Question 469099: Use the rational zero theorem and quotient polynomials to find all roots of the given equation
12x(^4)-52x(^3)+75x^-16x-5=0
(12x to the fourth minus 52x to the third plus 75x squared minus 16x-5=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
12x(^4)-52x(^3)+75x^-16x-5=0
(12x to the fourth minus 52x to the third plus 75x squared minus 16x-5=0)
...
p(x)=12x^4-52x^3+75x^2-16x-5=0
factors of p=-5: ±1, ±5
factors of q=12: ±1, ±2, ±3, ±4, ±6, ±12
possible rational zeros (p/q): ±1, ±1/2, ±1/3, ±1/4, ±1/6, ±1/1
..............................................±5, ±5/2, ±5/3, ±5/4, ±5/6, ±5/12
Using synthetic division to find roots:
0)......12....-52.....75.....-16.....-5
1)......12....-40.....35.......19.....14
1/2)..12....-46.....52.......10......0 (root=1/2)
p(x)=(x-1/2)(12x^3-46x^2+52x+10)
Try again
0)........12....-46....52.......10
-1)......12....-58....-8.......18
-1/6)..12....-48....60........0 (root=-1/6)
p(x)=(x-1/2)(x+1/6)(12x^2-48x+60)
12x^2-48x+60=0
x^2-4x+5=0
x=[4±sqrt(16-4*1*5)]/2
x=[4±√-4]/2=2±i
roots: -1/6, 1/2, and two complex roots, 2±i
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