Question

find all the zeros of f(x)=x^4+x^3-23x^2-3x
+60,if one of its zeors is root 3

Answer 1

\[-\sqrt{3}\]

Answer 2

when x is -5, 4, -1.732051, 1.7320508
actually it's two zeros with negative 3

Answer 3

4,-5

Answer 4

give me its solution

Answer 5

may be you should do it in trail and error method. i did like that.

Answer 6

If
\[x = \sqrt{3}\]
is a zero then it's conjugate is also a zero. So,
\[ x = - \sqrt{3}\]
is another zero.
Multiply the factors,
\[(x-\sqrt{3})(x+\sqrt{3}) = x^2-3\]
Using long division of polynomials will lead to
\[(x^2-3)(x^2+x-20) = (x^2-3)(x+5)(x-4)\]
Giving the last two zeros of -5 and 4.