Convert standard form to factor form Im so confused now 
I have no idea how to factor a standard form
i have example shown below
Standard form
$$x^4+6x^3-x^2-6x$$
Factor form
$$x(x+6)(x+1)(x-1)$$
Can you explain a little bit 
I have no idea
Thank you
 A: $$x^4+6x^3-x^2-6x=x^4-x^2+6x^3-6x=x^2(x^2-1)+6x(x^2-1)=$$ $$(x^2-1)x(x+6)=(x-1)(x+1) x(x+6)=x(x-1)(x+1)(x+6).$$
A: A rule of thumb to always try first for factoring four-termed polynomials is to check whether you can immediately apply the distributive rule by grouping terms in pairs (after a little rearrangement, perhaps). To see whether this works, for example, given the tetranomial $$a+b+c+d,$$ check whether $ad=bc.$ This is a necessary and sufficient condition for polynomials of this form.
In your case $$x^4+6x^3-x^2-6x,$$ it is immediately apparent that this holds as given, so that we may group the terms in pairs and bring out common factors. This gives us two (compound) terms with a common factor, and we apply the distributive rule again to complete the job.
A: In general it is quite difficult.
In this case, however, there are some things we can exploit. Look at the cofficients: $1, 6, -1, -6$. A pattern repeating (except for a sign change). That automatically means that $x^2-1 = (x+1)(x-1)$ is a factor (because that's what happens when you multiply a linear polynomial with $x^2+1$; if the sign hadn't changed and we had had $1, 6, 1, 6$ instead, then it would've been $x^2+1$). So we get
$$
x^4+6x^3-x^2-6x = (x-1)(x+1)(x^2+6x)
$$
The remaining second degree polynomial isn't difficult to factor the "hard way", but we can also clearly see that $x$ is a factor, so we get
$$
(x-1)(x+1)(x+6)x
$$
A: Observing your polynomial, notice that we can rewrite it as: $x^3(x+6)-x(x+6)$. Now pick up $(x+6)$ and obtain: $(x+6)(x^3-x)$. Repeat the process with $x$, so: $x(x+6)(x^2-1)$. There is a difference of cubes: $x(x+6)(x-1)(x+1)$.
A: The expression is $ x(x^3 + 6x^2-x-6) $ $\\$ if roots are integers they will be from positive or negative factors of constant i.e factors of 6 . $\\$ Because product of roots is $(-1)^n\,$constant term upon coffecient of highest degree i.e $x^3$ in this case.
