Without actual division, show that

Let:

$f(x)=x^{3}-3 x^{2}-13 x+15$

And,

$g(x)=x^{2}+2 x-3$

$=x^{2}+x-3 x-3$

$=x(x-1)+3(x-1)$

$=(x-1)(x+3)$

Now, $f(x)$ will be exactly divisible by $g(x)$ if it is exactly divisible by $(x-1)$ as well as $(x+3)$. For this, we must have:

$f(1)=0$ and $f(-3)=0$

Thus, we have

$f(1)=\left(1^{3}-3 \times 1^{2}-13 \times 1+15\right)$

$=(1-3-13+15)$

$=0$

And,

$f(-3)=\left[(-3)^{3}-3 \times(-3)^{2}-13 \times(-3)+15\right]$

$=(-27-27+39+15)$

$=0$

$f(x)$ is exactly divisible by $(x-1)$ as well as $(x+3)$. So, $f(x)$ is exactly divisible by $(x-1)(x+3)$. Hence, $f(x)$ is exactly divisible by $x^{2}+2 x-3$.