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What is the easy way to solve the problem?

The sum of the solutions to $6x^3+7x^2-x-2=0$ is: $$A) \ \frac{1}{6}$$ $$B) \ \frac{1}{3}$$ $$C) \ \frac{-7}{6}$$ $$D) -2$$ $$E) \text{ none of above}$$

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    https://en.wikipedia.org/wiki/Vieta's_formulas – 2017-02-22

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For a cubic equation of the form $ax^3 + bx^2 + cx +d$, the sum of all the roots is given by $\frac{-b}{a}$. Here, $b=7,a=6$ so sum of the roots $= \frac{-7}{6}$.

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We could factor the cubic polynomial as $6(x-x_1)(x-x_2)(x-x_3)$.

Then, note that the coefficient on the quadratic term is $-6(x_1+x_2+x_3)=7$

Hence, $x_1+x_2+x_3=-\frac 76$.

This result can be generalized for an $n$'th order polynomial and is part of Vieta's Foumulae as referenced in a comment by @Vadim123.

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    @ClaudeLeibovici Thank you Claude! Comment allez vous ce soir? – 2017-02-22
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    @ClaudeLeibovici Yes, it is. It's 11:32 p.m. and yesterday for you. – 2017-02-22
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Constant is -2. Using remainder theorem So you have 1,-1,2,-2.

By putting-1 in x you get to know that -1 satisfies the solution.

So -1is a solution and x+1 is a factor of above equation. After dividing 6x^2+7x^2-x-2 by x+1 we get,

6x^2 +x-2 as quotient.

Solving this quadratic equation you get 1/2 and -2/3 as solutions.

So -1 + 1/2 + -2/3= -7/6 none of the above.