McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
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Exercise 14 Page 217

Think of the process as multiplying two binomials in reverse.

(x+6)^2

Practice makes perfect
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. x^2+12x+36 In this case, we have 36. 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.) Let's also consider the coefficient of the linear term.

x^2+12x+36 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 12. Since a sum of two negative numbers is never positive, we should only consider positive constants.

Constants Product of Constants Sum of Constants
1 and 36 36 37
2 and 18 36 20
3 and 12 36 15
4 and 9 36 13
6 and 6 36 12
Now we can write the products of the constants and variables. 6x and 6x We found the factor constants whose product is 36 and also cause the sum of the products of x and the factor constants to equal 12x. x^2+12x+36 ⇔ (x+6)(x+6) Because both parentheses include the same expression, we can write our answer using an exponent. (x+6)(x+6)=(x+6)^2

Checking Our Answer

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We can check our answer by applying the Distributive Property and comparing the result with the given expression.
(x+6)^2
x^2+12x+36
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!