Math, asked by zk610428, 1 year ago

factors of z²+1/z²+2-2z-2/z​

Answers

Answered by MaheswariS
8

\underline{\textbf{Given:}}

\mathsf{z^2+\dfrac{1}{z^2}+2-2z-\dfrac{2}{z}}

\underline{\textbf{To find:}}

\textsf{Factors of}

\mathsf{z^2+\dfrac{1}{z^2}+2-2z-\dfrac{2}{z}}

\underline{\textbf{Solution:}}

\mathsf{Consider,}

\mathsf{z^2+\dfrac{1}{z^2}+2-2z-\dfrac{2}{z}}

\mathsf{Using the identity,}

\boxed{\mathsf{(a+b)^2=a^2+b^2+2ab}}

\mathsf{=\left(z+\dfrac{1}{z}\right)^2-2z-\dfrac{2}{z}}

\mathsf{=\left(z+\dfrac{1}{z}\right)^2-2\left(z+\dfrac{1}{z}\right)}

\mathsf{=\left(z+\dfrac{1}{z}\right)\;\left(z+\dfrac{1}{z}-2\right)}

\implies\boxed{\mathsf{z^2+\dfrac{1}{z^2}+2-2z-\dfrac{2}{z}=\left(z+\dfrac{1}{z}\right)\;\left(z+\dfrac{1}{z}-2\right)}}

\underline{\textbf{Find more:}}

1.x^2-(a-1/a)x+1 factorise the following  

https://brainly.in/question/4301404

2.Factorise x^4-(x-z)^4​  

https://brainly.in/question/14993176

Answered by plrohit2008
1

Answer:

(z + \frac{1}{z} - 2) (z + \frac{1}{z})

Step-by-step explanation:

[FIRST SQUARE (z + \frac{1}{z}) USING IDENTITY ]

(z + \frac{1}{z})² = z² + \frac{1}{z^{2} } + 2

[NOW (- 2z - \frac{2}{z}) IS MISSING, SO ADD THEM BOTH THE SIDES]

⇒ (z + \frac{1}{z}) × (z + \frac{1}{z})  - 2z - \frac{2}{z} =  z² + \frac{1}{z^{2} } + 2 - 2z - \frac{2}{z}

⇒ (z + \frac{1}{z}) × (z + \frac{1}{z})  - 2 (z + \frac{1}{z} ) = z² + \frac{1}{z^{2} } + 2 - 2z - \frac{2}{z}

[ HERE z + \frac{1}{z} AND - 2 ARE COMMON]

⇒ (z + \frac{1}{z} - 2) (z + \frac{1}{z}) = z² + \frac{1}{z^{2} } + 2 - 2z - \frac{2}{z}

HENCE FACTORS OF  z² + \frac{1}{z^{2} } + 2 - 2z - \frac{2}{z} ARE ⇒

(z + \frac{1}{z} - 2) (z + \frac{1}{z})

THANK YOU

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