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# Ex.2.5 Q5 Polynomials Solution - NCERT Maths Class 9

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## Question

Factorise:

(i)\begin{align}\left[ \begin{array}{l} 4 x^{2}+9 y^{2}+16 z^{2}+\\12 x y-24 y z-16 x z\end{array} \right]\end{align}

(ii)\begin{align}\left[ \begin{array}{l}2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+\\4 \sqrt{2} y z-8 x z\end{array} \right]\end{align}

Video Solution
Polynomials
Ex 2.5 | Question 5

## 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:

(i) \begin{align}\left[ \begin{array}{l} 4 x^{2}+9 y^{2}+16 z^{2}+\\12 x y-24 y z-16 x z\end{array} \right]\end{align}

This can be re-written as:

$\left[ \begin{array}{l}\left( {2{x^2}} \right) + {(3y)^2} + \left( { - 4{z^2}} \right) +\\ 2(2x)(3y) + 2(3y)( - 4z) +\\ 2( - 4z)(2x) + 2(2x)( - 4z)\end{array} \right]$

Which is of the form: \begin{align}\left[ \begin{array}{l}a^{2}+b^{2}+c^{2}+2 a b+2 b c+\\2 c a=(a+b+c)^{2}\end{array} \right]\end{align}

Here \begin{align}a=2 x , b=3 y, c=-4 z\end{align}

Hence \begin{align}\left[ \begin{array}{l}4 x^{2}+9 y^{2}+16 z^{2}+12 x y-\\24 y z-16 x z=(2 x+3 y-4 z)^{2}\end{array} \right]\end{align}

(ii) \begin{align}\left[ \begin{array}{l}2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+\\4 \sqrt{2} y z-8 x z\end{array} \right]\end{align}

This can be re-written as:

$\left[ \begin{array}{l}{( - \sqrt 2 x)^2} + {(y)^2} + {(2\sqrt 2 z)^2} + \\2( - \sqrt 2 x)(y) + 2(y)(2\sqrt 2 z) +\\ 2(2\sqrt 2 z)( - \sqrt 2 x)\end{array} \right]$

Which is of the form: \begin{align}\left[ \begin{array}{l}a^{2}+b^{2}+c^{2}+2 a b+2 b c+\\2 c a=(a+b+c)^{2}\end{array} \right]\end{align}

Here \begin{align}a=-2 \sqrt{2} x, b=y, c=2 \sqrt{2} z\end{align}

Hence  \begin{align}\left[ \begin{array}{l}2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-\\8 x z=(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}\end{array} \right]\end{align}

Video Solution
Polynomials
Ex 2.5 | Question 5

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