Lesson Video: Zeros of Polynomial Functions Mathematics • Third Year of Preparatory School

Video Transcript

In this video, we will learn how to find the zeros of a quadratic, cubic, or higher-degree polynomial function. These are the input values for 𝑥 for which 𝑓 of 𝑥 is equal to zero.

If 𝑓 of 𝑎 equals zero, then we say that 𝑎 is a zero, or root, of the function 𝑓. Consider, for example, the linear function 𝑓 of 𝑥 equals 𝑥 plus one. 𝑓 of negative one equals negative one plus one, which is equal to zero. So negative one is a zero, or root, of the function 𝑓. There are a few methods to find the roots of a function. If we are given the graph of the function 𝑓 of 𝑥, the roots of 𝑓 are the values of 𝑥 where the curve crosses the 𝑥-axis. In this example, the curve crosses the 𝑥-axis at 𝑥 equals negative one, 𝑥 equals zero, and 𝑥 equals two. From the graph, we can see that 𝑓 of negative one is equal to zero, 𝑓 of zero is equal to zero, and 𝑓 of two is equal to zero. Therefore, the zeros, or roots, of 𝑓 are negative one, zero, and two.

In set notation, we could write the set of zeros of 𝑓 in a pair of curly brackets, negative one, zero, and two. Some functions will of course have no zeros at all, for example, the function 𝑓 of 𝑥 equals one. In this function, every output value is always one, so no input will ever give an output of zero.

Another method to find the zeros of a polynomial function is algebraically, through factoring. Consider, for example, the quadratic function 𝑓 of 𝑥 equals 𝑥 squared plus five 𝑥 plus six. To find the zeros of the function, we set this expression equal to zero. This expression can be factorized by finding a pair of numbers that multiply to give six and add to give five. On inspection, two and three add together to make five and multiply together to make six. We can therefore factorize this expression into two binomials: 𝑥 plus two and 𝑥 plus three.

We’ve now expressed 𝑓 of 𝑥 as a product of two factors; let’s call them 𝑎 and 𝑏. If we set 𝑓 of 𝑥 equals zero, then 𝑎𝑏 is equal to zero. This implies that either 𝑎 or 𝑏 must be equal to zero or both of them. Therefore, if 𝑓 of 𝑥 is equal to zero, then either 𝑥 plus two equals zero or 𝑥 plus three equals zero. These linear equations can be easily solved for 𝑥, giving 𝑥 equals negative two and 𝑥 equals negative three. Therefore, the zeros of 𝑓 are negative two and negative three.

It’s worth noting that the roots of a quadratic can of course be found using the quadratic formula as well, which is especially useful if the roots are not integers. We can find the zeros of higher-order polynomials, such as cubics and quartics, using the same method. But factorizing may take more steps.

Consider, for example, the function 𝑓 of 𝑥 equals 𝑥 cubed plus 𝑥 squared minus four 𝑥 minus four. At first glance, it seems difficult to factorize this expression, since the terms do not individually share a common factor. However, the expression can be factorized by grouping. Notice that 𝑥 cubed is simply 𝑥 times 𝑥 squared and 𝑥 squared is just one times itself. Similarly, four 𝑥 is 𝑥 times four, and four is just one times itself. The first two terms in order contain a factor of 𝑥 and one. And likewise, the last two terms contain a factor of 𝑥 and one. The other factor in the first two terms is 𝑥 squared, and the other factor in the last two terms is four. Therefore, we can factorize the first two terms with a factor of 𝑥 squared and the last two terms with a factor of four, giving 𝑥 squared times 𝑥 plus one minus four times 𝑥 plus one.

Now, because we have chosen to factorize the first two terms and the last two terms with this common term of 𝑥 plus one, we can factorize these again with a common factor of 𝑥 plus one. This gives us 𝑥 squared minus four times 𝑥 plus one. Now notice that four is a square number. So 𝑥 squared minus four can itself be easily factorized again by using the difference of two squares. This gives us 𝑥 minus two times 𝑥 plus two times 𝑥 plus one. We are now left with a product of linear binomials. So we cannot factorize any further.

Now, we have a product of three terms equaling zero. So at least one of the terms must itself be equal to zero. Therefore, 𝑥 minus two equals zero or 𝑥 plus two equals zero or 𝑥 plus one equals zero. Therefore, in set notation, the zeros of 𝑓 are negative two, negative one, and two.

Let’s now consider some word examples of applying these techniques to determine the zeros of polynomials, starting with a linear function.

Find the set of zeros of the function 𝑓 of 𝑥 equals one-third times 𝑥 minus four.

To find the zeros of a function, we set the function equal to zero, which means that one-third times 𝑥 minus four equals zero. This is a linear equation in 𝑥, so it’s easily solved without factorization. We can start by multiplying both sides by three, giving 𝑥 minus four equals zero. Adding four to both sides gives 𝑥 equals four. Therefore, the set of zeros of the function 𝑓 is the single value four.

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

Find, by factoring, the zeros of the function 𝑓 of 𝑥 equals 𝑥 squared plus two 𝑥 minus 35.

To find the zeros of a function, we set the function equal to zero, which means that 𝑥 squared plus two 𝑥 minus 35 equals zero. We can solve this equation for 𝑥 to find the zeros of 𝑓. There are several ways to do this, but the question asks us to find the zeros by factoring. To factorize the expression, we need to find two numbers that add to give two and multiply to give negative 35. An easy way to do this is to list the pairs of factors of 35. These are one and 35 itself and five and seven. Clearly, one and 35 will not add or subtract to make two, but five and seven will. So these are the factors we need.

Since we need to form negative 35, one of these numbers will have to be negative and the other positive. If we take negative five and positive seven, they will add to make two. We can therefore factorize this to give 𝑥 minus five times 𝑥 plus seven equals zero. Since we have a product of two terms equaling zero, at least one of the terms must be equal to zero itself. Therefore, either 𝑥 minus five equals zero or 𝑥 plus seven equals zero. We can solve the first equation for 𝑥 by adding five to both sides, giving 𝑥 equals five, and the second equation by subtracting seven from both sides, giving 𝑥 equals negative seven. Therefore, the set of zeros of the function is negative seven and five.

It’s worth noting that with more complicated polynomial functions, we can always verify that these values are correct by substituting them back into the original equation of the function and showing that the output is indeed zero. Doing this with our first value, negative seven, we get negative seven squared plus two times negative seven minus 35, which is equal to 49 minus 14 minus 35, which is indeed equal to zero. And doing the same for our second value, 𝑓 of five is equal to five squared plus two times five minus 35, which is again equal to zero. Therefore, these two values are indeed zeros of the function 𝑓. We also know that they must be the only zeros of 𝑓, since a quadratic function can never have more than two zeros.

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

Find, by factoring, the zeros of the function 𝑓 of 𝑥 equals nine 𝑥 squared plus nine 𝑥 minus 40.

To find the zeros of a function, we set the function equal to zero, which means that nine 𝑥 squared plus nine 𝑥 minus 40 equals zero. We can solve this equation for 𝑥 to find the zeros of 𝑓.

The question asks us to find the zeros by factoring. But here we have the slight complication that the polynomial is nonmonic. So we will need to also choose the correct coefficients for the 𝑥-terms in the factors. To factorize a nonmonic quadratic, we can use the grouping method. We need to find a factor pair of 𝑎𝑐 that adds to give 𝑏, where 𝑎 is the coefficient of 𝑥 squared, 𝑏 is the coefficient of 𝑥, and 𝑐 is the constant coefficient. Taking the product, 𝑎𝑐 is nine times negative 40, which is negative 360. 𝑏 is of course just nine. So, we need to find a pair of numbers that will multiply to give negative 360 and add to give nine.

Since the numbers multiply to give a negative number, one of them must be negative and the other must be positive. The first pair of numbers we might think of that adds to make nine are negative one and 10. But clearly these numbers are too small in magnitude to multiply to negative 360. To find any other pair of numbers that will add to give nine, we can subtract one from the negative number and add one to the positive number. We can skip a few to, say, negative 11 and 20. These multiply to give negative 220. So we’re still a little short. If we proceed in this way, we eventually find that negative 15 and positive 24 add together to make nine and multiply together to make negative 360.

A perhaps faster way to do this is by listing all of the factor pairs of 360, picking one of the pair to be negative and seeing which pair add to make nine. In this case, however, 360 happens to be a highly divisible number. So this might actually take longer. Now that we have the factor pair, we can rewrite the original nine 𝑥 in the equation as negative 15𝑥 plus 24𝑥. The grouping method ensures that we now have the same common factor between the first two terms and the last two terms, in this case three 𝑥 minus five. We can rewrite the expression on the left-hand side as three 𝑥 times three 𝑥 minus five plus eight times three 𝑥 minus five. This same factor, three 𝑥 minus five, is also common to both of these terms. So we can rewrite this again as three 𝑥 plus eight times three 𝑥 minus five.

We now have a product of two binomials that are linear in 𝑥 and have no more common factors. So we cannot factorize any further. We have a product of two terms that is equal to zero. Therefore, at least one of the terms must be equal to zero itself. This means that either three 𝑥 plus eight is equal to zero or three 𝑥 minus five is equal to zero. We can solve for 𝑥 in the first equation by subtracting eight and dividing by three, giving 𝑥 equals negative eight over three. And we can solve for 𝑥 in the second equation by adding five and dividing by three, giving 𝑥 equals five over three. Therefore, the set of zeros of the function 𝑓 is negative eight over three and five over three.

We can verify that these are in fact the zeros of 𝑓 by substituting the values into the original polynomial. 𝑓 of negative eight over three is equal to nine times negative eight over three all squared plus nine times negative eight over three minus 40, which is equal to nine times 64 over nine plus negative 72 over three minus 40, which is equal to 64 minus 24 minus 40, which is indeed equal to zero. And likewise, we can take 𝑓 of five over three, which eventually simplifies to 25 plus 15 minus 40, which is also equal to zero. Therefore, these are indeed the zeros of 𝑓. And since 𝑓 is a quadratic function, we know that there cannot be any more zeros.

In the remaining examples, we will find the zeros of polynomials of degree three or higher.

Find the set of zeros of the function 𝑓 of 𝑥 equals negative nine 𝑥 to the fourth plus 225𝑥 squared.

To find the zeros of a function, we set the function equal to zero, which means that negative nine 𝑥 to the fourth plus 225𝑥 squared is equal to zero. We can solve this equation for 𝑥 to find the zeros of 𝑓. We can attempt a factorization first. Notice that we clearly have a common factor of 𝑥 squared in all of the terms on the left-hand side. We can also look at the coefficients to see if we have a common factor there. Nine is divisible by one, three, and itself. Starting from the largest divisor, nine, is 225 divisible by nine? And it is. Nine times 25 is equal to 225. So we also have a common factor of nine, as well as 𝑥 squared. Therefore, we can factor this with nine 𝑥 squared. Then, inside the parentheses, we have negative 𝑥 squared plus 25.

Now we need to see if we can factorize this any further. Inside the parentheses, we have the number 25 and negative 𝑥 squared. 25 is a square number, five squared, so we have a difference of two squares. Recall that if we have the difference of two squares, 𝑎 squared and 𝑏 squared, we can factorize with the product 𝑎 minus 𝑏 times 𝑎 plus 𝑏. We can therefore factorize the left-hand side again to give nine 𝑥 squared times five minus 𝑥 times five plus 𝑥. And we now have a product of binomials that are linear in 𝑥 and a monomial in 𝑥 squared. So we cannot factorize any further.

Since we have a product of three terms equal to zero, at least one of the terms must be equal to zero itself. Therefore, either nine 𝑥 squared equals zero, five minus 𝑥 equals zero, or five plus 𝑥 equals zero. We can solve for 𝑥 in the first equation by dividing through by nine and taking the square root, giving 𝑥 equals zero. For the second equation, we can simply add 𝑥 to both sides to give 𝑥 equals five. And for the third equation, we can simply subtract five from both sides to give 𝑥 equals negative five. Therefore, the set of zeros for 𝑓 is negative five, zero, five.

In the next example, we will determine the zeros of a cubic using factoring.

Find the set of zeros of the function 𝑓 of 𝑥 equals 𝑥 times 𝑥 squared minus 81 minus two times 𝑥 squared minus 81.

To find the zeros of a function, we set the function equal to zero, which means 𝑥 times 𝑥 squared minus 81 minus two times 𝑥 squared minus 81 is equal to zero. We can solve this equation for 𝑥 to find the zeros of 𝑓. We can see immediately that all of the terms on the left-hand side have a common factor of 𝑥 squared minus 81. So we can factorize the left-hand side with this term. This gives us 𝑥 minus two times 𝑥 squared minus 81 equals zero.

Now notice that 81 is a square number, nine squared. So, in this second term, we have a difference of two squares. When we have an expression of the form 𝑎 squared minus 𝑏 squared, we can factorize this to give 𝑎 minus 𝑏 times 𝑎 plus 𝑏. In our case, this means we can factorize 𝑥 squared minus 81 to give 𝑥 minus nine times 𝑥 plus nine. We now have a product of binomial terms, linear in 𝑥. So we cannot factorize any further. This is a product of three terms, which is equal to zero. Therefore, at least one of the terms must itself be equal to zero. Therefore, either 𝑥 minus two equals zero, 𝑥 minus nine equals zero, or 𝑥 plus nine equals zero.

We can solve the first equation by adding two to both sides to give 𝑥 equals two, the second equation by adding nine to both sides to give 𝑥 equals nine, and the third equation by subtracting nine from both sides to give 𝑥 equals negative nine. Therefore, the set of zeros of 𝑓 is negative nine, two, and nine.

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

The function 𝑓 of 𝑥 equals 𝑎 squared 𝑥 squared plus 54𝑥 plus 81 and the function 𝑔 of 𝑥 equals 𝑎𝑥 plus nine have the same set of zeros. Find 𝑎 and the set of zeros.

To find the zeros of a function, we set the function equal to zero. For 𝑓, we have 𝑎 squared 𝑥 squared plus 54𝑥 plus 81 equals zero. And for 𝑔, we have 𝑎𝑥 plus nine equals zero. We have two equations and two unknowns. One way to solve these simultaneous equations is to rearrange one equation to express one unknown in terms of the other and then substitute it into the second equation.

We can choose the simpler linear equation to rearrange to give 𝑥 in terms of 𝑎. By subtracting nine and dividing through by 𝑎, we have 𝑥 equals negative nine over 𝑎. This assumes that 𝑎 is nonzero, since we cannot divide by zero. However, if 𝑎 were equal to zero, the original equation would have no solutions, since we have nine is equal to zero, which is of course false for any value of 𝑥. Substituting this expression for 𝑥 into the equation for 𝑓 of 𝑥 gives us 𝑎 squared times negative nine over 𝑎 all squared plus 54 times negative nine over 𝑎 plus 81 equals zero. Taking the square and expanding the parentheses gives 𝑎 squared times 81 over 𝑎 squared minus 486 over 𝑎 plus 81 equals zero.

The 𝑎 squared factor will cancel with the 𝑎 squared denominator, leaving 81 minus 486 over 𝑎 plus 81 equals zero, which will simplify to negative 486 over 𝑎 plus 162 equals zero. We can rearrange for 𝑎 by subtracting 162, dividing by negative 486, and taking the reciprocal, which gives 𝑎 equals negative 486 over negative 162, which is equal to three. So we have the first part of our answer: the value of 𝑎 is three.

Substituting this value of 𝑎 back into the rearranged equation for 𝑔 gives us 𝑥 equals negative nine over three, which is equal to negative three. Since 𝑔 is a linear function, it has at most one zero. So this value of 𝑥 must be the only zero of both 𝑓 and 𝑔. Therefore, the complete set of zeros of both 𝑓 and 𝑔 is negative three.

It’s worth noting that the function 𝑓, a quadratic, could have up to two zeros. However, the requirement that it shares zeros with 𝑔 and that both 𝑓 of 𝑥 equals zero and 𝑔 of 𝑥 equals zero ensures that it has only one. This is because the value of 𝑎 for these conditions is three. Substituting this value into the equation for 𝑓 gives us nine 𝑥 squared plus 54𝑥 plus 81 equals zero. Every term in this equation is divisible by nine. So we can divide through by nine to give 𝑥 squared plus six 𝑥 plus nine equals zero. This can be factorized into two binomials, with 𝑥 plus three times 𝑥 plus three. We therefore have a repeated root of negative three, giving us the same answer as before.

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

Find the set of zeros of the function 𝑓 of 𝑥 equals 𝑥 cubed minus four 𝑥 squared minus 25𝑥 plus 100 equals zero, where all three zeros take integer values.

To find the zeros of a function, we set the function equal to zero. So we have 𝑥 cubed minus four 𝑥 squared minus 25𝑥 plus 100 equals zero. We are given that all three zeros of the function take integer values. So we may be able to factorize the polynomial via the grouping method. In the first two terms, we have a common factor of 𝑥 squared, which we can factorize with 𝑥 minus four. We also have a common factor of negative 25 in the last two terms. And this also factorizes with the term 𝑥 minus four. Therefore, we have a common factor of 𝑥 minus four between the first two terms and the last two terms. We can therefore factorize the polynomial to give 𝑥 minus four times 𝑥 squared minus 25. 25 is a square number, five squared. So, in this second term, we have a difference of two squares.

Recall that when we have an expression in the form 𝑎 squared minus 𝑏 squared, it can be factorized as 𝑎 minus 𝑏 times 𝑎 plus 𝑏. We can therefore factorize 𝑥 squared minus 25 as 𝑥 minus five times 𝑥 plus five. We now have a product of three binomials linear in 𝑥. So we cannot factorize any further. Since we have a product of three terms equal to zero, at least one of them must be equal to zero itself. Therefore, either 𝑥 minus four equals zero, 𝑥 minus five equals zero, or 𝑥 plus five equals zero. And we can solve these three equations for 𝑥 to give 𝑥 equals four, 𝑥 equals five, and 𝑥 equals negative five. Therefore, the set of zeros of 𝑓 is negative five, four, and five.

Let’s finish this video by summarizing some key points. The zeros or roots of a polynomial 𝑓 of 𝑥 are the values 𝑥 equals 𝑎 such that 𝑓 of 𝑎 is equal to zero. If 𝑥 minus 𝑎 is a factor of the polynomial 𝑓 of 𝑥, then 𝑎 is a zero of 𝑓. That is, 𝑓 of 𝑎 is equal to zero. Likewise, the reverse statement is true. If 𝑓 of 𝑎 is equal to zero, then the binomial 𝑥 minus 𝑎 is a factor of the polynomial 𝑓 of 𝑥. We can verify that 𝑥 equals 𝑎 is a zero of a given polynomial 𝑓 of 𝑥 by checking that 𝑓 of 𝑎 equals zero. There are many techniques we can use to help us determine the roots of polynomials, including the quadratic formula, factoring by grouping, and the difference of two squares.