Math, asked by gourav00748, 8 months ago

2x⁴+8x-8x²-2x give answer​

Answers

Answered by XxMissPaglixX
11

\huge\boxed{\underline{\mathbb{\red{❥A}\green{n}\pink{S}\orange{w}\blue{E}\pink{r:-}}}}

2 {x}^{4}  + 8x - 8 {x}^{2}  - 2x

 = 2 {x}^{4}  + (8x - 2x) - 8 {x}^{2}

 = 2 {x}^{4}  + 6x - 8 {x}^{2}

 = 2x( {x}^{3}  + 3 - 4x)(ans)

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Answered by Anonymous
8

We have to factorise :-

• 2x⁴ + 8x - 8x² - 2x

→ 2x⁴ - 8x² + 6x

→ 2x( x³ - 4x + 3)

Now factorise x³ - 4x + 3

• We can clearly see x -1 is a factor of above.

• by dividing x³ - 4x + 3 by x-1 , we get quotient

= x² + x - 3

Now we will factorise x² + x - 3

We will get its factors as

\sf{ \implies x = \dfrac{ -b \pm \sqrt{ b^2 - 4ac}}{2a} }

\sf{ \implies x = \dfrac{ -1 \pm \sqrt{ (1)^2 - 4(-3)(1)}}{2} }

\sf{ \implies x = \dfrac{ -1 \pm \sqrt{ 13}}{2} }

Or factors

\small{\sf{ = \left( x +   \left(\dfrac{ 1 + \sqrt{ 13}}{2} \right)\right) \: or \:  \left( x -  \left(\dfrac{  \sqrt{ 13} - 1}{2} \right)\right)} }

So our term

 \tiny{\sf{ = 2x.(x-1).\left( x -  \left(\dfrac{ \sqrt{ 13} - 1}{2} \right)\right).\left( x +  \left(\dfrac{ 1 + \sqrt{ 13}}{2} \right)\right)}}

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