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

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted March 6, 2013 at 3:56 AM via web

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Factor completely. 

 `2x(x^3)+5x(x^2)-8x-20`

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Wilson2014 | eNotes Employee

Posted March 6, 2013 at 9:34 PM (Answer #2)

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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)`

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mathsworkmusic | (Level 3) Associate Educator

Posted March 11, 2013 at 5:43 PM (Answer #4)

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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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