If $\left(x^{3}+a x^{2}+b x+6\right)$ has $(x-2)$ as a factor and leaves a remainder 3 when divided by

Solution:

Let:

$f(x)=x^{3}+a x^{2}+b x+6$

$(x-2)$ is a factor of $f(x)=x^{3}+a x^{2}+b x+6$

$\Rightarrow f(2)=0$

$\Rightarrow 2^{3}+a \times 2^{2}+b \times 2+6=0$

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

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

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

Now,

$x-3=0 \Rightarrow x=3$

By the factor theorem, we can say:

When $f(x)$ will be divided by $(x-3), 3$ will be its remainder.

$\Rightarrow f(3)=3$

Now,

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

$=(27+9 a+3 b+6)$

$=33+9 a+3 b$

Thus, we have:

$f(3)=3$

$\Rightarrow 33+9 a+3 b=3$

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

$\Rightarrow 3 a+b=-10 \quad \ldots(2)$

Subtracting (1) from (2), we get:

$a=-3$

By putting the value of $a$ in (1), we get the value of $b$, i.e., $-1$.

$\therefore a=-3$ and $b=-1$