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11) $2x^{3}-3x^{2}-18x + 27 = 0$

Question

  1. $2x^{3}-3x^{2}-18x + 27 = 0$

Explanation:

Step1: Group the terms

$(2x^{3}-3x^{2})-(18x - 27)=0$

Step2: Factor out common factors from each group

$x^{2}(2x - 3)-9(2x - 3)=0$

Step3: Factor out the common binomial factor

$(2x - 3)(x^{2}-9)=0$

Step4: Factor the difference - of - squares

$(2x - 3)(x + 3)(x - 3)=0$

Step5: Set each factor equal to zero and solve for x

If $2x-3=0$, then $2x=3$, $x=\frac{3}{2}$;
If $x + 3=0$, then $x=-3$;
If $x - 3=0$, then $x=3$.

Answer:

$x=\frac{3}{2},x=-3,x = 3$