Question

Find the roots of:
$1.\ &2x^3-7x^2+8x-3=0\\ 2. \ & x^3-x^2-4=0\\ 3. \ &6x^3+7x^2-9x+2=0 \end{align*}$

Answer 1

$\large\displaystyle\text{ \begin{gathered}\sf \pmb{1) \ 2x^3-7x^2+8x-3=0 } \end{gathered} }$

Synthetic division is used since the equation is of the third degree. The divisors of -3 are 1, -1, 3, +3. So:

  | 2  -7    8  -3

1 |      2   -5   3

  | 2   -5    3  0

1 |      2     -3  

    2   -3     0

So the factorization is (x-1)² (2x-3)=0. So:

                     $\bf{ x_1=x_2=1 \qquad x_2=\dfrac{3}{2} }$

$\large\displaystyle\text{ \begin{gathered}\sf \pmb{2) \ x^3-x^2-4=0 } \end{gathered} }$

Synthetic division is used since the equation is of the third degree. The divisors of -4 are 1, -1, 2, -2, 4, -4. So:

      |  1  -1  0  -4

  2  |     2  2    

         1  2  2  0

So the factorization is (x-2)(x²+x+2)=0 . When calculating the discriminant of the trinomial, it is concluded that it has no roots since the result is negative. So you only have one solution.

                   $\bf{ 1^2-4(2)(2)=1-16=-15 < 0 \quad \Longrightarrow \quad x=2 }$

$\large\displaystyle\text{ \begin{gathered}\sf \pmb{3) \ 6x^3+7x^2-9x+2=0 } \end{gathered} }$

Synthetic division is used since the equation is of the third degree. The divisors of 2 are 1, -1, 2, -2. So:

   | 6    7       9      2

-2 |      -12    10     -2

     6    -5     1       0

So the factorization is (x+2)(6x²-5x+1)=0 . The quadratic equation is solved by the general formula:

         $\bf{ x_{2, 3}&=\dfrac{5\pm \sqrt{(5)^2-4(6)(1)}}{2(6)}=\dfrac{5\pm \sqrt{25-24}}{12}=\dfrac{5\pm 1}{12} }}$

                     $\large\displaystyle\text{ \begin{gathered}\sf \begin{matrix} x_1=-2&\ \ \ \ \ \ x_{2}=\dfrac{6}{12} \qquad &\ \ \ x_3=\dfrac{4}{12}\\ &\ \ \ x_2=\dfrac{1}{2} \qquad &x_3=\dfrac{1}{3} \end{matrix} \end{gathered} }$