Question

let me see who will answer my question first.

Answer 1

1 and - 2 are the zeros of polynomial
P( x )= x³ + 10x² + px + q

so,
P( 1 ) = 0 and P(-2) = 0

( 1)³ + 10 (1)² + p ( 1) + q =0

P + q = -11 -------(1)

and

(-2)³ +10(-2)² +P(-2) + q = 0
-8 + 40 -2P + q =0
2P - q = 32 --------(2)

solve equations (1) and (2)

3P = 21
P = 7
put P = 7 in eqn (1)

q = -18

so, P = 7 and q = -18
now,
equation is
x³ + 10x² +7x -1 8=0

sum of roots = -10
1 -2 + ß = -10
ß = -9

hence 3rd root = -9

Answer 2

Given f(x)=x³+10x²+px+q

roots are 1,-2

f(1)=1+10+p+q
     =p+q+11  (1)
f(-2) =-8+40-2p+q
        =-2p+q+32  (2)

solving(1) and(2),

we get p=7 and q=-18

∴f(x)=x³+10x²+7x-18
 by solving we get 3rd zero as -9