Without actual division, show that (x^3-3 x^2-13 x+15 ) is exactl… - Vidyadip

Question

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

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Answer

Let $f(x)=x^3-3 x^2-13 x+15$
Now, $x^2+2 x-3$
$= x^2 + 3x - x - 3$
$= x (x + 3) - 1 (x + 3)$
$= (x + 3) (x - 1)$
Thus, $f(x)$ will be exactly divisible by $x^2+2 x-3$
$=(x+3)(x-1)$ if $(x+3)$ and $(x-1)$ are both factors of $f(x)$,
so by factor theorem, we should have $f(-3)=0$ and $f(1)=0$.
Now, $f(-3) = (-3)^3 - 3(-3)^2 - 13(-3) + 15$
$= -27 - 3 \times 9 + 39 + 15$
$= -27 - 27 + 39 + 15$
$= -54 + 54 = 0$
And, $f(1) = 1^3 - 3 \times 1^2 - 13 \times 1 + 15$
$= 1 - 3 - 13 + 15$
$= 16 - 16$
$= 0$
$\therefore$ $f(-3) = 0$ and $f(1) = 0$
So, $x^2 + 2x - 3$ divides $f(x)$ exactly.

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