Question

16
( x=pm sqrt{3} )
( x^{2}=3 )
( x^{2}-3=0 ) which is om ob roots.
Brividing ( left(x^{2}-3right) ) by to ( x^{2} e^{x} ) the ( e q^{u} )
( begin{array}{c}x^{2}-3 mid begin{array}{ccc}x^{2} & -3 x & -4 x^{4}-3 x^{3}-7 x^{2}+9 x+12 & 4 -frac{x^{4}}{5}-3 x^{2} & -3 x^{3}-4 x^{2}+9 x+12end{array} 1 & -3 x^{2}end{array} )
( frac{pm 3 x^{3}+9 x}{1-4 x^{2}+12} )
( +frac{-4 x^{2}+12}{0}= )
We get ( x^{2}-3 x-4=0 )
So ( Rightarrow(x-4)(x+1)=0 )
[
x=4,-1
]
Its zeros are ( pm sqrt{3}, 4,-1 )
Sum ob, thoes ( =+sqrt{3}-sqrt{3}+4-1 )
( =3 )

# 0 to 9. 1. Obtain all zeroes of x4 -3x3-7x2 + 9x +12 iftwo of its zeroes are +3. Sum of zeroes is che quadratic polynom

Solution