Question 239120
Whether you are working with just plain numbers or with expressions, you find LCM's the same way: You factor the numbers or expressions and use those factors to find the LCM.<br>
So we start by factoring and you always start factoring with the Greatest Common Factor (GCF). The GCF of {{{z^3+4z^2+4z}}} is z so we can factor it into:
{{{z^3+4z^2+4z = z(z^2+4z+4)}}}
Then the second factor fits the perfect square trinomial pattern, {{{a^2 + 2ab + b^2 = (a+b)^2}}}, where the "a" is "z" and the "b" is "2". So it will factor into {{{(z+2)^2}}}:
{{{z^3+4z^2+4z = z(z+2)^2}}}<br>
Now we'll factor {{{z^2-4z}}}. (Was this supposed to be {{{z^3-4z}}}? If so, the problem is a bit more interesting. Just in case, I'll solve using this, too, at the end.) Again we start with the GCF (which is z again):
{{{z^2-4z = z(z-4)}}}
And this is all the factoring we can do.<br>So we have
{{{z^3+4z^2+4z = z(z+2)^2}}}
and
{{{z^2-4z = z(z-4)}}}
To find the LCM you find the product of <i>every</i> different factor. In the case of any common factors, use the higher exponent on those factors. Between these two factored expressions, we have 3 different factors: z, z+2 and z-4. z is a factor common to both. In both factored expressions, the exponent on z is 1 so the "higher" exponent is 1. So the LCM is
{{{z(z+2)^2(z-4)}}}
If your answer is supposed to be simplified, you'll need to multiply this out.<br>
In case the second expression was supposed to be {{{z^3-4z}}}, then it factors this way:
{{{z^3-4z = z(z^2 - 4)}}}
The second factor is a difference of squares so it factors:
{{{z^3-4z = z(z+2)(z-2)}}}
So the LCM between 
{{{z^3+4z^2+4z = z(z+2)^2}}}
and
{{{z^3-4z = z(z+2)(z-2)}}}
would be
{{{z(z+2)^2(z-2)}}}
(Note how, with the z+2 factor, we used the higher exponent in the LCM).