Chapter 8.1, Problem 16CYU

To determine

To find: The sum or difference $\left(-3d^2-8+2d\right)+\left(4d-12+d^2\right)$

Answer to Problem 16CYU

  $-2d^2+6d-20$

Explanation of Solution

Given:

  $\left(-3d^2-8+2d\right)+\left(4d-12+d^2\right)$

Calculation:

The given expression is $\left(-3d^2-8+2d\right)+\left(4d-12+d^2\right)$

Open the parenthesis

  $-3d^2-8+2d+4d-12+d^2$

Group the like terms

  $-3d^2+d^2+2d+4d-12-8$

Combine the like terms

  $-2d^2+6d-20$