Without actual division, show that $\left(x^{3}-3 x^{2}-13 x+15\right)$ is exactly divisible by $\left(x^{2}+2 x-3\right)$
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 $x^{2}+2 x-3$.
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