Question

( pm sqrt{3} ) are zeroes ( Rightarrow(x+sqrt{3})(x-sqrt{3})=x^{2}-3 ) is a divisor
of ( x^{4}-3 x^{3}-7 x^{2}+9 x+12 )
[
begin{aligned}
q(x) &=x^{2}-3 x-4=x^{2}-4 x+x-4
&=x(x-4)+1(x-4)
q(x) &=(x+1)(x-4)
end{aligned}
]
zeroes of ( q(x) ) are -1,4
( therefore ) Remaining required zeroes are -1,4
Sum ( =-1+4=3 )

# Obtain all zeroes of x4-3x3-7x2 +9x +12 iftwo of its zeroes are + V3.Sum of zeroes is

Solution