8w + 32 = w (w - 2)(w + 4) ?

8w + 32 = w(w - 2)(w + 4)
8w + 32 = w(w^2 + 2w - 8)
8w + 32 = w^3 + 2w^2 - 8w
w^3 + 2w^2 - 8w - 8w - 32 = 0
w^3 + 2w^2 - 16w - 32 = 0
(w^3 + 2w^2) - (16w + 32) = 0
w^2(w + 2) - 16(w + 2) = 0
(w + 2)(w^2 - 16) = 0
(w + 2)(w^2 - 4^2) = 0
(w + 2)(w + 4)(w - 4) = 0

w + 2 = 0
w = -2

w + 4 = 0
w = -4

w - 4 = 0
w = 4

∴ w = -2 , ±4

Start by noting that the equation contains a cubic term, so by the fundamental theorem of algebra there will be three roots. The use of w traditionally can signify a complex number.

This problem is easily simplified by noting that a factor (w+4) is present on both sides. When you set this to zero, LHS = RHS = 0 and so (w+4) = 0 is a solution.

Once that simplification is made, you can cancel (w+4) from both sides for all values of w except w = -4 (cancelling zero with zero is a no-no)

What's left is a quadratic, and it can be solved by a formula I'm sure you know. Remember to put ²√-1 = i if it happens to show up!

Presuming you wish to solve for w.......

8w + 32 = w (w - 2)(w + 4) Note that the left-hand side = 8(w+4), so

8(w + 4) = w (w - 2)(w + 4) Divide across by (w +4), noting that w = -4 is one solution

8 = w^2 -2w

w^2 - 2w - 8 = 0

(w - 4)(w + 2) = 0

The other two solutions are w = 4 and w = -2

First, factor the left side:

8(w + 4) = w (w - 2)(w + 4)

Now, divide both sides by (w + 4)

8(w + 4) = w (w - 2)(w + 4)
-----------.....----------------------
..(w + 4)..........(w+4)

The (w+4) cancel out on both sides so you now have:

8 = w(w - 2)

Multiply the left side:
8 = w^2 - 2w

Bring the 8 to the right side
0 = w^2 - 2w - 8

Factor the right side
0 = (w - 4)(w + 2)

set each () equal to 0 and solve for w
w - 4 = 0.....w + 2 = 0
w = 4...........w = -2

Since we cancelled out the (w + 4) we also need to set that to 0:
w + 4 = 0
w = -4

Solution: -4, 4 and -2

-3 or 11
use the quadratic formula