Chapter 8.3, Problem 9CYU

To determine

To find: The product $\left(4y^2-3\right)\left(4y^2+7y+2\right)$

Answer to Problem 9CYU

  $16y^4+28y^3-4y^2-21y-6$

Explanation of Solution

Given:

  $\left(4y^2-3\right)\left(4y^2+7y+2\right)$

Calculation:

Multiply each term of the first polynomial with the each term of the second polynomial

  $\begin{array}{l}\left(4y^2-3\right)\left(4y^2+7y+2\right)\\ =4y^2\cdot 4y^2+4y^2\cdot 7y+4y^2\cdot 2-3\cdot 4y^2-3\cdot 7y-3\cdot 2\\ =16y^4+28y^3+8y^2-12y^2-21y-6\\ =16y^4+28y^3-4y^2-21y-6\end{array}$

So, the product of the polynomial is $16y^4+28y^3-4y^2-21y-6$