<s>[INST] tough factorisation problem How would you factorise this equation given that $x=7$ is a root of this equation 
$$x^3 - 67x + 126 = 0.$$
Any help would be thoroughly appreciated. 
 [/INST] $x^3-67x+126$ and the root is $x=7$. One writes $x=7$ as $x-7=0$ and convert the above equation in quadratic, by dividing $x^3-67x+126$ by $x-7$; so the quotient is $x^2+7x-18$. Now one factorises this $x^2+7x-18=x^2+9x-2x-18=x(x+9)-2(x+9)=(x-2)(x+9)$ then $\displaystyle(x-2)(x+9)=\frac{x^3-67x+126}{x-7}$, so $(x-2)(x+9)(x-7)$ is factorization of $x^3-67x+126$.
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