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

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Wilson2014's profile pic

Posted on

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)`

mathsworkmusic's profile pic

Posted on

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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