# Factor completely.  `2x(x^3)+5x(x^2)-8x-20`

When faced with factorizing an equation of higher order than two, we can look for integer roots by looking at integer factors of the zeroeth term (in this case -20).

Trying 1,-1,2,-2,4,-4,5,-5 it can be seen that there is a root between -2 and -3.

Using the method of bifurcation we try -2.5 next. This provides a root: (2x+5).

We can then use algebraic long division to factor out the equation:

`x^3 -4`

`2x+5 )` `overline (2x^4 + 5x^3 - 8x - 20)`

`-2x^4 + 5x^3`

`-8x - 20`

`- -8x -20`

`0`

So the equation factorizes (fully) to (2x+5)(x^3-4). There are two real roots (implying two additional complex roots as the equation is order 4).

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Method: Simplify first. Then factor by grouping.

`2x(x^3)+5x(x^2)-8x-20`

`=> 2x^4+5x^3-8x-20`

Factor by grouping. It becomes:

`x^3*(2x+5)-4*(2x+5)`

=> `(x^3 - 4)*(2x+5)`

This cannot be factored anymore.

Therefore the completely factored expression is

`(x^3-4)*(2x+5)`