Question

2+√3 and 2–√3 are the zeros of p(x)=x⁴–6x³–26x²+138x–35. Find the remaining zeros of p(x).

Answer 1

Hi ,

It is given that ,

p( x ) = x⁴ - 6x³ - 26x² + 138x - 35 ,

( 2 + √ 3 ) and ( 2 + √3 ) are zeroes of p(x).

[x-(2+√3)] , [x-(2-√3)] are factors of p(x).

[(x-2)-√3][(x-2)+√3] = ( x - 2 )² - ( √3 )²

= x² - 2x + 4 - 3

= x² - 2x + 1

x²-2x+1) x⁴-6x³-26x²+138x-35(x²-2x-35
**********x⁴+0+ x²
_______________________
************ - 2x³ - 27x² + 138x
************ - 2x³ + 8x² - 2x
_______________________
********************* - 35x² + 140x - 35
********************* -35x² + 140x - 35
___________________________
Remainder = 0

p( x ) = (x² - 4x + 1 )( x² - 2x - 35 )

= ( x² - 4x + 1 )( x² - 7x + 5x - 35 )

= ( x² - 4x + 1 )[ x( x - 7 ) + 5( x - 7 )]

= ( x² - 4x + 1 )( x - 7 ) ( x + 5 )

Therefore ,

x - 7 = 0 or x + 5 = 0

x = 7 or x = - 5

Required two zeroes are ,

-5 , 7

I hope this helps you.

: )