Chapter 8.8, Problem 30PPE

To determine

To find:the expression for the dimension of the trunk.

Answer to Problem 30PPE

The expression for the dimension of the trunk is y13

  $2x\left[\left(3x-2\right)\left(x+7\right)\right]$ .

Explanation of Solution

Given info:

The volume of thetrunk is $V=6x^3+38x^2-28x$

Formula used:

Show the general form of trinomial as follows:

  $a{x}^2+bx+c$

Here, $a,b,c$ are integers, $b$ is the value which depends on the multipliers $\left(n,m\right)$ of the product of $a$ and $c$ .

Show the factorization of trinomial form as follows:

  $\begin{array}{l}a{x}^2+bx+c\\ a{x}^2+nx+mx+c\\ x\left(ax+n\right)+\left(mx+c\right)\end{array}$

Calculation:

Find the dimension of the trunk by factorization rule as follows:

  $\begin{array}{l}=6x^3+38x^2-28x\\ =2x\left[\left(3x^2+19x-14\right)\right]\\ =2x\left[\left(3x^2+21x-2x-14\right)\right]\\ =2x\left[3x\left(\left(x+7\right)-2\left(x+7\right)\right)\right]\\ =2x\left[\left(3x-2\right)\left(x+7\right)\right]\end{array}$

Hence, the expression for the dimension of trunk is $2x\left[\left(3x-2\right)\left(x+7\right)\right]$

Conclusion:

Thus, the expression for the dimension of trunk is $2x\left[\left(3x-2\right)\left(x+7\right)\right]$