Question

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

Answer 1

a = 2

b = - 6

c = 0

Explanation:

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

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