How do you multiply #(3c+1)(5c-2)#?

2 Answers
Meave60
Apr 6, 2015

The answer is #15c^2-c-2#.

You use the "FOIL" method, in which you multiply the first term in each set of parentheses, then the outer terms, then the inner terms, then the last term in each set of parentheses. Then you add the results.

http://www.purplemath.com/modules/polymult2.htm

#(3c+1)(5c-2)# =

#15c^2-6c+5c-2# =

#15c^2-c-2#

Gió
Apr 6, 2015

Multiply each fator of the first bracket by each element of the second:
#3c(5c)-3c(-2)+1(5c)-1(2)=#
#15c^2-6c+5c-2=#
#15c^2-c-2#

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