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. 
 A: This is not as 'tough' as mentioned in the title. You just need to know some basic rules.
By factor theorem, if $7$ is a root of the polynomial, then $x-7$ is one of its factors. Now you can use two methods. First is the vanishing method. It is a powerful tool of trial that can be used to factorise almost any polynomial.

We know that we have to achieve $x-7$ as a factor. So, let's just break the terms such that we achieve the same.
The given expression $=(x^3-7x^2)+(7x^2-49x)-(18x-126)=x^2(x-7)+7x(x-7)-18(x-7)$
$=(x-7)(x^2+7x-18)$

This method of trial may appear to be quite absurd but actually I just needed to adjust the first term with $-7x^2$.The rest of it came automatically, step by step.
Another less technical but more logical method is to use the long division method.

Divide the polynomial by $x-7$. As it is a factor, no remainder will be left. Now, by division algorithm, the expression must be the product of the divisor and the quotient$=(x-7)(x^2+7x-18).$

A: Use the factor theorem: $r$ is a root of polynomial $p$ if and only if $x-r$ is a factor of $p$. Since 7 is a solution of $x^3 - 67x + 126 = 0$, 7 is a root of $x^3 - 67x + 126$ so $x-7$ divides $x^3 - 67x + 126$. You may use polynomial long division to factor out $x-7$, leaving a quadratic polynomial that you can factor in standard ways.
A: $x^3-67x+126$ 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 $x^3-67x+126$.
A: Using the fact  $$x^3-67x+126=x^2(x-7)+7x(x-7)-18(x-7)$$
So, the Quotient $\dfrac{x^3-67x+126}{x-7}=?$ which is a Quadratic Polynomial which can be factorized easily 
A: Just a note on the polynomial division here.
Once you have identified $x-7$ as a factor, note that if $x-a$ is a factor of $x^n+bx^{n-2}+\dots$ (i.e. there is a term missing) the factorisation will begin $(x-a)(x^{n-1}+ax^{n-2}+\dots)$
Also $126=7\times 18$
So you can write down $(x-7)(x^2+7x-18)$
This comes in handy quite often with cubics and other low degree polynomial examples.
A: As 7 is one of the Root of given expression therefore by factor theorem it is possible that x-7 is one of the factor of the expression dividing it by x-7 will give us other remaining expression which will be quadratic and after factorisation of it will give us factor of the expression
