# How do you evaluate (- 3x ^ { 3} y ^ { 4} z ^ { 2} ) ( x y z ^ { 2} ) ( - x ^ { 5} y ^ { 2} z )?

Jan 31, 2018

See a solution process below:

#### Explanation:

First, rearrange the expression as:

$\left(- 3 {x}^{3} {y}^{4} {z}^{2}\right) \left(x y {z}^{2}\right) \left(- {x}^{5} {y}^{2} z\right) \implies$

$\left(- 3 {x}^{3} {y}^{4} {z}^{2}\right) \left(x y {z}^{2}\right) \left(- 1 {x}^{5} {y}^{2} z\right) \implies$

$\left(- 3 \cdot - 1\right) \left({x}^{3} \cdot x \cdot {x}^{5}\right) \left({y}^{4} \cdot y \cdot {y}^{2}\right) \left({z}^{2} \cdot {z}^{2} \cdot z\right) \implies$

$3 \left({x}^{3} \cdot x \cdot {x}^{5}\right) \left({y}^{4} \cdot y \cdot {y}^{2}\right) \left({z}^{2} \cdot {z}^{2} \cdot z\right)$

Next, use this rule for exponents to rewrite the expression:

$a = {a}^{\textcolor{red}{1}}$

$3 \left({x}^{3} \cdot {x}^{\textcolor{red}{1}} \cdot {x}^{5}\right) \left({y}^{4} \cdot {y}^{\textcolor{red}{1}} \cdot {y}^{2}\right) \left({z}^{2} \cdot {z}^{2} \cdot {z}^{\textcolor{red}{1}}\right)$

Now, use this rule of exponents to complete the evaluation:

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$3 \left({x}^{\textcolor{b l u e}{3}} \cdot {x}^{\textcolor{red}{1}} \cdot {x}^{\textcolor{g r e e n}{5}}\right) \left({y}^{\textcolor{b l u e}{4}} \cdot {y}^{\textcolor{red}{1}} \cdot {y}^{\textcolor{g r e e n}{2}}\right) \left({z}^{\textcolor{b l u e}{2}} \cdot {z}^{\textcolor{g r e e n}{2}} \cdot {z}^{\textcolor{red}{1}}\right) \implies$

$3 {x}^{\textcolor{b l u e}{3} + \textcolor{red}{1} + \textcolor{g r e e n}{5}} {y}^{\textcolor{b l u e}{4} + \textcolor{red}{1} + \textcolor{g r e e n}{2}} {z}^{\textcolor{b l u e}{2} + \textcolor{g r e e n}{2} + \textcolor{red}{1}} \implies$

$3 {x}^{9} {y}^{7} {z}^{5}$

Jan 31, 2018

$3 {x}^{9} {y}^{7} {z}^{5}$

#### Explanation:

Take each different term and multiply them. Powers multiply by adding the exponents.
$- 3 \cdot - 1 = 3$
${x}^{3} \cdot {x}^{1} \cdot {x}^{5} = {x}^{9}$
${y}^{4} \cdot {y}^{1} \cdot {y}^{2} = {y}^{7}$
${z}^{2} \cdot {z}^{2} \cdot {z}^{1} = {z}^{5}$