How do you use synthetic division to find all the rational zeroes of the function f(x)= 2x^3 - 3x^2-11x+6f(x)=2x3−3x2−11x+6?

1 Answer
Vinícius Ferraz
Aug 20, 2015

3, -2, 1/23,−2,12

Explanation:

A rational zero must be ± frac{\text{a divisor of 6}}{\text{a divisor of 2}} = ± frac{{1,2,3,6}}{{1,2}} \in ± {1,2,3,6, 1/2, 2/2, 3/2, 6/2}±a divisor of 6a divisor of 2=±{1,2,3,6}{1,2}∈±{1,2,3,6,12,22,32,62}

f(1) = 2 - 3 - 11 + 6 < 0f(1)=2−3−11+6<0

f(2) = 16 - 12 - 22 + 6 < 0f(2)=16−12−22+6<0

f(3) = 54 - 27 - 33 + 6 = 0f(3)=54−27−33+6=0

By Briot-Ruffini,
enter image source here

\frac{f(x)}{x - 3} = 2x^2 + 3x - 2 = 0f(x)x−3=2x2+3x−2=0

\Delta = 9 + 4 * 2 * 2 = 25

x = frac{- 3 ± 5}{4}

Related questions
See all questions in Synthetic Division
Impact of this question
1630 views around the world
You can reuse this answer
Creative Commons License