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Given that x=2 is a root of p(x)=x^4-3x^3-18x^2+90x-100, find the other roots (real and nonreal).

 Oct 18, 2020
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Given that x=2 is a root of p(x)=x^4-3x^3-18x^2+90x-100, find the other roots (real and nonreal).

 

Hello Guest!

 

\(x_1=2\)

\((x^4-3x^3-18x^2+90x-100)\) : \((x-2)\) \(x^3 - x^2 - 20 x + 50\)

 \(\underline{x^4-2x^3}\)

          \(-x^3-18x^2\)

         \(\underline{-x^3\ +\ 2x^2}\)

                   \(-20x^2+90x\)

                  \(\underline{-20x^2+40x}\)

                                   \(50x-100\)

                                   \(\underline{50x-100}\)

                                                   0

\(x_2=-5\)

   \((x^3 -\ x^2 - 20 x + 50)\) : \((x+5)\) =  \(x^2-6x+10\) 

    \(\underline{x^3+5x^2} \)

          \(-6x^2-20x\)

          \(\underline{-6x^2-30x} \)

                         \(10x+50\)

                         \(\underline{ 10x+50} \)

                                       0

  \(x^2-6x+10=0\) 

  \(x=3\pm \sqrt{9-10}\)

 \(x_3=3+i\\ x_4=3-i\)

laugh  !

 
 Oct 18, 2020

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