Convert standard form to factor form

Question

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

Answer 1

$$x^4+6x^3-x^2-6x=x^4-x^2+6x^3-6x=x^2(x^2-1)+6x(x^2-1)=$$ $$(x^2-1)x(x+6)=(x-1)(x+1) x(x+6)=x(x-1)(x+1)(x+6).$$

Answer 2

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.

Answer 3

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 $$ (x-1)(x+1)(x+6)x $$

Answer 4

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: $x(x+6)(x-1)(x+1)$.

Answer 5

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.