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

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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 )

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