Question

Find all zeroes of the polynomial (2x4 – 9x3 + 5x2 + 3x – 1) if two of its zeroes are (2 + √3) and (2 – √3)​

Answer 1

Step-by-step explanation:

X = 2 + √3 and x = 2 - √3 are Zeros of p ( x )

( x - 2 - √3 ) ( x - 2 + √3 )

( using identity ( a - b ) ( a +b ) = a² - b²

then,

( x - 2 )² - ( √3 )²

x² + 4 - 4x - 3

x² + 1 - 4x .... ( i )

When we divide p ( x ) by ( i ) we get the other zeros

Other zeros are :- x = 1 and x = -1/2

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 .