How do you evaluate #(2n+7)(n-3)#?

2 Answers
smendyka
Jan 30, 2018

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

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

#2n^2 - 6n + 7n - 21#

We can now combine like terms:

#2n^2 + (-6 + 7)n - 21#

#2n^2 + 1n - 21#

#2n^2 + n - 21#

Jim G.
Jan 30, 2018

#2n^2+n-21#

Explanation:

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

#rArr(color(red)(2n+7))(n-3)#

#=color(red)(2n)(n-3)color(red)(+7)(n-3)#

#"distribute the brackets and collect like terms"#

#=2n^2-6n+7n-21#

#=2n^2+n-21#

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