How do you find the product (4y^2-3)(4y^2+7y+2)(4y23)(4y2+7y+2)?

1 Answer
Mar 12, 2017

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))(4y23)(4y2+7y+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))(4y2×4y2)+(4y2×7y)+(4y2×2)(3×4y2)(3×7y)(3×2)

16y^4 + 28y^3 + 8y^2 - 12y^2 - 21y - 616y4+28y3+8y212y221y6

We can now combine like terms:

16y^4 + 28y^3 + (8 - 12)y^2 - 21y - 616y4+28y3+(812)y221y6

16y^4 + 28y^3 - 4y^2 - 21y - 616y4+28y34y221y6