long division polynomials

[kpx001]

find x and the factors
2x+4; 2x^3-6x+4

.........2x^2-4x-1+8/2x+4
2x+4/2x^3+0x^2-6x+4
.........2x^3+8x^2
...................-8x^2-6x
...................--8x^2-8x
............................-2x+4
............................--2x-4
...................................8

how would i know what x is and is this factorable and what would it b.


also graphing x^3+x^2-3x+5 how many real zeros are there?
im guessing 2?

 

[pka]

For a polynomial with real coefficients complex roots occur in conjugate pairs.
Therefore a polynomial over the real field of odd degree must have an even number of complex roots. So your third degree polynomial would have one or three real roots.

 

[kpx001]

this is for "graphing x^3+x^2-3x+5 how many real zeros are there? " correct?
and also how would i find the x in the division problem?

 

[Gene]

A few mistakes. It should look like
(2x^3-6x+4)/(2x+4)

            x^2 -2x +1     
2x+4/2x^3 +0x^2 -6x +4
     2x^3 +4x^2       
          -4x^2 -6x
           4x^2 -8x
                +2x +4
                +2x +4
                     0

The new quotient is also factorable.

How would I know what x is, is this factorable and what would it be.

If 2x+4 is a factor then
2x+4=0
x=-2

2x^3-6x+4 =
2(-2)^3-6(-2)+4 =
-16+12+4=0
It checks.

BTW: You could have factored out a 2 from the numerator and denominator first.