McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Using the Distributive Property
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Exercise 33 Page 40

Try to think of the greatest common factor between the coefficients and between the variables separately.

gf(5f+g+15)

Practice makes perfect

We want to find the greatest common factor (GCF) of the terms in the given expression. To do so, we will consider coefficients and variables separately. 5 gf^2+ 1 g^2f+ 15 gf Let's start by finding the GCF of 5, 1 and 15. Factors of5:& 1and5 Factors of1:& 1 Factors of15:& 1,3,5 and 15 We found that the GCF of the coefficients is 1. To find the GCF of the variables, we need to identify the variables repeated in both terms, and write them with their minimum exponents. \begin{aligned} \textbf{Factors of }\bm{1^\text{st}}\textbf{ Variable:}&\ g,f,{\color{#FF0000}{gf}},f^2,gf^2\\ \textbf{Factors of }\bm{2^\text{nd}}\textbf{ Variable:}&\ g,f,{\color{#FF0000}{gf}}, g^2,g^2f\\ \textbf{Factors of }\bm{3^\text{rd}}\textbf{ Variable:}&\ g,f,{\color{#FF0000}{gf}} \end{aligned} We see that there is one repeated variable factor, gf. Thus, the GCF of the expression is 1* gf= gf. Now we can write the given expression in terms of the GCF. 5gf^2+g^2f+15gf ⇕ gf* 5f+ gf* g+ gf* 15 Finally, we will factor out the GCF. gf* 5f+ gf* g+ gf* 15 ⇕ gf(5f+g+15)

Checking Our Answer

Check your answer ✓
To check our answer, we can apply the Distributive Property and compare the result with the given expression.

gf(5f+g+15)
gf(5f)+gf(g)+gf(15)
5gf^2+g^2f+15gf

After applying the Distributive Property, the result is the same as the given expression. Therefore, we can be sure our solution is correct!