Question

find all the zeros of 3x³+16x²+23x+6 if two of its zeros are -2 and -3​

Answer 1

Answer:

using property, sum of zeroes = -b/a

Let third zero be y

-2-3+y = -16/3

y = 5-16/3

y=-1/3

Alter method,

if we use, product of zeroes = -d/a

(-2)(-3)y = -6/3

6y = -2

y = -1/3

Answer 2

Answer:

since -2 and -3 are two zeroes x+2 and x+3are two factors

divide 3x^3 + 16x^2 + 23x +6 by (x+2)(x+3)

(3x^3 + 16x^2 +23x+ 6)÷(x^2+5x+6)= 3x + 1

3x+1=0

x= -1/3

since power is three, there are three zeroes

-2, -3 and -1/3