Question

if 2 zeros of polynomial x⁴-6x³-26x²+138x-35 are 2±√3, find other zeros.

Answer 1

hi friend,

given 2 zeros are 2±√3

sum of zeros=2+√3+2-√3=4

product of zeros =4-3=1

so the quadratic polynomial from given 2 zeros are x²-4x+1-----(1)

given x⁴-6x³-26x²+138x-35 ----(2)

dividing (2) by (1), we get x²-2x-35

→x²-7x+5x-35

→x(x-7)+5(x-7)

→(x+5)(x-7)

so the other roots are -5 and 7


I hope this will help u ;)

Answer 2

Hi there !

Since 2+ √3 is a zero,
x-(2+√3) is a factor of the polynomial.

Since 2-√3 is a zero,
x-(2-√3) is a factor of the polynomial.

This means that (x-2-√3)(x-2+√3) are factors

(x-2-√3)(x-2+√3) [ using identity ]
= (x-2)² - (√3)²
= x² -4x +4 - 3
= x²- 4x+1

x²- 4x+1 is a factor of x⁴-6x³-26x²+138x-35


WE have to perform division algorithm ,
divide x⁴-6x³-26x²+138x-35 with x²- 4x+1

we get the answer as :-

x²- 2x - 35

Now split the middle terms !!

x²- 7x + 5x - 35

x [ x - 7 ] + 5 [ x - 7 ]

[ x+ 5 ] [ x - 7 ]

The other zeroes are :-

-5 and 7
..