In Exercises 133–136, factor each polynomial completely. Assume t...

Question

In Exercises 133–136, factor each polynomial completely. Assume that any variable exponents represent whole numbers.x³ⁿ + y¹²ⁿ

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Welcome back. I am so glad you're here. We are asked to perform complete factorization of the following polynomial expression. We are given N raised to the three P plus M raised to the 15 P. And our answer choices are answer choice. A open parentheses, N raised to the P plus M raised to the five P closed parentheses multiplied by open parentheses. N raised to the two P minus N raised to the PM. Raised to the five P plus M raised to the 10 P parentheses. Answer choice B open parentheses, N raise to the P minus M. Raise to the five P. Close parentheses multiplied by open parentheses. N raised to the two P plus N raised to the PM. Raised to the five P plus M raised to the 10 P closed parentheses. Answer choice C open parentheses. N raise to the P plus M. Raise to the five P closed parentheses multiplied by open parentheses and raised to the two P minus N raised to the PM. Raise to the five P plus M raised to the two P. Close parentheses. In answer choice, D open parentheses, N raised to the P minus M raise to the five P closed parentheses, multiplied by open parentheses. N raise to the two P plus N raised to the PM. Raised to the five P plus M raised to the two P closed parentheses. All right. Well, the first step when factoring any polynomial is that we look at our terms and see if they have any factors in common That we can factor out one variables in N1 variables in M. There are not any factors in common. The next thing we do is because we have two terms here that we're being asked to factor is we look, and we see if we have any situations such as a difference of squares or a difference of cubes or a sum of cubes, we're adding these two terms together. So the only possible of those three is that it could be a sum of cubes. So now we look at our terms individually and see if they are, in fact, cubes. Well, the first term here N raise to the three P. Well, that's essentially N raised to the P multiplied by N raise to the P multiplied by N raise to the P because we would add these three PS together. And that would give us N raises to the three P where we could put this in other terms, N raises to the P cubed. So this one is a cube. Now let's take a look at the second term, this M raised to the 15 P. Well, I'm not going to write out 15 M raised to the P S, but we could also write this one as M raised to the five P cubed because in this case, we would take the five P race to the higher power, the three multiply the five P and three and we would have 15 P, which is our original exponent. So this is a sum of cubes. Now, we recall from previous lessons that when we have a sum of cubes, which could be written as open parentheses. A cubed plus B cubed, closed parentheses. This can be factored as open parentheses. A plus B closed parentheses multiplied by open parentheses. A squared minus A B plus B squared closed parentheses. And essentially, all we need to do is identify our A and B terms and plug those in to our factored sum of cubes. All right. So our, a term that's the first term being cubed here. So our A cubed, our A term, excuse me, we're looking for a, our A term is going to be this N raised to the P and for our B cubed looking for our B term, that's the second one being cubed. And that's going to be in this case, M raised to the five P. So now everywhere there's an A in our factored sum of cubes, we are going to fill in N rays to the P and everywhere there's A B we're going to fill in M raised to the five P. So let's work on doing that. We have open parentheses no longer A but now N raised to the P, we have plus no longer B but now M raised to the five P closed parentheses multiplied by open parentheses. Now it's instead of A, it'll be N raised to the P squared minus instead of A, it's N raised to the P multiplied by, instead of B, it's M raised to the five P and then plus instead of B squared, it is M raised to the five P squared and closed parentheses. Now there's just a little bit of simplification that needs to happen. The two terms that need to be squared. We can do those. We're raising the, for the NRA to the P squared. We're taking a power race to a power. So we multiply those P multiplied by two will give us two P. And for this M race to the five P squared, we're again multiplying our exponents. So we have five P multiplied by two which will be 10 P. And so now we can write this final answer is open parentheses. N raised to the P plus M raised to the five P closed parentheses, multiplied by open parentheses. N raised to the two P minus N raised to the PM, raised to the five P plus M raised to the 10 P close parentheses. Now, looking at our answer choices, this matches with answer choice. A well done. We'll catch you on the next one.

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In Exercises 133–136, factor each polynomial completely. Assume that any variable exponents represent whole numbers.x³ⁿ + y¹²ⁿ

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