CONCEPT PREVIEW Work each problem.

Match each polynomial in Colu...

Question

CONCEPT PREVIEW Work each problem.

Match each polynomial in Column I with its factored form in Column II.

I                                              II
a. x² + 10xy + 25y²               A. (x + 5y) (x - 5y)
b. x² - 10xy + 25y²                B. (x + 5y)²
c. x² - 25y²                            C. (x - 5y)²
d. 25y² - x²                            D. (5y + x) (5y - x)

Accepted Answer

Hello, everyone. We are asked to factorize the following polynomial expression. Four M squared plus 28 MK plus K squared. Our answer choices are a, the parentheses, two M minus seven K, all of it squared. B the parentheses, two M plus seven K closed parentheses squared, C open parentheses. Four M plus seven K, close parentheses multiplied by the parentheses. M plus seven K D, the parentheses, four M plus 49 K multiplied by the parentheses M plus K. So looking at our expression, we again have four M squared plus MK plus 49 K squared. There's a few different ways you could approach this problem. I'm gonna approach it by separating my middle term into two terms. So I could separate this and rewrite this expression as four M squared plus MK plus 14 MK plus 49 K squared. I do this so I can factor by grouping. So I'm gonna group my first two terms together in parentheses. So four M squared plus 14 MK and group my second set of terms together 14 MK plus 49 K squared. I'm gonna see if I can end up with a common factor in my first set of parentheses. I noticed that I can factor out a two and an M from each term. So four M squared divided by two M is two M 14 MK divided by two M be a positive seven K closed parentheses. Looking at my second grouping, 14 MK plus 49 K squared. I noticed they have a common number of seven. So they, I'm gonna factor seven out as a common factor and they each have a K. So seven K is gonna be factored out as a GCF. And what's left in my parentheses will be 14 MK divided by seven K, which is two M plus K squared, divided by seven K will give me seven K. And luckily enough, my groupings, I do have a common binomial factor. So I'm going to rewrite this where I take my gcfs and put them in one set of parentheses. So two M plus seven and multiply it by the parentheses that both terms had in common, which is two M plus seven K. Oops, I'm sorry. My first set of parentheses also has a seven K noticing that these are identical. I can rewrite this as a binomial squared. So two M plus seven K closed parentheses squared. And when I check that matches answer choice B have a nice day.

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CONCEPT PREVIEW Work each problem. Match each polynomial in Column I with its factored form in Column II. I II a. x² + 10xy + 25y² A. (x + 5y) (x - 5y) b. x² - 10xy + 25y² B. (x + 5y)² c. x² - 25y² C. (x - 5y)² d. 25y² - x² D. (5y + x) (5y - x)