Question 469096: Use the rational zero theorem and quotient polynomials to find all roots of the given equation
x(^4)+ x(^3)+x(^2)+3x-6=0
(x to the fourth plus x to the third plus x squared plus 3x minus 6 equals 0)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Use the rational zero theorem and quotient polynomials to find all roots of the given equation
x(^4)+ x(^3)+x(^2)+3x-6=0
(x to the fourth plus x to the third plus x squared plus 3x minus 6 equals 0)
...
P(x)=x^4+x^3+x^2+3x-6
Factors of p=6: ±1, ±2, ±3, ±3, ±6
Factors of q=1: ±1
possible rational roots, p/q: ±1, ±2, ±3, ±3, ±6
..
Using synthetic division to find roots:
0)......1....1....1....3....-6
1)......1....2....3....6.....0 (root=1)
p(x)=(x-1)(x^3+2x^2+3x+6)
Try again:
0)......1....2....3....6
-1)....1....1....2....4
-2)....1....0....3....0 (root=-2)
p(x)=(x-1)(x+1)(x^2+3)
x^2+3=0
x=±√-3
zeros: -2, 1 and two complex roots, ±√-3
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