Lesson Explainer: Zeros of Polynomial Functions | Nagwa Lesson Explainer: Zeros of Polynomial Functions | Nagwa

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.

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.

It is worth noting that we can verify if a number is a zero of a polynomial by substituting it back into the function and evaluating. For example, in the previous example we found that the zeros of 𝑓(π‘₯)=π‘₯+2π‘₯βˆ’35 are π‘₯=5 and π‘₯=βˆ’7. We can verify both of these zeroes by evaluating 𝑓(5) and 𝑓(βˆ’7). We have 𝑓(5)=5+2(5)βˆ’35=25+10βˆ’35=0,𝑓(βˆ’7)=(βˆ’7)+2(βˆ’7)βˆ’35=49βˆ’14βˆ’35=0.

Since the function outputs zero at these values of π‘₯, we have confirmed they are zeros of the function. This is a useful check to make sure that our answers are correct.

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 next example, we will determine the value of a constant using the sets of zeros of two polynomial functions.

Example 6: Finding an Unknown Coefficient in a Quadratic and a Linear Function Given That They Have an Equal Set of Zeros

The function 𝑓(π‘₯)=π‘Žπ‘₯+54π‘₯+81 and the function 𝑔(π‘₯)=π‘Žπ‘₯+9 have the same set of zeros. Find π‘Ž and the set of zeros.

Answer

We recall that the set of zeros of a function β„Ž is the set containing all values 𝑏 such that β„Ž(𝑏)=0. We see that 𝑔(π‘₯) is a linear function; we can determine its set of zeros by solving the equation 𝑔(π‘₯)=0. We have π‘Žπ‘₯+9=0.

Subtracting 9 from both sides gives π‘Žπ‘₯=βˆ’9.

We want to divide through by π‘Ž, but we can only do this if π‘Ž is nonzero. Note that if π‘Ž=0, then 𝑔(π‘₯)=0π‘₯+9=9. Thus, 𝑔(π‘₯) would be a constant function and have no zeros. However, 𝑓(π‘₯)=0π‘₯+54π‘₯+81=54π‘₯+81.

We can see that this has a zero by solving 54π‘₯+81=0π‘₯=βˆ’8154.

Since 𝑓(π‘₯) and 𝑔(π‘₯) must have the same set of zeros, we can conclude that π‘Ž must be nonzero. We can now divide both sides of the linear equation by π‘Ž: π‘Žπ‘₯=βˆ’9π‘₯=βˆ’9π‘Ž.

Since 𝑔(π‘₯) is a linear function, this is its only root. We can also conclude that 𝑓(π‘₯) must have a repeated root of βˆ’9π‘Ž. We know that π‘“ο€Όβˆ’9π‘Žοˆ=0, so let’s substitute π‘₯=βˆ’9π‘Ž into 𝑓: π‘“ο€Όβˆ’9π‘Žοˆ=π‘Žο€Όβˆ’9π‘Žοˆ+54ο€Όβˆ’9π‘Žοˆ+81.

We can evaluate to get 0=π‘Žο€Ό81π‘Žοˆβˆ’54Γ—9π‘Ž+810=81βˆ’486π‘Ž+810=162βˆ’486π‘Ž.

We can now solve for π‘Ž; we add 486π‘Ž to both sides of the equation to get 486π‘Ž=162.

We then multiply both sides of the equation by π‘Ž to get 486=162π‘Ž.

Finally, we divide the equation through by 162 and evaluate to get π‘Ž=486162=3.

We have shown that these functions have a single zero at π‘₯=βˆ’9π‘Ž, so we can find this zero by substituting π‘Ž=3. We get π‘₯=βˆ’93=βˆ’3.

Hence, π‘Ž=3 and the set of zeroes of both functions is 𝑧(𝑓)={βˆ’3}.

In our final example, we will use factoring by grouping to determine the set of zeros of a cubic polynomial.

Example 7: Finding the Set of Zeros of a Cubic Function

Find the set of zeros of the function 𝑓(π‘₯)=π‘₯βˆ’4π‘₯βˆ’25π‘₯+100, where all three zeros take integer values.

Answer

Recall that π‘₯=π‘Ž is a zero of the function 𝑓 if 𝑓(π‘Ž)=0. To find the zeros of a function, we need to solve the equation 𝑓(π‘₯)=0.

Therefore, in this question, we need to solve the equation π‘₯βˆ’4π‘₯βˆ’25π‘₯+100=0.

Observe that 𝑓(π‘₯) is a cubic function, and recall that we can factor some higher-order polynomials by grouping. Since we are told that all three zeros of this cubic polynomial take integer values, then there is a chance we might be able to spot a pattern among the terms so that we can group them.

In fact, we can factor the first two terms to get π‘₯(π‘₯βˆ’4) and the last two terms to get βˆ’25(π‘₯βˆ’4). As these two expressions have a common factor of (π‘₯βˆ’4), we can take out this factor to get 𝑓(π‘₯)=π‘₯βˆ’4π‘₯βˆ’25π‘₯+100=π‘₯(π‘₯βˆ’4)βˆ’25(π‘₯βˆ’4)=ο€Ήπ‘₯βˆ’25(π‘₯βˆ’4).

Next, notice that π‘₯βˆ’25 is of the form π‘Žβˆ’π‘οŠ¨οŠ¨, which we recall is known as a difference of two squares and can be factored by using the formula π‘Žβˆ’π‘=(π‘Žβˆ’π‘)(π‘Ž+𝑏). This gives π‘₯βˆ’25=(π‘₯βˆ’5)(π‘₯+5).

We can now find the zeros of the original function, which are the solutions to the equation 𝑓(π‘₯)=0ο€Ήπ‘₯βˆ’25(π‘₯βˆ’4)=0(π‘₯βˆ’5)(π‘₯+5)(π‘₯βˆ’4)=0.

Setting each factor equal to zero, we get π‘₯=5, βˆ’5, and 4 as the zeros of the function. Writing these values in a set, we have that the set of zeros of 𝑓(π‘₯)=π‘₯βˆ’4π‘₯βˆ’25π‘₯+100 is {βˆ’5,4,5}.

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.
  • We can verify that π‘₯=π‘Ž is a zero of a given polynomial 𝑓(π‘₯) by checking that 𝑓(π‘Ž)=0. This can be a useful check to make sure that a number is a zero of the polynomial.
  • There are many techniques we can use to help us determine the roots of polynomials including the quadratic formula and factoring by grouping.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy