In Exercises 69–82, multiply using the rule for the product of th...

Question

In Exercises 69–82, multiply using the rule for the product of the sum and difference of two terms.

(1 − y⁵)(1 + y⁵)

Accepted Answer

Hey, everyone in this problem, we're asked to perform multiplication on the given expression. The expression we're given is two binomials. So we have the first one minus M to the exponent six multiplied by the second binomial one plus M to the exponent sixth. We're given four answer choices. Option A one minus two M to the exponent six plus M to the exponent 12. Option B 1 -M to the exponent 12. Option C one plus two M to the exponent six minus M to the exponent 12. In option D one plus M to the exponent 12, we're gonna start by rewriting what we were given. So the binomial one minus M to the exponent six multiplied by the binomial one plus M to the exponent six. And now we want to recognize that this is actually the difference of squares. So recall that when we have a difference of squares, if we have a binomial A minus B multiplied by the binomial A plus B, you can write this as the difference of squared A squared minus B squared. OK. So in our case, we have the left hand side, we have one minus M to the exponent six multiplied by one plus M to the exponent six. So in our case, A is equal to one and B is equal to M to the exponent six. And this difference of squares gives us a nice, easy way to write the product. So now we can write 1 -2 to the exponent six multiplied by one plus M to the exponent six S A squared. OK, we have A as one. So one squared minus B squared or B is M to the exponent six. So we have M to the exponent six squared. And this is a result of the multiplication. All that's left to do is simplify this expression. So we have one squared that's gonna give us one, we have added to the exponent six squared. OK. We call that if we have a value X to the exponent N raise to another exponent, say P this is equivalent to writing X to the exponent N multiplied by P. OK. So when we have an exponent raised to an exponent, we can multiply those exponents together. So M to the exponent 6 to the exponent two is gonna give us M to the exponent 12. Yes. So we have 1 -2 to the exponent 12. That is a result of our multiplication. And we've simplified our expression as much as we can. I wanna make a quick note here. If you didn't recognize that that was a difference of squares that's all right. OK. The, uh, the question asked you to do the multiplication and you could do the multiplication in whichever way you like to multiply by no meals. For example, you could use the foil method and you would work out to the exact same result. OK. This difference of squares is nice because it's quick and efficient. It saves you a little bit of that messy work in between where you can have signs mixed up and things like that. So it's really great to use if you recognize it. But if you didn't recognize it, that's OK too. You can work out that multiplication and you'll get the same result if we compare this to our answer choices and we found that when we perform the multiplication, we get a result of one minus M to the exponent which corresponds with answer choice. B Thanks everyone for watching. I hope this video helped see you in the next one.

In Exercises 69–82, multiply using the rule for the product of the sum and difference of two terms. (1 − y⁵)(1 + y⁵)

Key Concepts

Product of Sum and Difference

Difference of Squares

Algebraic Expressions

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