Question

if X³+ax²+bx+6has(x-2) as factor and leaves a reminder 3when divided by (x-3) find the values of a and b​

Answer 1

$p(x)=x^3+ax^2+bx+6=0$

(x - 2) is a factor of p(x). So p(2) = 0.

$\Longrightarrow\ p(2)=(2)^3+a(2)^2+b(2)+6=0 \\ \\ \Longrightarrow\ p(2)=8+4a+2b+6=0 \\ \\ \Longrightarrow\ p(2)=4a+2b+14=0 \\ \\ \\ 4a+2b+14=0 \\ \\ 4a+2b=-14 \\ \\ 2a+b=-7\ \ \ \ \ \ \ \ \ \ \longrightarrow\ \ \ (1)$

p(x) leaves remainder 3 on division by (x - 3). So p(3) = 3.

$\Longrightarrow\ p(3)=(3)^3+a(3)^2+b(3)+6=3 \\ \\ \Longrightarrow\ p(3)=27+9a+3b+6=3 \\ \\ \Longrightarrow\ p(3)=9a+3b+33=3 \\ \\ \\ 9a+3b+33=3 \\ \\ 9a+3b=3-33 \\ \\ 9a+3b=-30 \\ \\ 3a+b=-10\ \ \ \ \ \ \ \ \ \ \longrightarrow\ \ \ (2)$

(2) - (1)

$(3a+b)-(2a+b)=(-10)-(-7) \\ \\ 3a+b-2a-b=-10+7 \\ \\ a=\bold{-3}$

From (1),

$2a+b=-7 \\ \\ 2(-3)+b=-7 \\ \\ -6+b=-7 \\ \\ b=-7+6 \\ \\ b=\bold{-1}$

Thus,

$\Large\boxed{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a=\boxed{\bold{-3}}\ \ \ \ \ \ \ \ \ \ b=\boxed{\bold{-1}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$