Question

Using long division method, show that (x–3) is a factor of 6+x–4x^2
+x^3
.​

Answer 1

Step-by-step explanation:

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Answer 2

By long division method the given polynomial can be simplified to $\frac{x^3-4x^2+x+6}{x-3}=x^2-x-2$

Step-by-step explanation:

Given that x-3 is a factor of the polynomial $6+x-4x^2+x^3$

Rewritting the given polynomial as $x^3-4x^2+x+6$

To Solve the given polynomial by long division method :

        $x^2-x-2$

       _______________

x-3  )$x^3-4x^2+x+6$

        $x^3-3x^2$

      (-)__(+)___________

              $-x^2+x$

              $-x^2+3x$

      ____(+)__(-)___________

                      -2x+6

                       -2x+6

    _______ (+)__(-)_________

                            0

   ______________________

Hence $\frac{x^3-4x^2+x+6}{x-3}=x^2-x-2$