Answer:
a = 2
b = - 6
c = 0
Step-by-step explanation:
Since, -3,0 and 2 are the zeroes of the polynomial
[tex] Plug \:x = 0\: in \:p(X) \\
\implies p(0)= 0\\
\therefore 0= 0^3 +(a-1)(0)^2 +b(0)+c\\
\therefore 0= 0 +(a-1)\times 0 +b(0)+c\\
\therefore 0= 0 +0+0+c\\
\red{\bold{\therefore 0=c}}.....(1)\\\\
p(X)=x³+(a-1)x²+bx+c. \\
Plug \:x = -3\: in \:p(X) \\
[tex] \implies p(-3)= 0\\
\therefore 0= (-3)^3 +(a-1)(-3)^2 +b(-3)+c\\
\therefore 0= -27 +(a-1)\times 9 +b(-3)+c\\
\therefore 0= - 27 +9a-9 - 3b+c\\
\therefore 0= - 36+9a-3b+c\\
\therefore 36 =9a-3b+c... (2)\\
Plug \:c= 0\: in \:equation \: (2)\\
36= 9a - 3b + 0\\
36= 9a - 3b\\
36= 3(3a - b) \\
\purple {\bold{12 = 3a - b}} .... (3)\\\\
Plug \:x = 2\: in \:p(X) \\
\implies p(2)= 0\\
\therefore 0= 2^3 +(a-1)(2)^2 +b(2)+c\\
\therefore 0= 8 +(a-1)\times 4 +b(2)+c\\
\therefore 0= 8 +4a-4+2b+c\\
\therefore 0= 4 +4a+2b+c\\
\therefore - 4=4a+2b+c.....(4)\\
Plug \:c= 0\: in \:equation \: (4)\\
- 4 = 4a +2b+0\\
-4= 2(2a+b)\\
\orange{\bold{-2=2a + b}} .......(5)\\\\[/tex]
Adding equation (3) and (5), we find:
12 = 3a - b
-2 = 2a + b
_________
10 = 5a
10/5 = a
a = 2
Plug a = 2 in equation (5)
-2= 2(2) + b
-2 = 4 + b
-2 - 4 = b
-6 = b
b = - 6
