# Linear, Quadratic and Cubic Polynomials

## What is a Linear Polynomial?

A **linear polynomial** is a polynomial of degree **one, **i.e., the highest exponent of the variable is one.

Some examples:

\[\begin{array}{l}p\left( x \right):2x + 3\\q\left( y \right):\pi y + \sqrt 2 \\r\left( z \right):z + \sqrt 5 \\s\left( x \right): - 7x\end{array}\]

We note that a linear polynomial in one variable can have at the most two terms. In general, a linear polynomial in one variable will be of the form:

\[p\left( x \right):ax + b,\,\,a \ne 0\]

The constraint that *a* should not be equal to 0 is required because if *a* is 0, then this becomes a constant polynomial. Why do we use the adjective ** linear**? It turns out that when we draw the graph corresponding to a linear polynomial, we will get a straight line – hence the name linear.

### Solved Example:

**Example 1: **Is \(\frac{1}{x}\) a linear polynomial?

**Solution:** In general form, we can write it as \(1{x^{ - 1}} + 0\). Clearly, the degree of this polynomial is not one, it is **not** a linear polynomial.

**⚡Tip:** After converting any expression into the general form, if the *exponent* of the variable in any term is *not a whole number*, then it's *not a polynomial *either.

**Challenge 1:** What do you think about \(\frac{1}{{x + 1}}\). Is it a linear polynomial?

## What is a Quadratic Polynomial?

A **quadratic polynomial** is a polynomial of degree **two, **i.e., the highest exponent of the variable is two.

Some examples:

\[\begin{array}{l}p\left( x \right): & 3{x^2} + 2x + 1\\q\left( y \right): & {y^2} - 1\\r\left( z \right): & \sqrt 2 {z^2}\end{array}\]

We observe that a quadratic polynomial can have at the most three terms. In general, a quadratic polynomial will be of the form:

\[p\left( x \right):a{x^2} + bx + c,\,\,a \ne {\rm{0}}\]

The constraint that *a* should not be equal to 0 is required because if *a* is 0, then this becomes a linear polynomial.

### Solved Example:

**Example 2:** Which of the following are quadratic polynomials?

- \({y^2} + \sqrt 2 \)
- \(x + \frac{2}{x}\)
- \(\frac{\pi }{2}{x^2} + x\)

**Solution:** In (1) and (3), the degree of the polyomial is two. But In (2), the exponent of term \(\frac{2}{x}\) is not a whole number.

So, (1) and (3) are quadratic polynomials and (2) is not even a polynomial.

## What is a Cubic Polynomial?

A **cubic polynomial** is a polynomial of degree three, i.e., the highest exponent of the variable is three.

Some examples:

\[\begin{array}{l}p\left( x \right): & {x^3} - 6{x^2} + 11x - 6\\q\left( y \right): & 27{y^3} - 1\\r\left( z \right): & \pi {z^3} + {\left( {\sqrt 2 } \right)^{10}}\end{array}\]

We observe that a cubic polynomial can have at the most four terms. A cubic polynomial, in general, will be of the form

\[p\left( x \right):a{x^3} + b{x^2} + cx + d,\,\,a \ne {\rm{0}}\]

Once again, the constraint that *a* should not be equal to 0 is required because if *a* is 0, then this becomes a quadratic rather than a cubic polynomial.

**Challenge 2:** Is \({z^2} + \frac{{{z^6} + {z^7}}}{{{z^4}}} + 4\) a cubic polynomial?