Question 20643
<font face = "courier new" size = 3>find the roots of each polynomial equation:
x(cubed)-5x(squared)+2x+8=0
`<b><font size = 4>
x³ - 5x² + 2x + 8 = 0
`
Since the coefficient of x³ is 1, if it has any real rational roots,
they have to be a divisor of the absolute value of the constant term in
absolute value. So we can try +1, -1, +2, -2, +4, -4, +8, or -8
`
Try the easiest one first. Substitute x=1, you get 
`
(1)³ - 5(1)² + 2(1) + 8 = 0
` ` ` ` `1 - 5 + 2 + 8 = 0
` ` ` ` ` ` ` ` ` ` `6 = 0
`
Nope, 1 is not a root.  So try substituting x=-1, you get
`
(-1)³ - 5(-1)² + 2(-1) + 8 = 0
` ` ` ` -1 - 5 - 2 + 8 = 0
` ` ` ` ` ` ` ` ` ` `0 = 0
`
Yes that's a solution, so if x = -1 is a root, that means
`
x + 1 is a factor, because if you set that = 0 you'll get x = -1.
`
So divide by x+1, you use synthetic division, using -1 in the far left:
`
-1 | 1 -5  2 `8
` `|___-1_6_-8_ 
` ` `1 -6  8 `0
`
The numbers on the bottom mean x²-6x+8 with the last number 0 
being the remainder, which is how you know x+1 was a factor of 
the polynomial. So you now have the polynomial equation factored 
as
`
(x+1)(x²-6x+8) = 0
`
You can now factor the trinomial in the second parentheses as 
(x-2)(x-4) and get 
`
(x+1)(x-2)(x-4)=0
`
Set each of those three parenthetical expressions = 0, and get -1, 2, 
and 4.
`
Edwin
AnlytcPhil@aol.com