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.(x - 3)² - y²

Accepted Answer

Hello, everyone. The following expression is a special product called the difference of two squares. We're going to factorize it. The expression we're given is open parentheses. P minus four, closed parentheses squared minus V squared. Our answer choices are a open parentheses. P minus V minus two closed parentheses multiplied by open parentheses. P minus V plus two closed parentheses. B open parentheses, P minus V minus two closed parentheses multiplied by open parentheses. P plus V minus two closed parentheses. C open parentheses, P minus V minus four, closed parentheses multiplied by open parentheses. P plus V minus four closed parentheses. D open parentheses, P minus V minus four, closed parentheses multiplied by open parentheses. P minus V plus four, closed parentheses. Going back to my original expression, open parentheses, P minus V, I'm sorry, P minus four closed parentheses squared minus V squared. We are told we have the difference of two squares. In this case, the parentheses P -4, we're going to treat as one term. We're not gonna break them up until after we're done. Factoring the difference of two squares. Recall that when we factor the difference of two squares, we start with one binomial and they'll be factored into two binomials. So I open two sets of parentheses that are directly next to each other. The first term in each of my new binomials is the square root of the first term from my original binomial. So this would be the square root of the parentheses P - squared. This will leave us with P -4 and I'm gonna write it in parentheses still in both of my binomials. So the parentheses P -4 is the first term in each set of binomials. The second term in my new binomials is from the square root of the second term in my original binomial. So the square root of V squared is V. That is the second term in each set of parentheses. And then I recall that I need to have a plus sign between one set of the terms and a minus sign between the other set. So right now, my factorization is open parentheses. Then in parentheses, P minus four, closed parentheses plus V closed parentheses multiplied by the parentheses. P minus four, closed parentheses, minus V closed parentheses. The one last thing I want to do is see if there's anything I can distribute into my parentheses or if there's any other ways to simplify this. Since in neither of my binomials, do I have a sign in front of the parentheses? P -4? I can just drop the parentheses on P -4. So I'm going to rewrite this without the inner parentheses. So I open parentheses, P minus four plus V closed parentheses, multiplied by open parentheses. P -4 -3 closed parentheses. And the last thing I'm gonna do is rearrange the values inside the parentheses so that my variables are in alphabetical order and my constant is the last term. So open parentheses, so be P plus v - multiplied by the parentheses. P -V -4 closed parentheses. This cannot be factored any farther. This is our solution and recall that multiplication, it doesn't matter which order our factors are in as long as they're both there. So this does match answer choice C which has open parentheses. P minus V minus four closed parentheses multiplied by open parentheses. P plus V minus four have a nice day.

Struggling with College Algebra?

In Exercises 1–22, factor each difference of two squares. Assume that any variable exponents represent whole numbers.(x - 3)² - y²

Watch next

Factoring Polynomials practice set