(x-2)*(x^2+8x+16)=7(4+x)

To solve the equation \((x-2)(x^2+8x+16) = 7(4+x)\), we can start by expanding both sides.

First, let's expand the left side:

\[ (x-2)(x^2+8x+16) \]

This can be done by distributing \(x-2\) across \(x^2 + 8x + 16\):

\[ = x(x^2 + 8x + 16) - 2(x^2 + 8x + 16) \] \[ = (x^3 + 8x^2 + 16x) - (2x^2 + 16x + 32) \] \[ = x^3 + 8x^2 + 16x - 2x^2 - 16x - 32 \] \[ = x^3 + (8x^2 - 2x^2) + (16x - 16x) - 32 \] \[ = x^3 + 6x^2 - 32 \]

Now, let's expand the right side:

\[ 7(4+x) = 28 + 7x \]

Now we can set the expanded forms equal to each other:

\[ x^3 + 6x^2 - 32 = 28 + 7x \]

Next, we'll move all terms to one side of the equation:

\[ x^3 + 6x^2 - 7x - 32 - 28 = 0 \] \[ x^3 + 6x^2 - 7x - 60 = 0 \]

Now we have the polynomial equation \(x^3 + 6x^2 - 7x - 60 = 0\).

To solve this cubic equation, we can try to find rational roots using the Rational Root Theorem. Possible rational roots are the factors of \(-60\). We can test \(x = 2\):

\[ 2^3 + 6(2^2) - 7(2) - 60 = 8 + 24 - 14 - 60 = -42 \quad \text{(not a root)} \]

Let's try \(x = 3\):

\[ 3^3 + 6(3^2) - 7(3) - 60 = 27 + 54 - 21 - 60 = 0 \quad \text{(is a root)} \]

Since \(x = 3\) is a root, we can factor \(x - 3\) from the cubic polynomial. We can perform synthetic division of \(x^3 + 6x^2 - 7x - 60\) by \(x - 3\):

Using synthetic division:

3 |  1   6  -7  -60
   |      3  27   60
   -------------------
     1   9  20    0

The quotient is \(x^2 + 9x + 20\). Now factor \(x^2 + 9x + 20\):

\[ x^2 + 9x + 20 = 0 \] \[ (x + 4)(x + 5) = 0 \]

Therefore, \(x + 4 = 0\) or \(x + 5 = 0\). This gives us the roots:

\[ x = -4 \quad \text{and} \quad x = -5 \]

Thus, the complete set of solutions is:

\[ x = 3, \quad x = -4, \quad x = -5 \]

So the final solutions to the equation \((x-2)(x^2 + 8x + 16) = 7(4+x)\) are:

\[ \boxed{3, -4, -5} \]