How do you find the product of #(t+1)(t^2+2t+4)#?
1 Answer
Explanation:
You can use the distributive property to find:
#(t+1)(t^2+2t+4) = t(t^2+2t+4)+1(t^2+2t+4)#
#color(white)((t+1)(t^2+2t+4)) = (t^3+2t^2+4t)+(t^2+2t+4)#
#color(white)((t+1)(t^2+2t+4)) = t^3+2t^2+t^2+4t+2t+4#
#color(white)((t+1)(t^2+2t+4)) = t^3+3t^2+6t+4#
Alternatively, just focus on each power of
#t^3: " "t * t^2 = t^3#
#t^2: " "t(2t)+1(t^2) = 3t^2#
#t: " "t(4)+1(2t) = 6t#
#1: " "1(4) = 4#
Then add:
#(t+1)(t^2+2t+4) = t^3+3t^2+6t+4#
When looking at each power of
#t^3: " "1 * 1 = 1#
#t^2: " "1 * 2 + 1 * 1 = 3#
#t: " "1 * 4 + 1 * 2 = 6#
#1: " "1 * 4 = 4#
With practice you will probably find that you can normally do these calculations in your head as you write down the product:
#(t+1)(t^2+2t+4) = t^3+3t^2+6t+4#
