*Take the constant term of the divisor with the opposite sign and write it to the left.
*Write the coefficients of the dividend to the right.
Therefore, the solution is given by:
[tex]\begin{gathered} \frac{2x^3-13x^2+22x-8}{x-\frac{1}{2}}=2x^2-12x+16+\frac{0}{x-\frac{1}{2}} \\ \frac{2x^3-13x^2+22x-8}{x-\frac{1}{2}}=2x^2-12x+16 \end{gathered}[/tex]
Therefore, the other solutions can be found as follows:
[tex]2x^2-12x+16=0[/tex]
Divide both sides by 2:
[tex]x^2-6x+8=0[/tex]
The factors of 8 that sum to -6 are -2 and -4, So:
[tex]\begin{gathered} (x-2)(x-4)=0 \\ Hence\colon \\ x=2 \\ and \\ x=4 \end{gathered}[/tex]
