# How to factorize $x^3 - 7x + 6$?

How do you factorize this polynomial:

$${x^3 - 7x + 6}$$

Some online solver doesn't even work saying: using GCF method doesn't work, but sites like Mathway.com gave me the answer, is there a pre-step you need to do before factorizing?

the answer is $(x-1)(x-2)(x+3)$.

This is actually part (b) of a question, it said use the answer for part (a) i.e $x^3-8$ and factorize. I don't get the relationship, what does this hint actually shows?

• Would you mind including in your post the answer that you were given? – Michael Albanese Jul 30 '13 at 6:21
• Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Martin Sleziak Jul 30 '13 at 8:45

The method below will not always work, but it is another way to arrive at the same answer for this cubic.

This is difficult because it is a cubic, but if you consider the quadratic $x^2 - 7x + 6$, that factors nicely as $(x-1)(x-6)$. So we rewrite our given cubic as $$x^3 - 7x + 6 = (x^3 - x^2) + (x^2 - 7x + 6).$$ Factorising each bracket, we have \begin{align*} x^3 - 7x + 6 &= (x^3 - x^2) + (x^2 - 7x + 6)\\ &= x^2(x - 1) + (x - 1)(x - 6)\\ &= (x - 1)(x^2 + x - 6)\\ &= (x - 1)(x - 2)(x + 3). \end{align*}

As the degree is three, one of the factors must be linear. A linear factor with rational coefficients means there is a rational root and by the rational root theorem, it must be a divisor of $6$ ($\pm1,\pm2,\pm3,\pm6$). Can you see that one of these is indeed a root?

• Maybe I misunderstand your reasoning, but $X^3-2$ has no rational linear factor.But it is indeed true here that the rational root theorem restricts the quest for a possible rational root to the values you mention. Although it is not guaranteed to exist by the simple fact that the degree is $3$. – Julien Jul 30 '13 at 6:40
• @julien I think the intended meaning is 'As the degree is three, if there is a factorisation then one of the factors must be linear'. As opposed to, say, degree four factorising to two quadratic factors and no further. – AakashM Jul 30 '13 at 9:09
• @AakashM What I am saying is that the fact that the degree is $3$ does not bring any information towards proving there exists a rational root. – Julien Jul 30 '13 at 15:34

$x^3−8 = (x-2)(x^2+4+2x)$
LHS=$x^3-8-7x+14$
$=(x-2)(x^2+2x+4-7)$
$(x-2)(x-1)(x+3)$

• For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Jun 30 '15 at 17:58

We must add (-8) and then the polynomial becomes : X^3 -8 -7x + 14 and then we can factorise. I hope this helps .