Question 1206675
<br>
{{{4x^3+16x^2-22x-10=0}}}
{{{2x^3+8x^2-11x-5=0}}}<br>
The possible rational roots are 1, -1, 5, -5, 1/2, -1/2, 5/2, and -5/2.<br>
You could test each possible root using synthetic division, but that would be a long slow process.  Checking for zeros (roots) by evaluating the polynomial for each root is faster.<br><pre>
x=1:   2+8-11-5 = -6 --> not a root
x=-1: -2+8+11-5 = 12 --> not a root
x=5:   250+200-55-5 = 490 --> not a root
x=-5: -250+200+55-5 = 0 --> -5 is a root</pre>
Divide the polynomial by (x-(-5)) to find the remaining quadratic polynomial.<br><pre>

 -5 | 2   8 -11  -5
    |   -10  10   5  
    +---------------
      2  -2  -1   0</pre>
This tells us that<br>
{{{2x^3+8x^2-11x-5=(x+5)(2x^2-2x-1)}}}<br>
The quadratic factor is not factorable over the integers; find the other two roots using the quadratic formula.  The other two (irrational) roots are<br>
{{{(-b+-sqrt(b^2-4ac))/(2a)=(2+-sqrt(12))/(4)=(1+-sqrt(3))/2}}}<br>
ANSWERS: {{{-5}}},{{{(1+sqrt(3))/2}}}, {{{(1-sqrt(3))/2}}}<br>