Factor each polynomial. See Examples 5 and 6.

Question

27 z^{9} + 64 y^{12}

Accepted Answer

Hello, today we're going to be factoring the given expression. So what we are given is 88 to the power of 12 plus 25 B to the power of 15. Now there's a couple of changes that we can make to this expression. First, we know that eight is equal to two cubed. So we can replace eight with two cubed. Next, we can represent A to the power of 12 as A to the fourth cubed. Next 125 can be rewritten as five cubed and finally be to the power of 15 can be rewritten as B to the power of five cubed. So if we make these substitution to the given expression, what we end up with is two cubed times A to the power of four to the power of three plus five cubed times B to the power of five cubed. Now we can go ahead and combine these terms together by using the role of exponential distribution. And that states that if you have eight times B to the power of three, for instance, you can go ahead and distribute this exponent to each term in the product that's going to give you a cubed times B cubed. If we use this roll backwards, what we can do is combine two cubed and A to the power four cubed into two times eight to the power of four cubed. And if we use this roll backwards on the other quantity, five cubed times speed to the power of five cubed, we can combine that together to give us five B to the power of five cubed. So that's gonna be the simplified form of the expression. But notice that the expression is now a sum of cubed terms, we have a special factor for the sum of cubed terms. And that special factor states that if you have X cubed plus Y cubed, this is going to factor two, X plus Y times the quantity X squared minus X times Y plus Y squared. Here, we're going to let X equal to two times eight to the power of four. And we're going to let Y equal to five times B to the power of five. So if we plug in these values to our special factor, what we end up with is X plus Y, which is going to be to A, to the power of four plus five B to the fifth times the quantity X squared, which is going to be to A, to the power of four squared minus X times Y which is to A, to the power of four times five B to the power of five plus Y squared, which is going to be five B to the fifth squared. So there's a bit of simplification that we can go ahead and do with this quantity here as to A to the fourth plus five B to the fifth is already in its simplest form. First, let's go ahead and evaluate two times A to the fourth squared. And using our rule for exponential multiple distribution, we can distribute this to, to every term in the product. And this is going to give us two squared times A to the power of four times two, two squared is going to give us four and eight to the power of four times two is going to give us eight to the power of eight. So two times eight of the fourth squared can be replaced with four times eight to the power of eight. Next, let's go ahead and multiply to eight, the fourth with five B to the fifth. We're going to multiply two and 5 together and eight to the power of four and B to the power of fifth cannot be combined together. So we're going to represent it as A to the fourth times B to the fifth. And finally, two times five is going to give us 10. So this simplifies to 10, 8 to the fourth B to the fifth. Finally, let's simplify five B to the fifth squared. So again, we're going to use our exponential distribution and we can distribute this exponent to every term in the product that's going to give us five squared times B to the power of five times two and five squared is going to give us 25 B to the power five times two is going to give us B to the 10th. So now let's go ahead and, and distribute or substitute these three values into our factor. That's going to give us to A to the fourth plus five B to the fifth times to A to the fourth squared, which is +48 to the eighth minus to A to the fourth times five B to the fifth, which is going to be 10 A to the fourth B to the fifth plus five B to the fifth squared, which is going to be 25 B to the 10th. And this is going to be the completely factored version of the original expression. And with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to factor this expression. And I'll go ahead and see you all in the next video.

Factor each polynomial. See Examples 5 and 6. $27 z^{9} + 64 y^{12}$

Key Concepts

Sum of Cubes Formula

Identifying Perfect Powers

Polynomial Factoring Techniques

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