Question

If $(x^3+a{x}^2+bx+6)$ has $(x-2)$ as a factor and leaves a remainder $3$ when divided by $(x-3)$, find the values of $a$ and $b$.

Answer 1

$f\left(x\right)=x^3+a{x}^2+bx+6$

$(x-2)$ in a factor $\Rightarrow f\left(2\right)=0$

$f\left(2\right)\Rightarrow 2^3+a(2{)}^2+2b+6=0$

$\Rightarrow 8+4a+2b+b=0$

$\Rightarrow 4a+2b+14=0$

$\Rightarrow 2a+b+7=0$ -(1)

$f(x-3)=3$ (Remainder)

$\Rightarrow f\left(3\right)\Rightarrow 3^3+a(3{)}^2+b\times 3+6=3$

$\Rightarrow 27+9a+3b+b=3$

$\Rightarrow 9a+3b+30=0$

$\Rightarrow 3a+b+10=0$ _____ (2)

From (1) & (2)

$b=-2a-7$ & $b=-10-3a$

$-2a-7=-10-3a$

$3a-2a=-10+7$

$a=-3$

from (3)

$b=-2\left(a\right)-7=2(-3)-7$

$=6-7$

$b=-1$