Ex.2.5 Q5 Polynomials Solution - NCERT Maths Class 9

Go back to  'Ex.2.5'

Question

Factorise:

(i)\(\begin{align}{ 4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z} \end{align}\)

(ii)\(\begin{align} 2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z\end{align}\)

   

 Video Solution
Polynomials
Ex 2.5 | Question 5

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:

\(\begin{align}\text{(i)}\;\;4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z\end{align}\)

This can be re-written as:

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

Which is of the form: \(\begin{align}a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=(a+b+c)^{2}\end{align}\)

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

Hence \(\begin{align}4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z=(2 x+3 y-4 z)^{2}\end{align}\)

\(\begin{align}\text{(ii)}\;\;2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z\end{align}\)

This can be re-written as:

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

Which is of the form: \(\begin{align}a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=(a+b+c)^{2}\end{align}\)

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

Hence  \(\begin{align}2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z=(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}\end{align}\)