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