How do you find the product #(4y^2-3)(4y^2+7y+2)#?

1 Answer
Mar 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)(4y^2) - color(red)(3))(color(blue)(4y^2) + color(blue)(7y) + color(blue)(2))# becomes:

#(color(red)(4y^2) xx color(blue)(4y^2)) + (color(red)(4y^2) xx color(blue)(7y)) + (color(red)(4y^2) xx color(blue)(2)) - (color(red)(3) xx color(blue)(4y^2)) - (color(red)(3) xx color(blue)(7y)) - (color(red)(3) xx color(blue)(2))#

#16y^4 + 28y^3 + 8y^2 - 12y^2 - 21y - 6#

We can now combine like terms:

#16y^4 + 28y^3 + (8 - 12)y^2 - 21y - 6#

#16y^4 + 28y^3 - 4y^2 - 21y - 6#