In Exercises 1–68, factor completely, or state that the polynomia...

Question

In Exercises 1–68, factor completely, or state that the polynomial is prime. a²(x − y) + 4(y − x)

Accepted Answer

Hello, today we're gonna be completely factorizing the given expression. So what we are given is P squared multiplied by the quantity M minus N plus 81 multiplied by the quantity N minus M. Now, this form looks very similar to factoring a polynomial using factoring by grouping. However, we have the group M minus N and we have the group N minus M. Well, we can make these two groups look similar to each other by reordering the terms in the second group. See N minus M can be reordered to give us negative M plus N. And if we factor out a negative one from this quantity, what we have is negative one times the quantity M minus N which is going to match the group on the left hand side of the expression. So if we reorder the right hand group and factor out a negative one, we can rewrite the expression as P squared times the quantity M minus M plus negative one times times the quantity M minus N and negative one times 81 is going to give us negative 81. So we can go ahead and rewrite this portion of the expression as negative 81 multiplied by the quantity M minus N. Now what we have are two groups of M minus N. We can go ahead and combine this into a single group by adding the two coefficients together. And doing this is going to give us P squared minus multiplied by the quantity M minus N. Now this can still be factored. See here P squared can be rewritten as P times P or in other words, the quantity P raised to the power of two and 81 is a perfect square as it can be rewritten as nine times nine. Or in other words, it can be rewritten as nine squared. So what we have with P squared minus 81 is the difference of squares and the difference of square states that if you have the binomial in the form of A squared minus B squared, then this can factor to A plus B times A minus B. Here, we're going to allow A to equal to P and we're going to allow B to equal to nine. And so if we factor this left-hand portion using the difference of squares, we can factor the expression to give us P plus nine times P minus nine times the quantity M minus N. And this is going to be the completely factored version of the Gibbon expression. And with that being said, the answer to this problem is going to be a So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

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In Exercises 1–68, factor completely, or state that the polynomial is prime. a²(x − y) + 4(y − x)

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