Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
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Exercise 16 Page 539

Try to look for a pair of numbers that when multiplied gives the constant term, and when added is equal to the linear term of the expression.

(x-8)(x-9)

Practice makes perfect
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term. x^2+17x+72 In this case we have 72. 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 72 72
-1 and -72 72
2 and 36 72
-2 and -36 72
3 and 24 72
-3 and -24 72
4 and 18 72
-4 and -18 72
6 and 12 72
-6 and -12 72
8 and 9 72
-8 and -9 72

Next, let's consider the coefficient of the linear term. x^2+17x+72 For this term we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 17. Since 17 > 0, we will be considering sums of postive numbers only.

Factors Sum of Factors
1 and 72 73
2 and 36 38
3 and 24 27
4 and 18 22
6 and 12 18
8 and 9 17
We found the factors whose product is 72 and whose sum is 17. x^2+17x+72 ⇔ (x-8)(x-9)

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.
(x-8) (x-9)
x (x-9) - 8 (x-9)
x^2-9x - 8 (x-9)
x^2-9x-8x+72
x^2-17x+72
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!