Question

find all zero es of polynomial +2x4-9x3+5x2+3x-1)if twoof its zeros are (2+√3) and(2-√3)

Answer 1

2x⁴ - 9x³ + 5x² + 3x - 1 have two zeros (2 + √3) and (2-√3)
means ,{x - (2 + √3)} and {x - (2-√3)} are factors of 2x⁴ - 9x³ + 5x² + 3x - 1 .
∴2x⁴ - 9x³ + 5x² + 3x - 1 is divisible by (x - 2 - √3)(x - 2 + √3)
∴ 2x⁴ - 9x³ + 5x² + 3x -1 is divisible by x² - (2 + √3+2-√3)x + (2²-√3²)
∴ 2x⁴ - 9x³ + 5x² + 3x - 1 is divisible by x² -4x + 1
so, now 2x⁴ - 8x³ + 2x² - x³ + 4x² - x - x² + 4x + 1
= 2x²(x² - 4x + 1) -x(x² - 4x + 1) - 1(x² - 4x + 1)
= (2x² - x - 1)(x² - 4x + 1)

Hence, next two zeros occurs in (2x² - x - 1)
Use quadratic formula for finding zeros ,
x = {1 ± √(1+8)}/4 = (1 ± 3)/4 = 1 and -1/2

Hence, next two zeros are 1 and -1/2

[Note you can solve it by Euclid algorithm , try to solve by using them , it will build you concept :)]

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 .