6. $$ 2 x^{3}-x^{2}-162 x+81 $$

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To factor the cubic polynomial

2x³ – x² – 162x + 81

we can first try factoring by grouping.

Step 1. Group the terms:

(2x³ – x²) + (–162x + 81)

Step 2. Factor out the common factors in each group:

First group: Factor out x²:
    x²(2x – 1)

Second group: Factor out –81:
    –81(2x – 1)

Step 3. Now, notice that both groups contain the common factor (2x – 1):

(2x – 1)[x² – 81]

Step 4. Recognize that x² – 81 is a difference of two squares:

x² – 81 = (x – 9)(x + 9)

Thus, the completely factored form is:

(2x – 1)(x – 9)(x + 9)

If you need the roots, set each factor equal to zero:

2x – 1 = 0  ⟹  x = 1/2
  x – 9 = 0  ⟹  x = 9
  x + 9 = 0  ⟹  x = –9

So, the zeros of the polynomial are x = 1/2, x = 9, and x = –9.
The polynomial $$ 2x^{3} - x^{2} - 162x + 81 $$ factors to $$ (2x - 1)(x - 9)(x + 9) $$. The roots are $$ x = \frac{1}{2} $$, $$ x = 9 $$, and $$ x = -9 $$.

6. \( 2 x^{3}-x^{2}-162 x+81 \)

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