Question

Can anyone solve this question, I need it urgently.

Answer 1

Hi ,

Let p( x ) = x⁴ - 3x³ -x² + 9x- 6 ,

√3 and -√3 are zeroes of p(x ),

therefore ,

x - √3 and ( x + √3 ) are factors of p(x ),

( x - √3 ) ( x + √3 ) = x² - ( √3)² = x² - 3 is a

factor of p (x )

x² - 3 ) x⁴ - 3x³ - x² + 9x - 6 ( x² - 3x + 2
********x⁴ + 0 - 3x²
__________________
************-3x³ + 2x² + 9x
***********-3x³ + 0 + 9x
___________________
*****************2x² - 6
*****************2x² - 6
____________________
0



Therefore ,

p( x ) = ( x - √3 )(x+ √3 ) ( x² - 3x +2 )-----( 1 )

x² - 3x +2 = x² - x - 2x + 2

= x ( x - 1 ) - 2 ( x - 1 )

= ( x - 1 ) ( x - 2 ) ----( 2 )

other two zeroes are ,

x - 1 = 0 or x - 2 = 0

x = 1 or x = 2

I hope this helps you.

:)

Answer 2

Let x=√3. and x=-√3
x-√3=0. x+√3=0
(x-√3)(x+√3)=0
x²-3=0.----------g(x)
By dividing x⁴-3x³-x²+9x-6by g(X), we get(in pic)
q(X)=x²-3x+2
By factorising it we get
x²-2x-x+2
x(x-2)-1(x-2)
(x-1)(x-2)
x-1=0. x-2=0
x=1. x=2
Therefore other two zeros are 1 and 2.
HOPE IT WAS HELPFUL