Question

determine the zero of function y=2x3-17x2+23x+42

Answer 1

$y=2x^3-17x^2+23x+42\\\\Possible\ factors\ are:\ +or-\ factors\ of\ 42/factors\ of\ 2\\\\ie.,\ +or-\ :1,2,3,7,14,21,42,1/2,3/2,7/2,21/2\\\\By\ examining\ coefficients\ we\ find\ -1\ as\ a\ zero$

y = 2 (-1)³ - 1 7 (-1)² +23 (-1) + 42 = 0  =>  (x+1) is a factor.

Divide the given polynomial by (x+1).

or write  y = (x+1) (2x²+ax+42) = 2x³-17x²+23x+42 
               = > 2x³+x²(a+2)+x(a+42)+42 = 2x³-17x²+23x+42 
    a = -19  
 hence     y = (x+1) (2x²-19x+42) 
 
zeros or roots of quadratic equation 2x² - 19 x + 42 are
            [ +19 +- √(381- 336)] /2  = (19 +- 3√5) /2

zeroes are -1, (19+3√5)/2  ,  (19-3√5)/2