Difference between revisions of "1957 AHSME Problems/Problem 4"

(created page)
 
m (typo fix)
 
Line 9: Line 9:
 
== See also ==
 
== See also ==
  
{{AHSME box|year=1957|num-b=3|num-a=5}}
+
{{AHSME 50p box|year=1957|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 08:04, 25 July 2024

The first step in finding the product $(3x + 2)(x - 5)$ by use of the distributive property in the form $a(b + c) = ab + ac$ is:

$\textbf{(A)}\ 3x^2 - 13x - 10 \qquad  \textbf{(B)}\ 3x(x - 5) + 2(x - 5)\qquad \\ \textbf{(C)}\ (3x+2)x+(3x+2)(-5)\qquad \textbf{(D)}\ 3x^2-17x-10\qquad \textbf{(E)}\ 3x^2+2x-15x-10$

Solution

In order to find the product by using this form of the distributive property, we should express $(3x + 2)(x - 5)$ as $(3x + 2)(x) + (3x + 2)(-5)$. Thusly, $\boxed{\text{(C)}}$ is our answer, and we are done.

See also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png