How do you multiply #(p^4+3p^2-8)(p+1)#?

2 Answers
smendyka
Jun 19, 2017

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)(p^4) + color(red)(3p^2) - color(red)(8))(color(blue)(p) + color(blue)(1))# becomes:

#(color(red)(p^4) xx color(blue)(p)) + (color(red)(p^4) xx color(blue)(1)) + (color(red)(3p^2) xx color(blue)(p)) + (color(red)(3p^2) xx color(blue)(1)) - (color(red)(8) xx color(blue)(p)) - (color(red)(8) xx color(blue)(1))#

#p^5 + p^4 + 3p^3 + 3p^2 - 8p - 8#

Barney V.
Jun 20, 2017

#color(green)(p^5+p^4+3p^3+3p^2-8p-8#

Explanation:

#(p^4+3p^2-8)(p+1)#

#color(white)(aaaaaaaaaaaaa)##p^4+3p^2-8#
#color(white)(aaaaaaaaaaa)## xx underline(p+1)#
#color(white)(aaaaaaaaaaaaa)##p^5+3p^3-8p#
#color(white)(aaaaaaaaaaaaaaa)##p^4+3p^2-8#
#color(white)(aaaaaaaaaaaaa)##overline(p^5+p^4+3p^3+3p^2-8p-8)#

#color(white)(aaaaaaaaaaaaa)##color(green)(p^5+p^4+3p^3+3p^2-8p-8)#

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