# How to Find Zeros of Polynomial?

Zeros of the polynomial are points where the polynomial is equal to zero. Here you will learn how to find the zeros of a polynomial.

Zeros of a polynomial are also known as the roots of the equation and are often denoted by \(α\), \(β\), and \(γ\), respectively. Some of the methods used to find polynomial zeros are grouping, factorization, and the use of algebraic expressions.

## Step by step guide to zeros of a polynomial

The zeros of a polynomial are the values of \(x\) which satisfy the equation \(y = f(x)\). Where \(f(x)\) is a function of \(x\), and the zeros of the polynomial are the values of \(x\) for which the \(y\) value is equal to zero. The number of zeros of a polynomial depends on the degree of the equation \(y = f (x)\). All such domain values of the function whose range is equal to zero are called zeros of the polynomial.

**Note:** Graphically the zeros of the polynomial are the points where the graph of \(y = f(x)\) cuts the \(x\)-axis.

**How to find zeros of polynomial?**

The different types of equations and methods for finding their polynomial zeros are as follows:

**Linear Equation:**

A linear equation is of the form \(y=ax+b\). Zero of this equation can be calculated by substituting \(y = 0\), and on simplification we have \(ax+b=0\) or \(x=-\frac{b}{a}\).

**Quadratic Equation:**

There are two ways to factorize a quadratic equation:

- The quadratic equation of the form \(x^2+x(a+b)+ab=0\), can be factorized as \((x + a)(x + b) = 0\) and we have \(x = -a\), and \(x = -b\) as the zeros of the polynomial.
- For a quadratic equation of the form \(ax^2+bx+c=0\), which cannot be factorized, the zeros can be calculated using the formula method, and the formula is \(x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\).

**Cubic Equation: **

The cubic equation of the form \(ax^3+bx^2+cx+d\), can be factorized by applying the remainder theorem. As per remainder theorem, we can substitute any smaller value for the variable \(x = α\), and if the value of y reaches zero, \(y = 0\), then \((x – α)\) is one root of the equation. In addition, we can divide the cubic equation by \((x – α)\) using the long division to obtain a quadratic equation. Finally, the quadratic equation can be solved by factorization or the formula method to obtain the two required roots of the equation.

**Higher Degree Polynomial:**

The higher degree polynomial equation is of the form \(y=ax^n+bx^n-1+cx^n-2+…px+q\). These higher degree polynomials can be factorized using the remaining theorem to reach a quadratic equation. And the quadratic equation can be factorized to reach the two final factors required.

### Representing zeros of polynomial on graph

A polynomial expression in the form \(y = f (x)\) can be represented on a graph across the coordinate axis. The value of \(x\) is displayed on the \(x\)-axis and the value of \(f(x)\) or the value of \(y\) is displayed on the \(y\)-axis. A polynomial expression can be a linear expression, a quadratic expression, a cubic expression based on the degree of a polynomial. A linear expression represents a line, a quadratic equation represents a curve, and a higher degree polynomial represents a curve with uneven bends.

The zeros of a polynomial can be found from the graph by looking at the points where the graph line cuts the \(x\)-axis. The \(x\) coordinates of the points where the graph cuts the \(x\)-axis are the zeros of the polynomial.

### Zeros of Polynomial – Example 1:

Find zeros of the polynomial function \(f(x)=x^3-12x^2+20x\).

**Solution:**

First, take out \(x\) as common:

\(f(x)=x(x^2-12x+20)\)

Now by splitting the middle term:

\(f(x)=x(x^2-2x-10x+20)\)

So we get:

\(f(x)=x[x(x-2)-10(x-2)]\)

\(f(x)=x(x-2)(x-10)\)

Here

\(x=0\)

\(x-2=0 → x=2\)

\(x-10=0 →x=10\)

Therefore, the zeros of polynomial function is \(x = 0\) or \(x = 2\) or \(x = 10\).

## Exercises for Zeros of Polynomial

### Find the zeros of a polynomial.

- \(\color{blue}{f(x)=3x^3-19x^2+33x-9}\)
- \(\color{blue}{f(x)=x^2-10x+25}\)
- \(\color{blue}{f(x)=x^3+2x^2-25x-50}\)
- \(\color{blue}{f(x)=x^4+2x^{^3}-16x^2-32x}\)

- \(\color{blue}{x=3 , x=\frac{1}{3}}\)
- \(\color{blue}{x=5}\)
- \(\color{blue}{x=-2, x=-5, x=5}\)
- \(\color{blue}{x=0, x=-2, x=-4, x=4}\)

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