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.
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.
On
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.
On
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.
On
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$=(x-7)(x^2+7x-18).$
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