Question

Two zeroes of cubbic polynomial ax^3 +3x^2-bx-6 are -1and-2 find the third zero and values of a and b

Answer 1

α=-1 and β= -2
sum of zeroes = -b/a
α.+β = -b/a
-1-2 = -b/a
-3 = -b/ a
-3.-2/ -2
6/-2
product of zeroes =c/a
αβ = c/a
-1.-2= c/a
on comparing
a =-2
b =-6

Answer 2

Let f(x)=ax^3+3x^2-bx-6
since -1 is a zero therefore
f(-1)=0
Or, a×(-1)^3+3×(-1)^2-b×(-1)-6=0
Or, -a+3+b-6=0
Or, b-a=3......(1)
again f(-2)=0
Or, a×(-2)^3+3×(-2)^2-b×(-2)-6=0
Or, -8a+12+2b-6=0
Or, 2b-8a=-6
Or, b-4a=-3
Or, b=4a-3....(2)
Now from (1) we have
4a-3-a=3
Or, 3a=6
Or, a=2
from (2) we have
b=4×2-3
b=5
hence a=2 and b=5
now the polynomial becomes 2x^3+3x^2-5x-6
and the equation is
2x^3+3x^2-5x-6=0
Let other root be α
hence from the formula we have (-1)×(-2)×α=-(-6)/2
Or,2α=3
Or, α=3/2
hence other zero is 3/2