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Factor completely. `2x(x^3)+5x(x^2)-8x-20`
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Method: Simplify first. Then factor by grouping.
Factor by grouping. It becomes:
=> `(x^3 - 4)*(2x+5)`
This cannot be factored anymore.
Therefore the completely factored expression is
Posted by Wilson2014 on March 6, 2013 at 9:34 PM (Answer #2)
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:
`2x+5 )` `overline (2x^4 + 5x^3 - 8x - 20)`
`-2x^4 + 5x^3`
`-8x - 20`
`- -8x -20`
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).
Posted by mathsworkmusic on March 11, 2013 at 5:43 PM (Answer #4)
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