Chapter 14.4, Problem 20PPS

To determine

Tosimplify:the given equation.

Answer to Problem 20PPS

  $3n^2\left(n-4\right)+6n\left(3n^2+n-7\right)-4\left(n-7\right)=21n^3-6n^2-46n+28$

Explanation of Solution

Given:

The given equation: $3n^2\left(n-4\right)+6n\left(3n^2+n-7\right)-4\left(n-7\right)$

Calculation:

As thegiven equation is $3n^2\left(n-4\right)+6n\left(3n^2+n-7\right)-4\left(n-7\right)$ .

Simplify the equation by open the parenthesis, then

  $\begin{array}{l}3n^2\left(n-4\right)+6n\left(3n^2+n-7\right)-4\left(n-7\right)\\ =3n^3-12n^2+18n^3+6n^2-42n-4n+28\{\text{open the parenthesis}}\\ =21n^3-6n^2-46n+28\{\text{simplify}}\end{array}$

Hence, the value of the given simplification is $3n^2\left(n-4\right)+6n\left(3n^2+n-7\right)-4\left(n-7\right)=21n^3-6n^2-46n+28$ .