Question

Find the other two zeroes of p(x)=3x^4+6x-2x^2-10x-5

Answer 1

p(x)=3x4+6x3-2x2-10x-5

as we have given that two of its zeroes are root5/3 and -root5/3. so their factors are;

(x-root5/3) and (x+root5/3). the product of the factors ;

(x-root5/3) (x+root5/3)=x2-5/3

                                  =3x2-5/3=1/3(3x2-5)

                                   where k=1/3

now g(x)=3x2-5

dividing p(x) by g(x) we get

q(x)=x2+2x+1

by division algorithm;

p(x)=g(x)*q(x)+r(x)

      =(3x2-5)(x2+2x+1)

      =(3x2-5)(x2+x+x+1)

     =(3x2-5)(x(x+1)+1(x+1)

     (3x2-5)(x+1)(x+1)

the other two zeroes are -1 and-1