SOLUTION: My teacher gave us these problems: Solve each equation and state each zero using the Rational Root Theorem and the Conjugate Root theorem x^3+2x^2+3x+6=0 x^4-7x^2+12=0 3x

Question 560724: My teacher gave us these problems:
Solve each equation and state each zero using the Rational Root Theorem and the Conjugate Root theorem
x^3+2x^2+3x+6=0
x^4-7x^2+12=0
3x^3-7x^2-14x+24=0
I lost the notes for all of this and this is due tomorrow, I need help ASAP
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
x^3+2x^2+3x+6=0
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x^2(x+2)+3(x+2) = 0
(x+2)(x^2+3) = 0
x = -2 or x = i*sqrt(3) or x = -i*sqrt(3)
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x^4-7x^2+12=0
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(x-3)(x-4) = 0
x = 3 or x = 4
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3x^3-7x^2-14x+24=0
I graphed and found a zero at x = 4/3
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Use synthetic division to find the quadratic factor:
4/3)....3....-7....-14....24
.........3.....-3....-18...|..0
Quadratic: 3(x^2-x-6) = 0
(x-3)(x+2) = 0
x = 3 or x = -2 or x = 4/3
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Cheers,
Stan H.
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