How do you solve #(3k+2)(k-3)#?

Question

525 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-27T02:11:30

See a solution process below:

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

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

#(color(red)(3k) xx color(blue)(k)) - (color(red)(3k) xx color(blue)(3)) + (color(red)(2) xx color(blue)(k)) - (color(red)(2) xx color(blue)(3))#

#3k^2 - 9k + 2k - 6#

We can now combine like terms:

#3k^2 + (-9 + 2)k - 6#

#3k^2 + (-7)k - 6#

#3k^2 - 7k - 6#