What is the degree of 16x^4yz + 12x^2y^3z - 24x^3y^2z - 18xy^4z?

Mar 10, 2016

$6$

Explanation:

Defining Degree
Recall that the degree of a polynomial is the largest sum of the exponents of each of the variables in a term within the polynomial.

Finding the Degree
$1$. Start by locating the terms of the polynomial.

${\underbrace{16 {x}^{4} y z}}_{\textcolor{b l u e}{\text{term")+underbrace(12x^2y^3z)_color(blue)("term")-underbrace(24x^3y^2z)_color(blue)("term")-underbrace(18xy^4z)_color(blue)("term}}}$

$2$. Determine the degree of each term by adding the exponents of each variable within the term. Recall that the exponent on a term like $x$ or $y$ is $1$.

a) $16 {x}^{\textcolor{red}{4}} {y}^{\textcolor{p u r p \le}{1}} {z}^{\textcolor{g r a y}{1}} \textcolor{b r o w n}{\Rightarrow} \textcolor{red}{4} + \textcolor{p u r p \le}{1} + \textcolor{g r a y}{1} \stackrel{\textcolor{t e a l}{\text{degree }}}{\textcolor{b r o w n}{\Rightarrow}} 6$

b) $12 {x}^{\textcolor{red}{2}} {y}^{\textcolor{p u r p \le}{3}} {z}^{\textcolor{g r a y}{1}} \textcolor{b r o w n}{\Rightarrow} \textcolor{red}{2} + \textcolor{p u r p \le}{3} + \textcolor{g r a y}{1} \stackrel{\textcolor{t e a l}{\text{degree }}}{\textcolor{b r o w n}{\Rightarrow}} 6$

c) $24 {x}^{\textcolor{red}{3}} {y}^{\textcolor{p u r p \le}{2}} {z}^{\textcolor{g r a y}{1}} \textcolor{b r o w n}{\Rightarrow} \textcolor{red}{3} + \textcolor{p u r p \le}{2} + \textcolor{g r a y}{1} \stackrel{\textcolor{t e a l}{\text{degree }}}{\textcolor{b r o w n}{\Rightarrow}} 6$

d) $18 {x}^{\textcolor{red}{1}} {y}^{\textcolor{p u r p \le}{4}} {z}^{\textcolor{g r a y}{1}} \textcolor{b r o w n}{\Rightarrow} \textcolor{red}{1} + \textcolor{p u r p \le}{4} + \textcolor{g r a y}{1} \stackrel{\textcolor{t e a l}{\text{degree }}}{\textcolor{b r o w n}{\Rightarrow}} 6$

$3$. Since the degree for each term is $6$, the degree of the whole polynomial is also $6$. This can be written mathematically as:

color(green)(|bar(ul(color(white)(a/a)"deg"(16x^4yz+12x^2y^3z-24x^3y^2z-18xy^4z)=6color(white)(a/a)|)