(n+5) (n+4) = ?? someone help thank yoh

2 Answers
Mar 3, 2018

The result is #n^2+9n+20#.

Explanation:

You can use the distributive property twice. First, distribute #(n+5)# onto #n#, and then onto #4#, like this:

#color(white)=color(blue)((n+5))color(red)((n+4))#

#=color(blue)((n+5))color(red)n+color(blue)((n+5))color(red)4#

#=color(red)ncolor(blue)((n+5))+color(red)4color(blue)((n+5))#

Now, use the distributive in each of these smaller parts:

#color(white)=color(red)ncolor(blue)((n+5))+color(red)4color(blue)((n+5))#

#=color(red)ncolor(blue)n+color(red)ncolor(blue)5+color(red)4color(blue)((n+5))#

#=color(purple)(n^2)+color(blue)5color(red)n+color(red)4color(blue)((n+5))#

#=color(purple)(n^2)+color(blue)5color(red)n+color(red)4color(blue)n+color(red)4*color(blue)5#

#=color(purple)(n^2)+color(blue)5color(red)n+color(red)4color(blue)n+color(purple)20#

Lastly, combine the like terms:

#color(white)=color(purple)(n^2)+color(blue)5color(red)n+color(red)4color(blue)n+color(purple)20#

#=color(purple)(n^2)+color(purple)(9n)+color(purple)20#

This is the result. (It is called a quadratic.)

Mar 3, 2018

#color(red)(n^2)+color(blue)9color(red)n+color(blue)20#

Explanation:

To solve this, we must multiply each variable in one bracket by each variable in the other brackets.

This is called distributing:

#(color(red)n+color(blue)5)(color(red)n+color(blue)4)#

becomes:

#(color(red)n*color(red)n)+(color(red)n*color(blue)4)+(color(red)n*color(blue)5)+(color(blue)5*color(blue)4)#

#=color(red)(n^2)+color(blue)4color(red)n+color(blue)5color(red)n+color(blue)20#

Simplifying:

#->color(red)(n^2)+color(blue)9color(red)n+color(blue)20#

Thus, solved.