Question

How do you factor `y= 3x^5 -6x^2 -27x-18` ?

Answer 1

`3x^5-6x^2-27x-18`

`= 3(x+1)(x+1)(x-2)(x^2+3)`

`= 3(x+1)(x+1)(x-2)(x-sqrt(3)i)(x+sqrt(3)i)`

Explanation:

First separate out the common scalar factor `3` to find:

`y=3x^5-6x^2-27x-18=3(x^5-2x^2-9x-6)`

Next let `f(x) = x^5-2x^2-9x-6`

Notice that if the signs of the coefficients are reversed on the terms of odd degree then their sum is zero. That is:

`-1-2+9-6 = 0`

Hence `f(-1) = 0`, so `x = -1` is a zero and `(x+1)` a factor:

`x^5-2x^2-9x-6 = (x+1)(x^4-x^3+x^2-3x-6)`

Notice that if we reverse the signs of the coefficients of the terms of odd degree in the remaining quartic factor then the sum is zero. That is:

`1+1+1+3-6 = 0`

Hence we have another factor `(x+1)`:

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

Let `g(x) = x^3-2x^2+3x-6`

By the rational root theorem, any rational zeros of `g(x)` must be expressible in the form `p/q` for integers `p`, `q` with `p` a divisor of the constant term `-6` and `q` a divisor of the coefficient `1` of the leading term.

That means that its only possible rational zeros are:

`+-1`, `+-2`, `+-3`, `+-6`

By examining the sum of the coefficients we can rule out `+-1`, but we find:

`g(2) = 8-8+6-6 = 0`

So `(x-2)` is a factor:

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

The remaining quadratic factor has no linear factors with Real coefficients since `x^2+3 >= 3 > 0` for all `x in RR`. If we allow Complex coefficients then it factors as `(x-sqrt(3)i)(x+sqrt(3)i)`

Putting it all together:

`3x^5-6x^2-27x-18`

`= 3(x+1)(x+1)(x-2)(x^2+3)`

`= 3(x+1)(x+1)(x-2)(x-sqrt(3)i)(x+sqrt(3)i)`