There are many ways to solve these types of problems but this is a special case.
First, we must pull out of the equation any common factors among the terms. Each term in this case has an x2 factor that we can pull out (x2*x2-16*x2). This gives us a quadratic equation that’s more organized: x2 (x2-16). Now we can put the x2 to the side and work with the x2-16. As you can see, this quadratic equation has only two terms. This puts it into a category of special quad equations that have a solution that looks like this: (x-a)(x+a) where a is a factor of a perfect square. When using FOIL, the –ax and +ax terms cancel and thus we achieve an equation such as the one we are working with of the form x2±b where b is the product of the two a’s (a perfect square). Our b is 16 and its factors (a) are 4 and 4. Using this fact we split x2-16 into (x-4)(x+4).
We now go back and put everything we have together and our result should look like this:
All that’s left to do is to “T off” the terms into three groups: x2, (x-4), and (x+4). We set each one equal to 0 and solve for x. The reason we do this is because if one of these terms is equal to 0, then the others will be too because anything multiplied by 0 is 0 (i.e. if (x-4) is zero, since each term is connected by multiplication, we multiply that with the other terms and get 0).
So we get three answers:
x2=0 when x=0
x-4=0 when x=4
x+4=0 when x=-4
All three of our x values, 0, 4 and -4, are real and are therefore the answers. If you’re unsure that these are the answers, plug them back into the original equation and solve.
Checking never hurt anybody and will eliminate any doubts you may have. Hope this helped!