If 2 zeroes of the polynomial x^4-6x^3-26x^2+138x-35 are 2+root3, find other zeroes.
can anyone write it's ans and tell.
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Answer:
7, -5
Step-by-step explanation:
let other roots be a,b
roots are 2+root3,2-root3,a,b
sum of roots= - coefficient of x^3 /coefficient of x^4
2+root3+2-root3+a+b=6/1
4+a+b=6
a+b=2; b=2-a
product of roots= constant/ coefficient of x^4
(2+root3)(2-root3)(a)(b)= -35
(4-3)(a)(b)= -35
ab= -35
a(2-a)=-35
a^2-2a-35=0
(a-7)(a+5)=0
a=-5, a=7
therefore if a=-5 ; b=7 (or)
a=7 ; b= -5
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