How do you find the product of #(k+4)(k^2+3k-6)#?

2 Answers
Apr 12, 2017

Answer:

See the entire 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)(k) + color(red)(4))(color(blue)(k^2) + color(blue)(3k) - color(blue)(3))# becomes:

#(color(red)(k) xx color(blue)(k^2)) + (color(red)(k) xx color(blue)(3k)) - (color(red)(k) xx color(blue)(6)) + (color(red)(4) xx color(blue)(k^2)) + (color(red)(4) xx color(blue)(3k)) - (color(red)(4) xx color(blue)(6))#

#k^3 + 3k^2 - 6k + 4k^2 + 12k - 24#

We can now group and combine like terms:

#k^3 + 3k^2 + 4k^2 + 12k - 6k - 24#

#k^3 + (3 + 4)k^2 + (12 - 6)k - 24#

#k^3 + 7k^2 + 6k - 24#

Apr 12, 2017

Answer:

#=k^3+7k^2+6k-24#

Explanation:

Each term in the first bracket must be multiplied by each term in the second bracket. This means there will be 6 terms at first.

#(color(red)(k) + color(blue)(4))(k^2+3k-6)#

#=color(red)(k)(k^2+3k-6) + color(blue)(4)(k^2+3k-6)#

#=color(red)(k^3+3k^2-6k +color(blue)(4k^2+12k-24)#

Now add any like terms:

#=k^3+7k^2+6k-24#