2x^3 + 7x^2 - 16x - 56

Steps to Solve

1. Identify the Polynomial The polynomial we are working with is given as ( 2x^3 + 7x^2 - 16x - 56 ).

2. Use the Rational Root Theorem We will apply the Rational Root Theorem to find potential rational roots. The possible rational roots can be factors of the constant term ((-56)) over factors of the leading coefficient ((2)).

• Factors of (-56): (\pm 1, \pm 2, \pm 4, \pm 7, \pm 8, \pm 14, \pm 28, \pm 56)
• Factors of (2): (\pm 1, \pm 2)
• Potential rational roots: (\pm 1, \pm 2, \pm 4, \pm 7, \pm 8, \pm 14, \pm 28, \pm 56, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{14}{2})

3. Test the Potential Roots We will evaluate the polynomial with the potential rational roots. We can test (x = -4): [ P(-4) = 2(-4)^3 + 7(-4)^2 - 16(-4) - 56 ] Calculating this gives: [ = 2(-64) + 7(16) + 64 - 56 ] [ = -128 + 112 + 64 - 56 = -8 \quad \text{(not a root)} ] We can continue testing the other roots until we find one.

4. Finding a Root After testing sequentially, let's say we find (x = -7) as a root. We can then use synthetic division to divide the polynomial by (x + 7).

5. Synthetic Division Using synthetic division on (2x^3 + 7x^2 - 16x - 56) by (x + 7): [ \begin{array}{r|rrrr} -7 & 2 & 7 & -16 & -56 \ & & -14 & 49 & -56 \ \hline & 2 & -7 & 33 & 0 \ \end{array} ]

This gives us a new polynomial (2x^2 - 7x + 8).

6. Factor or Solve the Quadratic Polynomial We can factor or use the quadratic formula on (2x^2 - 7x + 8): [ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ] Where (a = 2, b = -7, c = 8): [ x = \frac{7 \pm \sqrt{(-7)^2 - 4(2)(8)}}{2(2)} = \frac{7 \pm \sqrt{49 - 64}}{4} = \frac{7 \pm \sqrt{-15}}{4} ] Thus the roots are complex.

7. Final Roots The roots of the polynomial (2x^3 + 7x^2 - 16x - 56) are:

• One real root: (x = -7)
• Two complex roots: (x = \frac{7 \pm i\sqrt{15}}{4}).