First we should observe that the given radicand is factored, so we don’t need to worry about our step 0 of factoring. To actually simplify the radical we want to group each of the (multiplicative) terms of the radicand into “perfect cube” pieces (since the root-value is , just like we did in the numeric case), and then the leftover “remainder” piece (just like we did with the prime factors in the numeric radical instance.)
The good news is that the simplification process is the same as the numeric examples we’ve seen (albeit with a looser definition of ‘prime factor’). It should not be a total surprise that the factored form treats each (multiplicative) factor as a ‘prime factor’ in terms of simplifying the radical, similar to how we simplify numeric radicals. Indeed, we’ve already seen that these kinds algebraic factors (like “”) are analogous to ‘prime-numbers’ when it comes to decomposing polynomials to their ‘most basic parts’, so it shouldn’t be surprising that they end up being treated similarly in other situations (like radicals) as well!
Unfortunately, the bad news is that the result isn’t quite as straightforward as in the case of the numeric radical case. To motivate the next part, we will first proceed with an example.
This example seems straight forward. According to our rules of simplifying it seems like . Indeed, if we do that then we get as an answer and, plugging that back in to the original equation, we get . So far so good right? But what about ? If we plug that in we will also get , so is irrelevantis also a solutionworks, but isn’t a solution since square roots are only positive . So, why didn’t our method give us this answer as well?
If we look closely at the previous example it may become apparent that the reason that is also a solution, is that the in the original equation kills the negative sign. Thus, since is a solution, and the even power nullifies the effect of the negative sign, then is also a solution (that is to say; since and , there is no difference between the two once we square them). But understanding why the solution is valid is not quite the same thing as figuring out why we didn’t find it the first time. After all if our methodology were correct, it should have given us all the valid solutions the first time around right?
So what’s the deal with even roots?
To understand why we failed to detect the solution of to , it is helpful to realize that this question is equivalent to solving the equality . At it’s heart, then, the answer to why we couldn’t detect the as a solution to comes down to the fact that a square root is intended to be a tool to isolate in the equality , but we also want the mechanism for this (the “square root” itself) to be a function.
And this immediately presents a problem. In our example of we determined that both and are solutions. So, if we use a square root to ‘undo’ the power on we get . But we know that and are both solutions. So, on the one hand, the right hand side could be either or ... and in fact it would need to be both if we want to make sure to get all the valid solutions. On the other hand we want the process of using square roots to be a function, and a function can only output a single answer. So we have a pretty big problem, either we lose half our answers, or we square rooting isn’t a function!
So, what do we do? There are a number of legitimate ways to address this issue, but it turns out mathematicians decided long ago that being a function is just way too useful to give up easily. As a result, they decided to simply pick a default answer. Since positives are generally easier to deal with than negatives, we decide by convention to have the square root return the positive valued answer.
Now, you might object at this point by saying that this doesn’t actually get around our original problem... and you’d be right! On the one hand if we relax the requirement on the square root operation to give both positive and negative values, it is no longer a function (which is all kinds of bad). On the other hand, if we only use the positive value result, we lose half of the ‘valid answers’ - which is also bad.
For the moment however, we sacrifice half our answers on the altar of function worship - meaning that we decide to make the square root (or really, any even root) to only output the positive answer. We’ll discuss how to account for the negative answers in the next segment, as it turns out there is still a way to recover the half of answers that we potentially lost by picking this “default answer” for even roots.
In the meantime however, this choice also impacts the process of simplifying algebraic radicals.
Ok... but really, what’s the deal with the even roots?
In order to simplify algebraic roots, we must treat even root values differently than odd root values. For even root values, when we try to simplify a term by evaluating the root, we can represent the fact that the output is only the positive value by using another function that only outputs positive values - the absolute value. For example, to evaluate something like we wouldn’t want to write as it is not clear that we are ‘choosing’ the positive output. Instead we write .
In fact, since this entire issue stems from the fact that a negative raised to an even power becomes positive, this comes up whenever we simplify any even root-value radicals. A good rule of thumb is that everything that is pulled out of an even radical gets an absolute value. Unfortunately, it doesn’t end there though. Consider, for example, the following: . To solve this, we can do the same method we did with algebraic radicals above, with the added step that - since the root value is 4 (and thus even), anything we pull out gets absolute values.
| Rewrite constant as prime factors. | ||
| Break apart terms as multiples of 4 | ||
| Simplify out multiples of 4, and apply absolute values. | ||
Now, the above isn’t wrong, it is true that . Unfortunately, for reasons that will be much clearer in calculus, absolute values are actually incredibly annoying to deal with - indeed, we will get a little glimpse of the annoyance when we discuss absolutely values explicitly in the future. But as a consequence, although the equality is correct, it is not considered simplified or finished.
To see why, first recall (or maybe learn, if you hadn’t seen this before) that “the product of absolute values is the absolute value of the product” - in other words, you can split up products of absolute values as so: . But with very little thought, we should be able to figure out what is - and no, it’s not a trick question... it’s just ! Importantly, this means we can simply remove the absolute value around the , because we already know what it is, so we could rewrite our equality as . But, it turns out, this isn’t the best we can do!
Remember that this whole mess occurs because raising numbers to even powers obliterates negative signs. So the inside the absolute values is having the same result - the power is even, and thus even if itself is negative, is going to be positive, and so the absolute value isn’t doing anything! This means we can actually rewrite the original problem as: . Notice however, that the absolute values were originally necessary - because the root value was even. It was only the fact that we could justify, for each term inside those absolute values, that we didn’t actually need the absolute values (because the inside was already positive) that let us then remove the absolute values afterward.
With this, we have the final idea for simplifying even root-values. When simplifying even roots, you put absolute values around everything that comes out of the radical. Once you have finished pulling everything out that you can, you then go through everything inside the absolute values and remove the absolute values from anything that you can - typically because the term must be positive for some reason (most commonly because it is being raised to an even power).
As a final demonstration, consider the following:
| Factor anything not factored yet. | ||
| Break apart factors | ||
At this point we can simplify the radical by pulling out the parts that are multiples of the root value. But now we need to note that the root-value is even, so we also need to put absolute values around the parts we pull out.
| Factor anything not factored yet. | ||
| Break apart factors | ||
| Simplify terms, apply abs values. | ||
Now, we have simplified the radical and have a correct equivalent expression. But the problem asked us to fully simplify the radical, and so we aren’t done. In particular, we still need to look at the terms inside the absolute value, and determine which can be pulled outside of the absolute value without changing anything.
In this case, we can rewrite our absolute value as: . At this point, hopefully it is clear that we can remove the absolute values around the (since ). Moreover, the term is being raised to an even power, which means it must be positive - so that tells us that . On the other hand, is being raised to an odd power, and so it isn’t necessarily positive - indeed, if we plug in in this case, we see , so can definitely be negative. With these observations we can see that the absolute value can be re-written as follows:
| Factor anything not factored yet. | ||
| Break apart factors | ||
| Simplify terms, apply abs values. | ||
| Split absolute values. | ||
| Simplifying absolute values. | ||
| Reordering Terms. | ||
So, the final answer to our problem is: