### 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 π₯
for which π of π₯ is equal to zero.

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

In set notation, we could write the
set of zeros of π 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 π of π₯ 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 π of π₯ equals π₯ squared plus five π₯ 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: π₯ plus two and π₯ plus three.

Weβve now expressed π of π₯ as a
product of two factors; letβs call them π and π. If we set π of π₯ equals zero,
then ππ is equal to zero. This implies that either π or π
must be equal to zero or both of them. Therefore, if π of π₯ is equal to
zero, then either π₯ plus two equals zero or π₯ plus three equals zero. These linear equations can be
easily solved for π₯, giving π₯ equals negative two and π₯ equals negative
three. Therefore, the zeros of π are
negative two and negative three.

Itβs 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
π of π₯ equals π₯ cubed plus π₯ squared minus four π₯ 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 π₯ cubed is simply π₯
times π₯ squared and π₯ squared is just one times itself. Similarly, four π₯ is π₯ times
four, and four is just one times itself. The first two terms in order
contain a factor of π₯ and one. And likewise, the last two terms
contain a factor of π₯ and one. The other factor in the first two
terms is π₯ squared, and the other factor in the last two terms is four. Therefore, we can factorize the
first two terms with a factor of π₯ squared and the last two terms with a factor of
four, giving π₯ squared times π₯ plus one minus four times π₯ plus one.

Now, because we have chosen to
factorize the first two terms and the last two terms with this common term of π₯
plus one, we can factorize these again with a common factor of π₯ plus one. This gives us π₯ squared minus four
times π₯ plus one. Now notice that four is a square
number. So π₯ squared minus four can itself
be easily factorized again by using the difference of two squares. This gives us π₯ minus two times π₯
plus two times π₯ 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, π₯ minus two equals zero
or π₯ plus two equals zero or π₯ plus one equals zero. Therefore, in set notation, the
zeros of π are negative two, negative one, and two.

Letβs 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 π of π₯ equals one-third times π₯ minus four.

To find the zeros of a function, we
set the function equal to zero, which means that one-third times π₯ minus four
equals zero. This is a linear equation in π₯, so
itβs easily solved without factorization. We can start by multiplying both
sides by three, giving π₯ minus four equals zero. Adding four to both sides gives π₯
equals four. Therefore, the set of zeros of the
function π 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 π of π₯ equals π₯ squared plus two π₯ minus 35.

To find the zeros of a function, we
set the function equal to zero, which means that π₯ squared plus two π₯ minus 35
equals zero. We can solve this equation for π₯
to find the zeros of π. 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 π₯ minus five times π₯ 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 π₯ minus five
equals zero or π₯ plus seven equals zero. We can solve the first equation for
π₯ by adding five to both sides, giving π₯ equals five, and the second equation by
subtracting seven from both sides, giving π₯ equals negative seven. Therefore, the set of zeros of the
function is negative seven and five.

Itβs 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, π 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 π. We also know that they must be the
only zeros of π, 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 π of π₯ equals nine π₯ squared plus nine π₯ minus 40.

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

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 π₯-terms in the factors. To factorize a nonmonic quadratic,
we can use the grouping method. We need to find a factor pair of
ππ that adds to give π, where π is the coefficient of π₯ squared, π is the
coefficient of π₯, and π is the constant coefficient. Taking the product, ππ is nine
times negative 40, which is negative 360. π 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 weβre 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 π₯ in the equation as negative 15π₯ plus 24π₯. 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 π₯ minus five. We can rewrite the expression on
the left-hand side as three π₯ times three π₯ minus five plus eight times three π₯
minus five. This same factor, three π₯ minus
five, is also common to both of these terms. So we can rewrite this again as
three π₯ plus eight times three π₯ minus five.

We now have a product of two
binomials that are linear in π₯ 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 π₯
plus eight is equal to zero or three π₯ minus five is equal to zero. We can solve for π₯ in the first
equation by subtracting eight and dividing by three, giving π₯ equals negative eight
over three. And we can solve for π₯ in the
second equation by adding five and dividing by three, giving π₯ equals five over
three. Therefore, the set of zeros of the
function π is negative eight over three and five over three.

We can verify that these are in
fact the zeros of π by substituting the values into the original polynomial. π 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 π 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 π. And since π 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 π of π₯ equals negative nine π₯ to the fourth plus 225π₯ squared.

To find the zeros of a function, we
set the function equal to zero, which means that negative nine π₯ to the fourth plus
225π₯ squared is equal to zero. We can solve this equation for π₯
to find the zeros of π. We can attempt a factorization
first. Notice that we clearly have a
common factor of π₯ 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 π₯ squared. Therefore, we can factor this with
nine π₯ squared. Then, inside the parentheses, we
have negative π₯ 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 π₯ 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, π squared and π squared, we can factorize with the
product π minus π times π plus π. We can therefore factorize the
left-hand side again to give nine π₯ squared times five minus π₯ times five plus
π₯. And we now have a product of
binomials that are linear in π₯ and a monomial in π₯ 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 π₯ squared
equals zero, five minus π₯ equals zero, or five plus π₯ equals zero. We can solve for π₯ in the first
equation by dividing through by nine and taking the square root, giving π₯ equals
zero. For the second equation, we can
simply add π₯ to both sides to give π₯ equals five. And for the third equation, we can
simply subtract five from both sides to give π₯ equals negative five. Therefore, the set of zeros for π
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 π of π₯ equals π₯ times π₯ squared minus 81 minus two times π₯ squared
minus 81.

To find the zeros of a function, we
set the function equal to zero, which means π₯ times π₯ squared minus 81 minus two
times π₯ squared minus 81 is equal to zero. We can solve this equation for π₯
to find the zeros of π. We can see immediately that all of
the terms on the left-hand side have a common factor of π₯ squared minus 81. So we can factorize the left-hand
side with this term. This gives us π₯ minus two times π₯
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 π squared minus π squared, we can factorize this to give π minus π times π
plus π. In our case, this means we can
factorize π₯ squared minus 81 to give π₯ minus nine times π₯ plus nine. We now have a product of binomial
terms, linear in π₯. 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 π₯ minus two
equals zero, π₯ minus nine equals zero, or π₯ plus nine equals zero.

We can solve the first equation by
adding two to both sides to give π₯ equals two, the second equation by adding nine
to both sides to give π₯ equals nine, and the third equation by subtracting nine
from both sides to give π₯ equals negative nine. Therefore, the set of zeros of π
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 π of π₯ equals π
squared π₯ squared plus 54π₯ plus 81 and the function π of π₯ equals ππ₯ plus nine
have the same set of zeros. Find π and the set of zeros.

To find the zeros of a function, we
set the function equal to zero. For π, we have π squared π₯
squared plus 54π₯ plus 81 equals zero. And for π, we have ππ₯ 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 π₯ in terms of π. By subtracting nine and dividing
through by π, we have π₯ equals negative nine over π. This assumes that π is nonzero,
since we cannot divide by zero. However, if π 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 π₯. Substituting this expression for π₯
into the equation for π of π₯ gives us π squared times negative nine over π all
squared plus 54 times negative nine over π plus 81 equals zero. Taking the square and expanding the
parentheses gives π squared times 81 over π squared minus 486 over π plus 81
equals zero.

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

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

Itβs worth noting that the function
π, a quadratic, could have up to two zeros. However, the requirement that it
shares zeros with π and that both π of π₯ equals zero and π of π₯ equals zero
ensures that it has only one. This is because the value of π for
these conditions is three. Substituting this value into the
equation for π gives us nine π₯ squared plus 54π₯ plus 81 equals zero. Every term in this equation is
divisible by nine. So we can divide through by nine to
give π₯ squared plus six π₯ plus nine equals zero. This can be factorized into two
binomials, with π₯ plus three times π₯ 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 π of π₯ equals π₯ cubed minus four π₯ squared minus 25π₯ 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 π₯ cubed minus four π₯
squared minus 25π₯ 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 π₯ squared, which we can factorize with π₯ minus four. We also have a common factor of
negative 25 in the last two terms. And this also factorizes with the
term π₯ minus four. Therefore, we have a common factor
of π₯ minus four between the first two terms and the last two terms. We can therefore factorize the
polynomial to give π₯ minus four times π₯ 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 π squared minus π squared, it can be factorized as π minus
π times π plus π. We can therefore factorize π₯
squared minus 25 as π₯ minus five times π₯ plus five. We now have a product of three
binomials linear in π₯. 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 π₯ minus four
equals zero, π₯ minus five equals zero, or π₯ plus five equals zero. And we can solve these three
equations for π₯ to give π₯ equals four, π₯ equals five, and π₯ equals negative
five. Therefore, the set of zeros of π
is negative five, four, and five.

Letβs finish this video by
summarizing some key points. The zeros or roots of a polynomial
π of π₯ are the values π₯ equals π such that π of π is equal to zero. If π₯ minus π is a factor of the
polynomial π of π₯, then π is a zero of π. That is, π of π is equal to
zero. Likewise, the reverse statement is
true. If π of π is equal to zero, then
the binomial π₯ minus π is a factor of the polynomial π of π₯. We can verify that π₯ equals π is
a zero of a given polynomial π of π₯ by checking that π of π 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.