Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Mid-Chapter Quiz
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Exercise 15 Page 224

Think of the process as multiplying two binomials in reverse.

(n-5)^2

Practice makes perfect
We are asked to factor a trinomial. Before we do that, let's rewrite it in standard form, in which the terms are ordered from highest exponent to lowest. - 10n+25+n^2 ⇕ n^2-10n+25 To factor a trinomial with a leading coefficient of one, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term. n^2-10n+ 25

In this case, we have 25. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative).

Factor Constants Product of Constants
1 and 25 25
- 1 and - 25 25
5 and 5 25
- 5 and - 5 25

Next, let's consider the coefficient of the linear term. n^2-10n+ 25 ⇕ n^2+( - 10n)+ 25 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 10.

Factors Sum of Factors
1 and 25 26
- 1 and - 25 - 26
5 and 5 10
- 5 and - 5 - 10
We found the factors whose product is 25 and whose sum is - 10. n^2+( - 10n)+ 25 ⇔ (n-5)(n-5) Since both expressions inside parentheses are the same, we can write our answer using an exponent. (n-5)(n-5) ⇔ (n-5)^2

Checking Our Answer

Check your answer âś“
We can check our answer by applying the Distributive Property and comparing the result with the given expression.
(n-5)^2
(n-5)(n-5)
n(n-5)-5(n-5)
n^2-5n-5(n-5)
n^2-5n-5n+25
n^2-10n+25
- 10n+25+n^2
After applying the Distributive Property and simplifying, the result is the same as the given expression. Therefore, we can be sure our solution is correct!