In Exercises 1–22, factor each difference of two squares. Assume ...

Question

In Exercises 1–22, factor each difference of two squares. Assume that any variable exponents represent whole numbers.

x¹⁴ - y⁴

Accepted Answer

Hello, everyone. The following expression is a special product called the difference of two squares. We're gonna factorize it. The expression we're given is U raised to the fourth power minus V raised to the 14th power. Our answer choices are a, the parentheses, U minus V multiplied by the parentheses, U plus V multiplied by the parentheses. U squared minus V raised to the 12th power B, the parentheses U squared minus V raised to the seventh power multiplied by the parentheses. U squared plus V raised to the seventh power C the parentheses U minus V multiplied by the parentheses U plus V multiplied by the parentheses U squared plus V raised to the 12th power D, the parentheses U squared minus V raised to the square root of 14, multiplied by the parentheses U squared plus V raised to the square root of 14th power. Our original expression again was you to the 4th race to the 4th power -3 raised to the 14th power. We are told these are both perfect squares. If you recall that when you have a power raised to a power, you multiplied your exponents. This means that if I wanted to, I could rewrite you squared squared and that would be the same as you raised to the fourth power. So you squared being in parentheses raised to another square is equivalent to you raised to the fourth power. I can do the same thing with the V. If I know that raising a power to a power means I multiply my exponents, I could say, OK, if I want to break this into two exponents, I could do 14 divided by two. So that would give me in parentheses. V raise to the 7th power closed parentheses squared. Now I can see my two squares a little clearer. I knew it was the difference of two squares. But now I can see the squares recall that when you do factor the difference of two squares, you're going to end up with two binomials. So I'm opening two sets of parentheses next to each other that are currently empty. And we're going to fill those as our solution. Binomials. The first term in each binomial comes from the square root of the first term in the difference of two squares. So the square root of U squared squared will be U squared. So the first term in each binomial is U squared. Next, the last term in our two new binomials is from the square root of the last term in the difference of two squares. So the square root of V raised to the seventh power squared, which would be V raise to the seventh power. So that's the last term each binomial. And I recall, I need to put opposite signs in between the terms. So one needs a plus sign and one needs a minus sign. At this step. In factorization, I have parentheses U squared minus V raised to the seventh power multiplied by the parentheses U squared plus V raised to the seventh power since seven is an odd number, it is not a perfect square. And thus, I can't do the difference of two squares again, had this been an eight, I absolutely would have. But the exponent is a seven. So I cannot, this is as factored as this expression can get. So the parentheses, U squared minus V raised to the seventh power multiplied by the parentheses U squared plus V raised to the seventh power is our solution. And answer choice B have a nice day.

In Exercises 1–22, factor each difference of two squares. Assume that any variable exponents represent whole numbers. x¹⁴ - y⁴

주요 개념

Difference of Squares

Factoring

Exponents and Powers

다음 시청