Let's tackle each expression step by step using the Distributive Property.
1. Expression: [tex]\( 4(2x + 5) \)[/tex]
- Using the Distributive Property, we distribute [tex]\( 4 \)[/tex] to both terms inside the parentheses:
- First term: [tex]\( 4 \cdot 2x = 8x \)[/tex]
- Second term: [tex]\( 4 \cdot 5 = 20 \)[/tex]
- Therefore, [tex]\( 4(2x + 5) = 8x + 20 \)[/tex]
2. Expression: [tex]\( 3(3x + 1) \)[/tex]
- Using the Distributive Property, we distribute [tex]\( 3 \)[/tex] to both terms inside the parentheses:
- First term: [tex]\( 3 \cdot 3x = 9x \)[/tex]
- Second term: [tex]\( 3 \cdot 1 = 3 \)[/tex]
- Therefore, [tex]\( 3(3x + 1) = 9x + 3 \)[/tex]
3. Expression: [tex]\( 15x + 25 \)[/tex]
- To factor this expression using the Distributive Property in reverse, we look for a common factor in [tex]\( 15x \)[/tex] and [tex]\( 25 \)[/tex]:
- Both terms have a common factor of [tex]\( 5 \)[/tex]:
- First term: [tex]\( 15x = 5 \cdot 3x \)[/tex]
- Second term: [tex]\( 25 = 5 \cdot 5 \)[/tex]
- Therefore, [tex]\( 15x + 25 = 5(3x + 5) \)[/tex]
Now, let's fill in the blanks as given in the question:
[tex]\[
\begin{array}{l}
4(2 x+5)= \ \boxed{8} \ x+ \ \boxed{20} \ \\
3(3 x+1)= \ \boxed{9} \ x+ \ \boxed{3} \ \\
15 x+25= \ \boxed{5} \ \boxed{3 x} + \ \boxed{5} \ \\
\end{array}
\][/tex]
This is how each expression is completed using the Distributive Property!
