Finding Zeros (Roots) of a Polynomial

The zeros (or roots) of a polynomial are the values of x that make the polynomial equal to zero. These values are used to understand the polynomial’s behavior, such as its graph, turning points, and intersections with the x-axis. 

For example, let us consider the polynomial P(x) = x² – 4

The zeros of this polynomial are x = 2 and x = -2 because substituting these values into the equation yields x² – 4 = 0

Here, we will explore the different methods for finding the zeros of polynomials

Setting the Polynomial Equal to Zero

We start by equating the polynomial to zero.

For example, if a polynomial is of the form P(x) = x³ – 6x² + 11x – 6, we solve P(x) = 0

Identifying the Type of Polynomial and Using the Appropriate Method

Then, we will identify whether it is a linear, quadratic, cubic, or higher-degree polynomial. Depending on the degree of the polynomial, different methods like factoring, quadratic formula, and synthetic division are followed.

Linear Polynomial 

A linear polynomial is a first-degree polynomial with the general form ax + b = 0. Here, the highest power of the variable is 1. Such equations always have exactly one real zero. 

To find the zero, we will isolate x by solving the equation ${x=-\dfrac{b}{a}}$

For example, for the linear polynomial P(x) = 7x + 5, we can find its zero by solving:

7x + 5 = 0

⇒ 7x + 5 – 5 = 0 – 5

⇒ 7x = -5

⇒ ${\dfrac{7x}{7}=-\dfrac{5}{7}}$

⇒ ${x=-\dfrac{5}{7}}$

Thus, the zero of 7x + 5 is ${x=-\dfrac{5}{7}}$

Quadratic Polynomial 

A quadratic or second-degree polynomial has the general form ax² + bx + c = 0

Here, the highest power of the variable is 2. Quadratic equations can have up to two real zeros or complex zeros. 

To solve these polynomials, we can use factoring, completing the square, or the quadratic formula:

${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

By Factoring

Let us consider the polynomial P(x) = x² – 5x + 6

Factoring the Polynomial

⇒ P(x) = x² – 3x – 2x + 6

⇒ P(x) = x(x – 3) – 2(x – 3)

⇒ P(x) = (x – 3)(x – 2)

Setting Each Factor Equal to 0

(x – 2) = 0 and (x – 3) = 0

Solving for x

⇒ x = 2 and x = 3

Thus, the zeros of P(x) are x = 2 and x = 3

By Quadratic Formula

When it is difficult to factor a quadratic polynomial, we use the quadratic formula:

${x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Let us find the zeros of P(x) = 2x² – 4x – 6 …..(i)

Comparing With the Standard Equation 

As we know, the standard equation is y = ax² + bx + c …..(ii)

Comparing equations (i) and (ii), we get

a = 2, b = -4, and c = -6

Substituting the Values in the Quadratic Formula

${x=\dfrac{-\left( -4\right) \pm \sqrt{\left( -4\right) ^{2}-4\times 2\times \left( -6\right) }}{2\times 2}}$

Solving for x 

⇒ ${x=\dfrac{4\pm \sqrt{64}}{4}}$

⇒ x = 3 and x = -1

Thus, the zeros are x = 3 and x = -1

Higher-Degree Polynomials

For polynomials of degree 3 or higher, finding zeros becomes more complex and may have real or complex zeros. 

Apart from factoring, here we use the synthetic division method, the rational zero theorem, or graphing to find the zeros. 

Note: Sometimes, we may get complex zeros (i.e., have an imaginary part). Complex zeros always come in pairs.

By Rational Root Theorem

The Rational Root Theorem predicts possible rational zeros of a polynomial using the factors of the constant term divided by the factors of the leading coefficient.

It states that possible rational zeros of a polynomial are of the form ${\dfrac{p}{q}}$

Here,

• p is a factor of the constant term
• q is a factor of the leading coefficient

Let us find all the possible rational zeros of P(x) = 2x³ + x² – 4x + 1

Finding Factors of the Constant Term and Leading Coefficient

Given,

The constant term is 1

The leading coefficient is 2

Here,

The factors of 1 are ±1 

The factors of 2 are ±1 and ±2

Finding the Possible Zeros

Thus, the possible zeros are ±1 or ${\pm \dfrac{1}{2}}$

Verifying the Possible Zeros

Now, we test each possible zero by substituting it in the polynomial P(x). If any of the four real zeros are rational, then they are one of the factors of P(x).

Verifying for x = -1, 

P(-1) = 2(-1)³ + (-1)² – 4(-1) + 1 ≠ 0

Verifying at x = ${-\dfrac{1}{2}}$,

${P\left( -\dfrac{1}{2}\right)}$ = ${2\left( -\dfrac{1}{2}\right) ^{3}+\left( -\dfrac{1}{2}\right) ^{2}-4\left( -\dfrac{1}{2}\right) +1}$ ≠ 0

Verifying for x = ${\dfrac{1}{2}}$

${P\left( \dfrac{1}{2}\right)}$ = ${2\left( \dfrac{1}{2}\right) ^{3}+\left( \dfrac{1}{2}\right) ^{2}-4\left( \dfrac{1}{2}\right) +1}$ ≠ 0

Verifying for x = 1,

P(1) = 2(1)³ + (1)² – 4(1) + 1 = 0

Thus, x = 1 is one of the zeros of P(x). 

Now, we divide P(x) by (x – 1) using synthetic division to get other zeros.

By Synthetic Division

Synthetic division further simplifies the polynomial and verifies possible zeros, along with the Rational Root Theorem. If the remainder is zero, the tested value is a zero of the polynomial.

Let us find the factors for P(x) = 2x³ + x² – 4x + 1

As we know (x – 1) is one of the zeros of P(x)

Now, dividing P(x) by (x – 1), we get

Here, the remainder is 0

Thus, P(x) can be factored as P(x) = (x – 1)(2x² + 3x – 1)

To find additional zeros, we then use the quadratic formula.

${x=\dfrac{-\left( 3\right) \pm \sqrt{\left( 3\right) ^{2}-4\times 2\times \left( -1\right) }}{2\times 2}}$

⇒ ${x=\dfrac{-3\pm \sqrt{17}}{4}}$

⇒ ${x=\dfrac{-3+\sqrt{17}}{4}}$ and ${x=\dfrac{-3-\sqrt{17}}{4}}$

Thus, the zeros are ${x=1}$, ${x=\dfrac{-3+\sqrt{17}}{4}}$, and ${x=\dfrac{-3-\sqrt{17}}{4}}$

The table below summarizes the possible combinations of real and complex zeros for polynomials:

Degree	Zeros	Possible Combinations
1	1	1 Real Zerot
2	2	2 Real Zeros, or 2 Complex Zeros
3	3	3 Real Zeros, or 1 Real and 2 Complex Zeros
4	4	4 Real Zeros, 2 Real and 2 Complex Zeros, or 4 Complex Zeros

From the Graph

A polynomial function’s graph provides a visual way of finding its real zeros. On the graph, the x-axis represents the input values (x), and the y-axis represents the output (f(x)) of the polynomial function. 

The zeros of the polynomial can be found by the points where the graph intersects the x-axis.

Graphing a Linear Polynomial

For linear polynomials, the graph represents a straight line. 

Here, the graph of the function f(x) = x + 7 shows a line that intersects the x-axis at only 1 point. 

Graphing a Quadratic Polynomial

For quadratic polynomials, the graph forms a parabolic curve. 

Here, the graph of the function f(x) = x² + 6x + 8 shows a parabola that intersects the x-axis at 2 points.

Thus, f(x) = x² + 6x + 8 has 2 zeros at x = -4 and x = -2

Graphing a Cubic Polynomial

Higher-degree polynomials represent curves with more complex shapes and turning points. 

Here is the graph of the polynomial f(x) = x³– 6x² + 11x – 6 

The graph intersects the x-axis at three points, corresponding to the real zeros of the polynomial. Thus, the zeros are x = 1, x = 2, and x = 3

From the Sum and Product of Zeros

The zeros of a polynomial can also be calculated from the Sum and Product of Zeros.

Quadratic Polynomial

If α and β are the zeros of the quadratic equation ax² + bx + c = 0, then

• α + β = ${-\dfrac{b}{a}}$ …..(i)
• αβ = ${\dfrac{c}{a}}$ …..(ii)

Here, 

• a = coefficient of x² 
• b = coefficient of x
• c = constant term

Now, by solving the equations (i) and (ii), the zeros of the polynomials can be verified. 

Let us verify if x = -1 and 3 are the zeros of the polynomial f(x) = 2x² – 4x – 6 

Comparing it with the standard form of the cubic polynomial, we get

a = 2, b = -4, and c = -6

If α and β are the zeros (here, α = -1 and β = 3), then

α + β = ${-\dfrac{b}{a}}$ = 2 …..(i)

αβ = ${\dfrac{c}{a}}$ = -3 …..(ii)

Now, verifying the zeros, we get

α + β = -1 + 3 = 2 …..(iii)

αβ = (-1)(3) = -3 …..(iv)

Since (i) = (iii) and (ii) = (iv)

The zeros are x = -1 and x = 3, verified.

Cubic Polynomial

If α, β, and ɣ are the zeros of the quadratic equation ax³ + bx² + cx + d = 0, then

• α + β + ɣ = ${-\dfrac{b}{a}}$ …..(i)
• αβ + βɣ + ɣα = ${\dfrac{c}{a}}$ …..(ii)
• αβɣ = ${-\dfrac{d}{a}}$ …..(iii)

Here,

• a = coefficient of x³ 
• b = coefficient of x² 
• c = coefficient of x
• d = constant term

Now, by solving the equations (i), (ii), and (iii), the zeros of the polynomials can be verified.

Let us verify if x = 1, 2, and 3 are the zeros of the polynomial f(x) = x³ – 6x² + 11x – 6

Comparing it with the standard form of the cubic polynomial, we get

a = 1, b = -6, c = 11, and d = -6

If α, β, and ɣ are the zeros (here, α = 1, β = 2, and ɣ = 3), then

α + β + ɣ = ${-\dfrac{b}{a}}$ = 6 …..(i)

αβ + βɣ + ɣα = ${\dfrac{c}{a}}$ = 11 …..(ii)

αβɣ = ${-\dfrac{d}{a}}$ = 6 …..(iii)

Now, verifying the zeros, we get

α + β + ɣ = 1 + 2 + 3 = 6 …..(iv)

αβ + βɣ + ɣα = (1)(2) + (2)(3) + (3)(1) = 2 + 6 + 3 = 11 …..(v)

αβɣ = (1)(2)(3) = 6 …..(vi)

Since (i) = (iv), (ii) = (v), and (iii) = (vi)

The zeros are x = 1, x = 2, and x = 3, verified.

Solved Examples

E.g.5.