin{align}(3 a-7 b-c)^{2} &=(3 a)^{2}+(-7 b)^{2}+(-c)^{2}+2(3 a)(-7 b)+2(-7 b)(-c)+2(-c)(3 a) \\ &=9 a^{2}+49 b^{2}+c^{2}-42 a b+14 b c-6 c a \end{align}\]

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

Identity: \(\begin{align}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\end{align}\)

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

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

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

Identity: \(\begin{align}(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \end{align}\)

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

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