How do you factor $x^3 - 7x - 6$ ?

Question

1538 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-25T22:10:30

$x^3-7x-6 = (x-1)(x-3)(x+2)$

Given:

$x^3-7x-6$

Notice that if you reverse the signs of the coefficients on the terms of odd degree, then their sum is $0$ . That is:

$-1+7-6 = 0$

Hence $x=-1$ is a zero and $(x+1)$ a factor:

$x^3-7x-6 = (x+1)(x^2-x-6)$

To factor the remaining quadratic, find a pair of factors of $6$ which differ by $1$ . The pair $3, 2$ works and hence we find:

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

Putting it all together:

$x^3-7x-6 = (x-1)(x-3)(x+2)$

How do you factor x^3 - 7x - 6x^3 - 7x - 6?