Question

If $p\left( u \right) = {u^2} + 3u + 4,q\left( u \right) = {u^2} + u - 12$ and $r\left( u \right) = u - 2$ , then find the degree of $p\left( u \right)q\left( u \right)r\left( u \right)$.

Answer 1

Hint:
It is given in the question that $p\left( u \right) = {u^2} + 3u + 4,q\left( u \right) = {u^2} + u - 12$ and $r\left( u \right) = u - 2$
So, we have to find the degree of $p\left( u \right)q\left( u \right)r\left( u \right)$ .
First, we have to find the degree of $p\left( u \right),q\left( u \right),r\left( u \right)$ and after that we will find the degree of $p\left( u \right)q\left( u \right)r\left( u \right)$ and we will get the answer.

Complete step by step solution:
It is given in the question that $p\left( u \right) = {u^2} + 3u + 4,q\left( u \right) = {u^2} + u - 12$ and $r\left( u \right) = u - 2$
So, we have to find the degree of $p\left( u \right)q\left( u \right)r\left( u \right)$ .
Degree of $p\left( u \right) = 2$
Degree of $q\left( u \right) = 3$
Degree of $r\left( u \right) = 1$
$\therefore $Degree of $p\left( u \right)q\left( u \right)r\left( u \right)$ $ = $ $p\left( u \right) + q\left( u \right)r + \left( u \right)$
$\therefore $Degree of $p\left( u \right)q\left( u \right)r\left( u \right)$ $ = $ 2+2+1

$\therefore $Degree of $p\left( u \right)q\left( u \right)r\left( u \right)$ $ = 5$

Note:
Degree: Degree: Degree of the function can be defined as the highest power of the variable.
Types of equation as per degree:
The equation with degree 1 is called linear equation.
For example: In equation $ax + b$ , the degree of variable x is 1.
The equation with degree 2 is called quadratic equation.
For example: In equation $a{x^2} + bx + c$ , the degree of variable x is 2.
The equation with degree 3 is called the cubic equation.
For example: In equation $a{x^3} + b{x^2} + cx + d$ , the degree of variable x is 3.
The equation with degree 4 is called a biquadratic equation.
For example: In equation $a{x^4} + b{x^3} + c{x^2} + dx + e$ , the degree of variable x is 4.