Chapter 5.3, Problem 47WE

To determine

To find: The factor of the polynomial $96w{x}^3{y}^2{z}^2-144w^{3}x{y}^2{z}^2$ .

Answer to Problem 47WE

The factor of the given polynomial is $48wx{y}^2{z}^2\left(2x^2-3w^2\right)$ .

Explanation of Solution

Given information:

The polynomial is $96w{x}^3{y}^2{z}^2-144w^{3}x{y}^2{z}^2$ .

The given polynomial is $96w{x}^3{y}^2{z}^2-144w^{3}x{y}^2{z}^2$ .

Factorization of polynomials is the process in which a polynomial split into the product of factors.

Split the terms of the given polynomial into prime factors.

  $\begin{array}{c}96w{x}^3{y}^2{z}^2-144w^{3}x{y}^2{z}^2=\left(2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\right)\left(w\cdot x\cdot x\cdot x\cdot y\cdot y\cdot z\cdot z\right)\\ -\left(2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\right)\left(w\cdot w\cdot w\cdot x\cdot y\cdot y\cdot z\cdot z\right)\end{array}$

Find the GCF of each of the term of the given polynomial.

Now, take common out 48 from the $96$ and $144$ to obtain the GCF of the $96$ and $144$ .

Therefore, the GCF of coefficient of $96$ and $144$ is $48$ .

Apply the same process with $w{x}^3{y}^2{z}^2$ and $w^{3}x{y}^2{z}^2$ to obtain their GCF.

Now, take common out $wx{y}^2{z}^2$ from the $w{x}^3{y}^2{z}^2$ and $w^{3}x{y}^2{z}^2$ to obtain the GCF of the $w{x}^3{y}^2{z}^2$ and $w^{3}x{y}^2{z}^2$ .

Therefore, the GCF of $w{x}^3{y}^2{z}^2$ and $w^{3}x{y}^2{z}^2$ is $wx{y}^2{z}^2$ .

The factor of given polynomial can be written as follows:

  $96w{x}^3{y}^2{z}^2-144w^{3}x{y}^2{z}^2=48wx{y}^2{z}^2\left(2x^2-3w^2\right)$

Therefore, the factor of the given polynomial is $48wx{y}^2{z}^2\left(2x^2-3w^2\right)$ .

Conclusion:

Thus, the factor of the given polynomial is $48wx{y}^2{z}^2\left(2x^2-3w^2\right)$ .