Question

-3,0 and 2 are the zeroes of the polynomial p(X)=x³+(a-1)x²+bx+c ,find a,c.​

Answer 1

Answer:

a = 2

b = - 6

c = 0

Step-by-step explanation:

Since, -3,0 and 2 are the zeroes of the polynomial

$Plug \:x = 0\: in \:p(X) \\ \implies p(0)= 0\\\therefore 0= 0^3 +(a-1)(0)^2 +b(0)+c\\\therefore 0= 0 +(a-1)\times 0 +b(0)+c\\\therefore 0= 0 +0+0+c\\\red{\bold{\therefore 0=c}}.....(1)\\\\p(X)=x³+(a-1)x²+bx+c. \\Plug \:x = -3\: in \:p(X) \\[tex] \implies p(-3)= 0\\\therefore 0= (-3)^3 +(a-1)(-3)^2 +b(-3)+c\\\therefore 0= -27 +(a-1)\times 9 +b(-3)+c\\\therefore 0= - 27 +9a-9 - 3b+c\\\therefore 0= - 36+9a-3b+c\\\therefore 36 =9a-3b+c... (2)\\Plug \:c= 0\: in \:equation \: (2)\\36= 9a - 3b + 0\\36= 9a - 3b\\36= 3(3a - b) \\\purple {\bold{12 = 3a - b}} .... (3)\\\\Plug \:x = 2\: in \:p(X) \\\implies p(2)= 0\\\therefore 0= 2^3 +(a-1)(2)^2 +b(2)+c\\\therefore 0= 8 +(a-1)\times 4 +b(2)+c\\\therefore 0= 8 +4a-4+2b+c\\\therefore 0= 4 +4a+2b+c\\\therefore - 4=4a+2b+c.....(4)\\Plug \:c= 0\: in \:equation \: (4)\\- 4 = 4a +2b+0\\-4= 2(2a+b)\\\orange{\bold{-2=2a + b}} .......(5)$

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