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Find all rational zeros of f. Then​ (if necessary) use the depressed equation to find all roots of the equation

​f(x)equals=0.

6x^3-7x^2-6x-1=0

 

So lost on how to do this thought i had it but will not be able to finish from any steps and got all of them wrong on test. the answer for this would be appreciated thank you so much

 Mar 16, 2020
 #1
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Find all rational zeros of f. Then​ (if necessary) use the depressed equation to find all roots of the equation

​f(x)equals=0.

f(x)=6x^3-7x^2-6x-1=0

 

Hello mharrigan920!

 

\(6x^3-7x^2-6x-1=0\)

\(x^3-\frac{7}{6}x^2-x-\frac{1}{6}=0\) 

\(x=-\frac{1}{3}\ ?\)      supposed (graph).

 

\((x^3-\frac{7}{6}x^2-x-\frac{1}{6}):(x+\frac{1}{3})\) = \(x^2-\frac{9}{6}x-\frac{1}{2}\)

\(\ \underline{x^3+\frac{1}{3}x^2}\)

       \(-\frac{9}{6}x^2\ -x\)

       \(\underline{-\frac{9}{6}x^2-\frac{1}{2}x}\)

                   \(-\frac{1}{2}x-\frac{1}{6}\)

                   \(\underline{-\frac{1}{2}x-\frac{1}{6}}\)

                    \(- - - -\)

\(x_1=-\frac{1}{3}\)

 

\(x^2-\frac{3}{2}x-\frac{1}{2}=0\\ x=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ x=\frac{3}{4}\pm \sqrt{\frac{9}{16}+\frac{1}{2}}\)

\(x=\frac{3}{4}\pm \sqrt{\frac{17}{16}}\)

\(x_2=\frac{3-\sqrt{17}}{4}\approx -0,280776\\ x_2=\frac{3+\sqrt{17}}{4}\approx 1,780776\)

laugh  !

 Mar 17, 2020

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