Question

Find all the zeros of 2x4 – 9x3 +5x2 +3x -1, if two of its zeroes are 2+√3 and 2-√3

Answer 1

Given p(x)=2x^4-9x^3+5x^2+3x-1

If two of its zeros are (2+√3) and (2-√3) so [x-(2+√3)] [x-(2-√3)]

So we get g(x) = (x-2-√3) (x-2+√3) = x²-4x+1

On dividing p(x) by g(x) we get      

  2x²-x-1

  2x²-2x+x-1

  2x(x-1)+1(x-1)

  (2x+1)(x-1)

So its zeros are -1/2 , 1

So, all of its zeros are 2+√3 , 2-√3 , -1/2 , 1

Long division method is in the attachment.

Hope it helps.

Thanks!!!

Answer 2

Answer:

All zeroes of p ( x ) are 1 , - 1 / 2 , 2 + √ 3 and 2 - √ 3

Step-by-step explanation:

Given :

p ( x ) = 2 x⁴ - 9 x³ + 5 x² + 3 x - 1

Two zeroes of p ( x ) :

2 + √ 3 and 2 - √ 3

p ( x ) = ( x - 2 - √ 3 ) ( x - 2 + √ 3 ) × g ( x )

g ( x ) = p ( x ) ÷ ( x² - 4 x + 1 )

On dividing ( 2 x⁴ - 9 x³ + 5 x² + 3 x - 1 ) by ( x² - 4 x + 1 )

$\displaystyle{\begin{tabular}{c | c | c}(x^2-4x+1) & \overline{2x^4-9x^3+5x^2+3x-1}} & 2x^2-x-1\\ & 2x^4-8x^3+2x^2 \ \ \ \ \ \ \ \ \ \ \ \\ \\ & ( - ) ( + ) ( - ) \ \ \ \ \ \ \ \\ \\ & \overline{-x^3+3x^2+3x} \\ \\ & - x^3+4x^2-x\\ \\ & ( + ) ( - ) ( + ) \\ \\ & \overline{-x^2 + 4x-1} \\ \\ & -x^2 + 4x-1 \\ \\ & ( + ) ( - ) ( + ) \\ \\ & \overline{0}\endtabular$

We got ,

g ( x ) = 2 x² - x - 1

By splitting mid term

2 x² - 2 x + x - 1

2 x ( x - 1 ) + ( x - 1 )

( x - 1 ) ( 2 x + 1 )

Now zeroes of g ( x )

x = 1 or x = - 1 / 2 .