# Linear, Quadratic and Cubic Polynomials

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## 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?

1. $${y^2} + \sqrt 2$$
2. $$x + \frac{2}{x}$$
3. $$\frac{\pi }{2}{x^2} + x$$

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?

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