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.

p³ − pq² + p²q − q³

Accepted Answer

Hello, everyone. We've been asked to perform a complete factorization of the given expression or categorize it as a prime polynomial. The expression we are given is M cubed minus four M N squared plus M squared, N minus four N cubed. Our answer choices are a open parentheses. M plus N closed parentheses multiplied by open parentheses. M minus two N closed parentheses multiplied by open parentheses. M plus two N closed parentheses. B open parentheses, M plus N closed parentheses multiplied by open parentheses. M plus two, N closed parentheses multiplied by open parentheses. M plus two N closed parentheses. C open parentheses, M plus N closed parentheses multiplied by open parentheses. M minus two N closed parentheses multiplied by open parentheses. M minus two N closed parentheses. D the given polynomial is prime rewriting the original expression we were given M cubed minus four M N squared plus M squared, N minus four N cubed. I recall that the first thing I check when I want to do a factorization is if there is a greatest common factor looking at my four terms. Here, there is nothing that is in common with all four terms. So we can move on from that. I also recall that when I have an expression with four terms, typically I try to factor by grouping. So I'm gonna group my first two terms together in a set of parentheses. So M cubed minus four M N squared is now in a set of parentheses. And then I'm gonna group my last two terms together in a set of parentheses. So M squared, N minus four N cubed are in a second set of parentheses. Looking at my first set of parentheses, the M cubed minus four M N squared. I want to see if there is a common factor I could take out of both of those terms since both of those terms have the variable M I could take out an M from both. So I have M multiplied by open parentheses. And now I'm going to divide each of those terms by M M cubed divided by M results in M squared negative four M N squared divided by M is negative four N squared closed parentheses. Looking at my second set of parentheses. I have M squared N -4 n cubed. There's an N in both of those terms. So I'm gonna factor that out as plus N then multiply it by a set of parentheses. I will do M squared N divided by N which results in M squared. The negative four N cubed divided by N is negative four N squared. Close parentheses at this step in factorization. I have M multiplied by open parentheses. M squared minus four M N squared, closed parentheses plus N multiplied by open parentheses. M squared minus four N squared, closed parentheses. I have now created a binomial greatest common factor because each term now has the binomial M squared minus four N squared in common. So I'm gonna factor that out. I'm gonna pull it to the front still in parentheses. M squared minus four N squared, closed parentheses. And I'm gonna multiply that by another set of parentheses in which I'm gonna place M plus N. So the G C F S we had from our first step are now grouped together in another set of parentheses. I want to see if this can be factored any further. I'm noticing that M squared minus four N squared is the difference of two squares. So I'm gonna factor that. I recall that I will open two sets of parentheses when I have the difference of two squares. And this will be two new binomials. So the first term in my new binomials comes from the square root of the first term in my difference of two squares. So the square root of M squared, which is M is the first term in my new binomials. The second term in those binomials comes from the square root of the second term in the difference of two squares. So the square root of four N squared, the squared of four is two square root of N squared is N. So two N is the second term in each binomial. One of them needs a plus sign and one of them needs a minus sign. And then I also still need to rewrite the parentheses M plus N that we had from the previous step. Currently, I have this completely factored about as much as I can. Um It does not directly match any of my answer choices at this point. What I have is open parentheses. M plus two N closed parentheses multiplied by open parentheses. M minus two N closed parentheses multiplied by open parentheses. M plus N closed parentheses recall that with multiplication, you can rearrange your factors and when I rearrange my factors, so I have parentheses, M plus N closed parentheses multiplied by the parentheses. M minus two N closed parentheses multiplied by the parentheses. M plus two N closed parentheses. It now matches answer choice. A have a nice day.

In Exercises 1–68, factor completely, or state that the polynomial is prime. p³ − pq² + p²q − q³

Key Concepts

Factoring Polynomials

Grouping Method

Prime Polynomials

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