How do you find the product #(2y-11)(y^2-3y+2)#?

1 Answer
May 9, 2017

Answer:

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)(2y) - color(red)(11))(color(blue)(y^2) - color(blue)(3y) + color(blue)(2))# becomes:

#(color(red)(2y) xx color(blue)(y^2)) - (color(red)(2y) xx color(blue)(3y)) + (color(red)(2y) xx color(blue)(2)) - (color(red)(11) xx color(blue)(y^2)) + (color(red)(11) xx color(blue)(3y)) - (color(red)(11) xx color(blue)(2))#

#2y^3 - 6y^2 + 4y - 11y^2 + 33y - 22#

We can now group and combine like terms:

#2y^3 - 6y^2 - 11y^2 + 4y + 33y - 22#

#2y^3 + (-6 - 11)y^2 + (4 + 33)y - 22#

#2y^3 + (-17)y^2 + 37y - 22#

#2y^3 - 17y^2 + 37y - 22#