Show that $(x + 3)$ is a factor of $x^3 - 3x^2 - 10x + 24$ and hence factorise $x^3 - 3x^2 - 10x + 24$ fully. - Scottish Highers Maths - Question 3 - 2015

Now that we know $(x + 3)$ is a factor, we can perform polynomial long division or synthetic division to factor the cubic polynomial completely.

1. Divide $x^3 - 3x^2 - 10x + 24$ by $(x + 3)$:

   • The first term gives $x^2$

   • Multiply $(x + 3)$ by $x^2$ to get $x^3 + 3x^2$

   • Subtract this from the original polynomial:

     $(x^3 - 3x^2 - 10x + 24) - (x^3 + 3x^2) = -6x + 24$

   • Next term gives $-6$

   • Multiply $(x + 3)$ by $-6$ to get $-6x - 18$

   • Subtract:

     $(-6x + 24) - (-6x - 18) = 42$

Thus, we find the quotient:

$x^2 - 6$

2. Therefore, we can express:

   $x^3 - 3x^2 - 10x + 24 = (x + 3)(x^2 - 6)$

3. Now we can further factor $x^2 - 6$:

   x^2 - 6 = (x - rac{ ext{sqrt}(6)}{1})(x + rac{ ext{sqrt}(6)}{1})

Finally, we have the complete factorization:

$x^3 - 3x^2 - 10x + 24 = (x + 3)(x - ext{sqrt}(6))(x + ext{sqrt}(6))$