Chapter 4.4, Problem 21WE

(a).

To determine

To find: the GCF of the given monomials.

Answer to Problem 21WE

The GCF of $110h^3{k}^{2}r$ and $-88h^2{k}^2{r}^2$ is $22h^2{k}^{2}r$ .

Explanation of Solution

Given information :

The given monomials:

  $110h^3{k}^{2}r,-88h^2{k}^2{r}^2$

Calculation :

The prime factorization of the coefficients are:

  $\begin{array}{l}110=2\cdot 5\cdot 11\\ 88=2^3\cdot 11\end{array}$

The GCF of $110$ and $-88$ is

  $2\cdot 11=22$

Compare the powers of each variable occurring in both monomials. Use the power with the least exponent.

Compare $h^3$ and $h^2$ . Use $h^2$ .

Compare $r$ and $r^2$ . Use $r$ .

Use $k^2$ .

  $\text{GCF}=22\cdot h^2\cdot k^2\cdot r=22h^2{k}^{2}r$

Therefore, the GCF of $110h^3{k}^{2}r$ and $-88h^2{k}^2{r}^2$ is $22h^2{k}^{2}r$ .

(b).

To determine

The LCM of the given monomials.

Answer to Problem 21WE

The LCM of $110h^3{k}^{2}r$ and $-88h^2{k}^2{r}^2$ is $440h^3{k}^2{r}^2$ .

Explanation of Solution

Given information :

The given monomials:

  $110h^3{k}^{2}r,-88h^2{k}^2{r}^2$

Calculation :

As per problem,

The prime factorization of the coefficients are:

  $\begin{array}{l}110=2\cdot 5\cdot 11\\ 88=2^3\cdot 11\end{array}$

The LCM of $110$ and $-88$ is

  $2^3\cdot 5\cdot 11=440$

Compare the powers of each variable occurring in either monomial. Use the power with the greatest exponent.

Compare $h^3$ and $h^2$ . Use $h^3$ .

Compare $r$ and $r^2$ . Use $r^2$ .

Use $k^2$ .

  $\text{LCM}=440\cdot h^3\cdot k^2\cdot r^2=440h^3{k}^{2}r$

Therefore, the LCM of $110h^3{k}^{2}r$ and $-88h^2{k}^2{r}^2$ is $440h^3{k}^2{r}^2$ .