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

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Solution

ANSWER:
Let:
$f (x) = x^{3} - 3 x^{2} - 13 x + 15$ And,
$g (x) = x^{2} + 2 x - 3$
$\to x^{2} + x - 3 x - 3 = (x - 1) (x + 3)$
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:
$f (1) = 1^{3} - 3 \times 1^{2} - 13 \times 1 + 15 = 1 - 3 - 13 + 15 = 0$
And,
$f (- 3) = (- 3)^{3} - 3 \times (- 3)^{2} - 13 \times - 3 + 15 = - 27 - 27 + 39 + 15 = 0$
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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