Lesson Explainer: Zeros of Polynomial Functions Mathematics

In this explainer, we will learn how to find the set of zeros of a quadratic, cubic, or higher-degree polynomial function.

Polynomial functions appear all throughout science and in many real-world applications. For example, a ball thrown in the air will follow a parabolic arc that can be modeled by a quadratic equation. In particular, the height of the ball from the ground will be a quadratic function. Therefore, if we want to determine how long it would take the ball to hit the ground, we will need to find the values where a quadratic function is equal to zero.

The input values of π‘₯ where a function 𝑓 outputs zero are called the zeros (or roots) of the function, and we can write this more formally as follows.

Definition: Zeros or Roots of a Function

If 𝑓(π‘Ž)=0, then we say that π‘Ž is a zero (or root) of the function 𝑓.

For example, for the function 𝑓(π‘₯)=π‘₯+1, we can see that 𝑓(βˆ’1)=βˆ’1+1=0, so βˆ’1 is a root of this function.

There are a few methods of finding the roots of a function. For example, since we are looking for the values of π‘₯ where 𝑓(π‘₯)=0, we can sketch the graph 𝑦=𝑓(π‘₯), and then the points on the curve where the output is zero are the roots; these will be the π‘₯-intercepts. To see this, consider the following graph of 𝑦=𝑓(π‘₯).

From the graph, we can see that 𝑓(βˆ’1)=0, 𝑓(0)=0, and 𝑓(2)=0; these are the roots of the function. Since there can be multiple roots of a function, we usually write them in a set called the set of roots of the function; in this case, the set of roots of 𝑓(π‘₯) is {βˆ’1,0,2}.

Not all functions have zeros. Consider the function 𝑓(π‘₯)=1. In this function, every output is always 1, so no input can ever give us 0. In fact, the constant function of the form 𝑓(π‘₯)=𝑐 will only have zeros if 𝑐=0. When a function does not have any zeros, we can say the set of zeros is βˆ….

There are pros and cons for finding the zeros of a function graphically versus algebraically. In this explainer, we will focus on finding algebraic roots of polynomials since these will give us exact values for the roots.

We can find the zeros of polynomials by factoring. For example, consider the function 𝑓(π‘₯)=π‘₯+5π‘₯+6. We can factor this by finding a pair of numbers that multiply to give 6 and add to give 5; we see that 2Γ—3=6 and 2+3=5. Hence, π‘₯+5π‘₯+6=(π‘₯+2)(π‘₯+3).

Therefore, the zeros of the function are the solutions of the equation (π‘₯+2)(π‘₯+3)=0.

On the left-hand side of this equation, we have a product that must be equal to zero. We note that for a product to be equal to zero, one of the factors must be equal to zero. Hence, either π‘₯+2=0π‘₯+3=0.or

We can solve both of these equations separately to get π‘₯=βˆ’2 and π‘₯=βˆ’3 as the zeros of the function. It is worth noting that we could have used the quadratic formula to find the roots of the equation, but we can only use this when we have a quadratic function.

We recall that we can factor some higher-order polynomials by grouping. For example, in the function 𝑓(π‘₯)=π‘₯+π‘₯βˆ’4π‘₯βˆ’4, we can factor the first two terms to get π‘₯(π‘₯+1) and the last two terms to get βˆ’4(π‘₯+1). Since these share a factor, we can take this out of the expression to get 𝑓(π‘₯)=π‘₯+π‘₯βˆ’4π‘₯βˆ’4=π‘₯(π‘₯+1)βˆ’4(π‘₯+1)=(π‘₯+1)ο€Ήπ‘₯βˆ’4.

We can factor π‘₯βˆ’4 by using the difference between squares, which we recall states that we can factor a difference of squares as follows: π‘Žβˆ’π‘=(π‘Žβˆ’π‘)(π‘Ž+𝑏). This gives π‘₯βˆ’4=(π‘₯βˆ’2)(π‘₯+2).

Therefore, the zeros of this function are the solutions to the equation 𝑓(π‘₯)=0(π‘₯+1)ο€Ήπ‘₯βˆ’4=0(π‘₯+1)(π‘₯+2)(π‘₯βˆ’2)=0.

Solving each factor to be equal to zero, we get π‘₯=βˆ’1,βˆ’2,2 as the zeros of the function.

There are other methods of factoring a polynomial. One such method is to use a fact called the factor theorem together with polynomial division. The factor theorem tells us that if 𝑓 is a polynomial and 𝑓(π‘Ž)=0, then (π‘₯βˆ’π‘Ž) must be a factor of the polynomial. It is worth noting that the reverse statement is also true: if (π‘₯βˆ’π‘Ž) is a factor of a polynomial 𝑓, then 𝑓(π‘Ž)=0.

We can use this to help us find the roots of a polynomial. If 𝑓(π‘₯) is a polynomial and we know that 𝑓(π‘Ž)=0, then we know that π‘₯βˆ’π‘Ž is a factor of 𝑓(π‘₯). In particular, this means we can use polynomial division to divide 𝑓(π‘₯) by π‘₯βˆ’π‘Ž, allowing us to factor 𝑓(π‘₯).

Let’s see some examples of applying these techniques to determine the zeros of polynomials. We will start with a linear function.

Example 1: Finding the Zeros of a Linear Function

Find the set of zeros of the function 𝑓(π‘₯)=13(π‘₯βˆ’4).

Answer

We recall that π‘₯=π‘Ž is a zero of the function 𝑓 if 𝑓(π‘Ž)=0. Therefore, to find the zeros for this function, we need to solve the equation 𝑓(π‘₯)=0.

This is the equation 13(π‘₯βˆ’4)=0.

Multiplying through by 3 gives 3Γ—13(π‘₯βˆ’4)=3Γ—0π‘₯βˆ’4=0.

We then add 4 to both sides of the equation π‘₯βˆ’4+4=0+4π‘₯=4.

We see that the only zero of the function 𝑓 is 4. Hence, the set of zeros of the function is {4}.

In our second example, we will determine the zeros of a quadratic function by factoring.

Example 2: Finding the Zeros of a Monic Quadratic Function by Factoring

Find, by factoring, the zeros of the function 𝑓(π‘₯)=π‘₯+2π‘₯βˆ’35.

Answer

We recall that the zeros of a function 𝑓 are the input values such that 𝑓(π‘₯)=0. Therefore, to find the zeros of the given quadratic, we need to solve the equation π‘₯+2π‘₯βˆ’35=0.

There are several different methods of doing this. For example, we can use the quadratic formula. However, we will fully factor the quadratic. Recall that to factor a quadratic π‘₯+𝑏π‘₯+π‘οŠ¨, we need to find two numbers that multiply to give 𝑐 and add to give 𝑏. In our case, we have 𝑏=2 and 𝑐=βˆ’35. We can list the possible factor pairs of βˆ’35:

βˆ’135
βˆ’57
βˆ’75
βˆ’351

Of these, we see that βˆ’5+7=2. We can use these to rewrite the term 2π‘₯ in the quadratic as 2π‘₯=7π‘₯βˆ’5π‘₯.

Using this to rewrite the equation gives us π‘₯+2π‘₯βˆ’35=0π‘₯+7π‘₯βˆ’5π‘₯βˆ’35=0.

We can then factor the first two terms and last two terms separately: π‘₯+7π‘₯βˆ’5π‘₯βˆ’35=0π‘₯(π‘₯+7)βˆ’5(π‘₯+7)=0.

We then take out the shared factor of π‘₯+7 to get (π‘₯+7)(π‘₯βˆ’5)=0.

For the product of two numbers to be equal to zero, one of the factors must be equal to zero. Hence, either π‘₯+7=0 or π‘₯βˆ’5=0. We can solve each equation separately.

First, π‘₯+7=0.

We subtract 7 from both sides of the equation to get π‘₯+7βˆ’7=0βˆ’7π‘₯=βˆ’7.

Second, π‘₯βˆ’5=0.

We add 5 to both sides of the equation to get π‘₯βˆ’5+5=0+5π‘₯=5.

Therefore, the zeros of the quadratic are βˆ’7 and 5.

In our next example, we will find the roots of a nonmonic quadratic by factoring.

Example 3: Finding the Zeros of a Nonmonic Quadratic Function by Factoring

Find, by factoring, the zeros of the function 𝑓(π‘₯)=9π‘₯+9π‘₯βˆ’40.

Answer

We recall that we say π‘₯=π‘Ž is a zero of 𝑓 when 𝑓(π‘Ž)=0. Therefore, to find the zeros for this function, we need to solve the equation 𝑓(π‘₯)=09π‘₯+9π‘₯βˆ’40=0.

We could do this by using the quadratic formula; however, we will use the grouping method. We recall that to factor a quadratic π‘Žπ‘₯+𝑏π‘₯+π‘οŠ¨, we need to find a factor pair of π‘Žπ‘ that add to give 𝑏. In this quadratic, we have π‘Ž=9, 𝑏=9, and 𝑐=βˆ’40, so we need to find a pair of numbers that multiple to give 9Γ—(βˆ’40)=βˆ’360 and add to give 9. We notice that (βˆ’15)Γ—24=βˆ’360βˆ’15+24=9.

To apply the grouping method to factor this quadratic, we use these two numbers to rewrite the second term in the quadratic as follows: 9π‘₯=βˆ’15π‘₯+24π‘₯.

Hence, we can rewrite the quadratic equation as 9π‘₯+9π‘₯βˆ’40=09π‘₯βˆ’15π‘₯+24π‘₯βˆ’40=0.

We then factor the first two terms and second two terms separately: 3π‘₯(3π‘₯βˆ’5)+8(3π‘₯βˆ’5)=0.

We can take out the shared factor of 3π‘₯βˆ’5: (3π‘₯βˆ’5)(3π‘₯+8)=0.

For a product to be equal to zero, one of the factors must be equal to zero. Hence, either 3π‘₯βˆ’5=03π‘₯+8=0.or

We can solve each equation separately. First, 3π‘₯βˆ’5=0.

We add 5 to both sides of the equation to get 3π‘₯βˆ’5+5=0+53π‘₯=5; then, we divide through by 3, which gives us π‘₯=53.

Second, 3π‘₯+8=0.

We subtract 8 from both sides of the equation to get 3π‘₯+8βˆ’8=0βˆ’83π‘₯=βˆ’8; then, we divide through by 3, which gives us π‘₯=βˆ’83.

Therefore, the zeros of the function are 53 and βˆ’83.

Thus far, we have only found the zeros of polynomials of degree 2 or less. In the remaining examples, we will find the zeros of polynomials of degree 3 or more.

Example 4: Finding the Zeros of a Quartic by Factoring

Find the set of zeros of the function 𝑓(π‘₯)=βˆ’9π‘₯+225π‘₯οŠͺ.

Answer

We recall that the zeros or roots of a function are the input values that cause the function to output zero. Therefore, to find the zeros of 𝑓 we need to solve the equation 𝑓(π‘₯)=0; this is the equation βˆ’9π‘₯+225π‘₯=0.οŠͺ

We can solve this equation by factoring. We note that both terms share a factor of 9π‘₯; taking out this factor gives 9π‘₯ο€Ύβˆ’9π‘₯+225π‘₯9π‘₯=09π‘₯ο€Ήβˆ’π‘₯+25=0.οŠͺ

If we then take out a factor of βˆ’1, we have βˆ’9π‘₯ο€Ήπ‘₯βˆ’25=0.

We can factor further by recalling the difference between squares result, which states that π‘₯βˆ’π‘Ž=(π‘₯βˆ’π‘Ž)(π‘₯+π‘Ž).

Applying this with π‘Ž=√25=5 gives π‘₯βˆ’25=(π‘₯βˆ’5)(π‘₯+5).

We can then rewrite the equation: βˆ’9π‘₯ο€Ήπ‘₯βˆ’25=0βˆ’9π‘₯(π‘₯βˆ’5)(π‘₯+5)=0.

Since π‘₯=π‘₯β‹…π‘₯, this is now the product of linear factors, so we cannot factor any further. We can now find the zeros of the equation by recalling that if a product is equal to zero, then one of the factors must be equal to zero. We can solve each of the factors being equal to zero separately. First, βˆ’9π‘₯=0.

We divide through by βˆ’9: π‘₯=0.

Then, we take the square root of both sides of the equation, noting that zero is the only root: π‘₯=0.

Second, π‘₯βˆ’5=0.

We add 5 to both sides of the equation: π‘₯βˆ’5+5=0+5π‘₯=5.

Third, π‘₯+5=0.

We subtract 5 from both sides of the equation, which gives us the final zero of the function: π‘₯+5βˆ’5=0βˆ’5π‘₯=βˆ’5.

It is worth noting that since we were not asked to factor the polynomial fully, we could have found the values of π‘₯ that make each factor in the expression βˆ’9π‘₯ο€Ήπ‘₯βˆ’25=0 equal to zero directly. We would then have π‘₯βˆ’25=0π‘₯=25, which we can solve by taking the square root of both sides of the equation, where we remember we will get a positive and a negative square root: π‘₯=±√25=Β±5.

Similarly, we have βˆ’9π‘₯=0π‘₯=0.

Writing these as a set gives us that the set of zeros of this function is {βˆ’5,0,5}.

Example 5: Finding the Zeros of a Partially Factored Cubic Function

Find the set of zeros of the function 𝑓(π‘₯)=π‘₯ο€Ήπ‘₯βˆ’81ο…βˆ’2ο€Ήπ‘₯βˆ’81ο…οŠ¨οŠ¨.

Answer

We recall that π‘₯=π‘Ž is a zero of the function 𝑓 if 𝑓(π‘Ž)=0. Therefore, to find the zeros for this function, we need to solve the equation 𝑓(π‘₯)=0.

This is the equation π‘₯ο€Ήπ‘₯βˆ’81ο…βˆ’2ο€Ήπ‘₯βˆ’81=0.

Since both terms on the left-hand side of this equation share a factor of π‘₯βˆ’81, we will do this by factoring. Taking out the shared factor, we have ο€Ήπ‘₯βˆ’81(π‘₯βˆ’2)=0.

We can factor this further by recalling that the difference between squares tells us that for any constant π‘Ž, π‘₯βˆ’π‘Ž=(π‘₯βˆ’π‘Ž)(π‘₯+π‘Ž). Setting π‘Ž=9, we have π‘₯βˆ’81=(π‘₯βˆ’9)(π‘₯+9). Therefore, we can rewrite the equation as (π‘₯βˆ’9)(π‘₯+9)(π‘₯βˆ’2)=0.

Since this is three linear factors, we have factored the expression fully. For a product to be equal to zero, one of the factors must be equal to zero. Hence, we have π‘₯βˆ’9=0,π‘₯+9=0,π‘₯βˆ’2=0.or

We can solve each equation separately. First, π‘₯βˆ’9=0.

We add 9 to both sides of the equation to get π‘₯βˆ’9+9=0+9π‘₯=9.

Second, π‘₯+9=0.

We subtract 9 from both sides of the equation, which gives us π‘₯+9βˆ’9=0βˆ’9π‘₯=βˆ’9.

Third, π‘₯βˆ’2=0.

We add 2 to both sides of the equation to get π‘₯βˆ’2+2=0+2π‘₯=2.

Therefore, we have three solutions: π‘₯=9, βˆ’9, or 2.

It is worth noting that since we were not asked to factor the polynomial fully, we could have found the values of π‘₯ that make each factor in the expression ο€Ήπ‘₯βˆ’81(π‘₯βˆ’2)=0 equal to zero directly. We would then have π‘₯βˆ’81=0π‘₯=81, which we can solve by taking the square root of both sides of the equation, where we remember we will get a positive and a negative square root: π‘₯=±√81=Β±9.

Similarly, we have π‘₯βˆ’2=0π‘₯=2.

These are the zeros of the function. Writing these in a set, we have that the set of zeros of 𝑓(π‘₯) is {βˆ’9,2,9}.

In our final example, we will use the factor theorem to determine the roots of a cubic polynomial.

Example 6: Finding the Zeros of a Cubic Function

Given that 𝑓(π‘₯)=π‘₯+3π‘₯βˆ’13π‘₯βˆ’15 and 𝑓(βˆ’1)=0, find the other roots of 𝑓(π‘₯).

Answer

We start by recalling that the roots of a function 𝑓 are the values of π‘Ž such that 𝑓(π‘Ž)=0. Since we are given that 𝑓(βˆ’1)=0, we know that βˆ’1 is a root of 𝑓(π‘₯). One way of finding the roots of a polynomial is to factor the function fully.

Since we are given a root of the function, we can do this by recalling part of the factor theorem: if 𝑓(π‘₯) is a polynomial and 𝑓(π‘Ž)=0, then (π‘₯βˆ’π‘Ž) is a factor of 𝑓(π‘₯).

In our case, 𝑓 is a cubic polynomial, and we are told 𝑓(βˆ’1)=0. So, setting π‘Ž=βˆ’1, the factor theorem tells us that (π‘₯βˆ’(βˆ’1))=π‘₯+1 is a factor of 𝑓(π‘₯). There are then a few different methods we can use to factor the cubic polynomial; we will show two of these.

First, we can factor by using the grouping method. Since we know that π‘₯+1 is a factor of 𝑓, we can write 𝑓(π‘₯)=(π‘₯+1)ο€Ήπ‘Žπ‘₯+𝑏π‘₯+π‘ο…οŠ¨. Distributing the parentheses gives us π‘₯+3π‘₯βˆ’13π‘₯βˆ’15=π‘Žπ‘₯+𝑏π‘₯+𝑐π‘₯+π‘Žπ‘₯+𝑏π‘₯+𝑐=π‘Žπ‘₯+(π‘Ž+𝑏)π‘₯+(𝑏+𝑐)π‘₯+𝑐.

Equating the coefficients of π‘₯ gives π‘Ž=1. We can then substitute this into the expression to get π‘₯+3π‘₯βˆ’13π‘₯βˆ’15=π‘₯+(1+𝑏)π‘₯+(𝑏+𝑐)π‘₯+𝑐.

Equating the coefficients of π‘₯ gives 3=1+𝑏, which we can solve for 𝑏 by subtracting 1 from both sides of the equation 3βˆ’1=1+π‘βˆ’12=𝑏.

Similarly, equating the constant terms gives 𝑐=βˆ’15.

We can substitute these values into the equation to see π‘₯+3π‘₯βˆ’13π‘₯βˆ’15=π‘₯+(1+2)π‘₯+(2βˆ’15)π‘₯βˆ’15=π‘₯+3π‘₯βˆ’13π‘₯βˆ’15.

This confirms that we found the values of the coefficients correctly. We have 𝑓(π‘₯)=(π‘₯+1)ο€Ήπ‘₯+2π‘₯βˆ’15.

We can factor this further by noticing that 5Γ—(βˆ’3)=βˆ’15 and 5βˆ’3=2. This means that π‘₯+2π‘₯βˆ’15=(π‘₯+5)(π‘₯βˆ’3).

Hence, we have shown 𝑓(π‘₯)=(π‘₯+1)(π‘₯+5)(π‘₯βˆ’3).

For 𝑓(π‘₯)=0, we have (π‘₯+1)(π‘₯+5)(π‘₯βˆ’3)=0, and for a product to be zero, one of the factors must be equal to zero; hence, π‘₯=βˆ’1, βˆ’5, or 3. Since the question asked for the other two roots, we have π‘₯=3 and π‘₯=βˆ’5.

Alternatively, we could have used polynomial division to factor π‘₯+1 from 𝑓(π‘₯). We have

Thus, 𝑓(π‘₯)=(π‘₯+1)ο€Ήπ‘₯+2π‘₯βˆ’15.

We could then factor the quadratic in the same way and solve to find that the two other roots of 𝑓(π‘₯) are π‘₯=3 and π‘₯=βˆ’5.

Let’s finish by recapping some of the important points of this explainer.

Key Points

  • The zeros or roots of a polynomial 𝑓(π‘₯) are the values π‘₯=π‘Ž such that 𝑓(π‘Ž)=0.
  • If 𝑓 is a polynomial and 𝑓(π‘Ž)=0, then (π‘₯βˆ’π‘Ž) is a factor of 𝑓. The same statement is true in reverse: if (π‘₯βˆ’π‘Ž) is a factor of the polynomial 𝑓, then 𝑓(π‘Ž)=0.
  • There are many techniques we can use to help us determine the roots of polynomials including the quadratic formula, factoring by grouping, and polynomial division.

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