How do you simplify #(3+ 2c ) ( 7- 4c )#?

2 Answers
smendyka
Jul 29, 2017

See a solution process below:

Explanation:

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#(color(red)(3) + color(red)(2c))(color(blue)(7) - color(blue)(4c))# becomes:

#(color(red)(3) xx color(blue)(7)) - (color(red)(3) xx color(blue)(4c)) + (color(red)(2c) xx color(blue)(7)) - (color(red)(2c) xx color(blue)(4c))#

#21 - 12c + 14c - 8c^2#

We can now combine like terms:

#21 + (-12 + 14)c - 8c^2#

#21 + 2c - 8c^2#

Or in standard form:

#-8c^2 + 2c + 21#

Jim G.
Jul 29, 2017

#21+2c-8c^2#

Explanation:

#"simplify means to multiply the factors together"#

#"each term in the second factor is multiplied by each term"#
#"in the first factor"#

#rArr(color(red)(3+2c))(7-4c)#

#=color(red)(3)(7-4c)color(red)(+2c)(7-4c)#

#"distributing the brackets gives"#

#=(color(red)(3)xx7)+(color(red)(3)xx-4c)+(color(red)(2c)xx7)+(color(red)(2c)xx-4c)#

#=21+(-12c)+14c+(-8c^2)#

#=21+2c-8c^2#

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