### Video Transcript

In this video, weβll learn how to
find the set of zeros of a quadratic, cubic, or higher-degree polynomial
function. We begin by recalling what we
actually mean by the zeros of a function. Finding the zeros of a function is
the same as finding the roots. And these are the values of π₯ that
make the polynomial equal to zero.

If we think about this graphically,
we know that when the function π¦ equals π of π₯ is set equal to zero, weβre
interested in the places where our graph intersects the π₯-axis. And so the roots or zeros of the
function are the π₯-intercepts of the graph. But of course, we donβt always want
to draw the graphs of our functions. So weβre going to ask ourselves
what other techniques we have to solve a polynomial π of π₯ is equal to zero. Of course, for factorable
functions, we know we can factor the expression and find the roots that way. So letβs see what that might look
like.

Find, by factoring, the zeros of
the function π of π₯ equals π₯ squared plus two π₯ minus 35.

Remember the zeros of the function
are the same as the roots. And we find the roots of a function
by solving the equation π of π₯ is equal to zero. And so we find the zeros of our
function by setting π₯ squared plus two π₯ minus 35 equal to zero and solving for
π₯. Now the question actually tells us
the method to use. Weβre going to factor the
expression π₯ squared plus two π₯ minus 35.

Firstly, we see that this
expression is a quadratic, and so itβs going to factor into a pair of parentheses,
as shown. These will consist of a pair of
binomials. And the first term in each binomial
must be π₯ since π₯ multiplied by π₯ gives us π₯ squared. And then to find the numerical
part, we want to find two numbers who have a product; that means they multiply to
make negative 35 and they have a sum of two. Weβll begin by finding two numbers
that have a product of 35. And weβll consider their signs in a
moment.

Of course, we have one times
35. Well, we have five times seven. Since weβre trying to make negative
35 and we know a negative multiplied by a positive gives a negative, we know that
one number in our factor pair will have to be negative. So how will we ensure that these
numbers add to make two? Well, if we use negative five and
positive seven, we know negative five times positive seven is negative 35 but
negative five plus seven is two. And so weβre going to put negative
five and seven into our parentheses. And so the expression π₯ squared
plus two π₯ minus 35 is equal to π₯ minus five times π₯ plus seven.

So how does this help us solve the
equation π₯ squared plus two π₯ minus 35 is equal to zero? Well, we know that π₯ minus five
and π₯ plus seven are multiplying one another. And when they do, the answer is
zero. So what can we say about one or
other of these binomials? We know that for the product of two
numbers to be equal to zero, either one or other of the numbers must itself be equal
to zero. So in this case, that must mean
that either π₯ minus five is equal to zero or π₯ plus seven is equal to zero. Then we solve as we would any other
linear equation. In this case, we add five to both
sides. And we get π₯ equals five. And for our second equation, we
subtract seven. So π₯ equals negative seven.

Remember, the zeros of the function
are the values of π₯ that make the polynomial equal to zero. So the zeros of our function are
simply the numbers negative seven and five. Now, of course, we know that the
zeros are the values of π₯ that make the function equal to zero. So we can check our solutions by
substituting them into the expression π₯ squared plus two π₯ minus 35. In either case, we do get zero as
required. So we know that weβve done our
calculations correctly.

In our next example, weβll look at
how to find a set of zeros from a cubic function.

Find the set of zeroes of the
function π of π₯ equals π₯ times π₯ squared minus 81 minus two times π₯ squared
minus 81.

We might begin by recalling that
the zeros of a polynomial function are the values of π₯ that make it equal to
zero. And so we need to set the
expression π₯ times π₯ squared minus 81 minus two times π₯ squared minus 81 equal to
zero and solve for π₯.

Now, if we were to distribute our
parentheses, weβd probably spot that we have a cubic. Thatβs a polynomial whose order is
three. And so we might be thinking that we
need to distribute the parentheses and go from there. However, if we look really
carefully, we see that the two terms share a common factor. They share a factor of π₯ squared
minus 81. And so to solve this equation,
weβre going to factor by removing that common factor of π₯ squared minus 81.

If we divide the first term by π₯
squared minus 81, that leaves us simply with π₯. And then if we divide the second
term by π₯ squared minus 81, we get negative two. And so we can see that the
expression π₯ times π₯ squared minus 81 minus two times π₯ squared minus 81 is equal
to π₯ squared minus 81 times π₯ minus two. And so if we can fully factor an
expression when finding the zeros, itβs really helpful because we now need to ask
ourselves, which values of π₯ make this expression equal to zero?

And of course, since weβre
multiplying π₯ squared minus 81 by π₯ minus two, and that gives us zero, we can say
that either π₯ squared minus 81 must be equal to zero or π₯ minus two must be equal
to zero. Weβll now solve each of these
equations in turn. Weβll solve our first equation by
adding 81 to both sides. And that tells us that π₯ squared
is equal to 81. Weβre now going to take the square
root of both sides. But we do need to be a little bit
careful. Weβll need to take both the
positive and negative square root of 81.

And so the solutions to the
equation π₯ squared minus 81 equals zero are π₯ equals nine and π₯ equals negative
nine. The other equation is a little bit
more straightforward. Weβre just going to add two to both
sides. And so we find another solution to
the equation π of π₯ equals zero and, therefore, a zero of our function to be
two. And we found the solutions to the
equation π of π₯ equals zero, but the question asks us to find the set of zeros of
the function. And so we use these squiggly
brackets to represent the set containing the elements negative nine, two, and
nine.

And thatβs the answer to our
question. The set of zeros of the function π
of π₯ equals π₯ times π₯ squared minus 81 minus two times π₯ squared minus 81
contains the elements negative nine, two, and nine. Note, of course, that we could go
back to our original function to check whether these answers are correct. We would need to substitute each
one in turn and double-check that we do indeed get an answer of zero.

In our next example, weβll look at
how we can find the roots of a cubic equation by factoring using the factor
theorem.

Given that π of π₯ is equal to π₯
cubed plus three π₯ squared minus 13π₯ minus 15 and π of negative one equals zero,
find the other roots of π of π₯.

The roots of π of π₯ or the zeros
of the function of the values of π₯ that make the function equal to zero. And so we need to find the values
of π₯ such that the equation π₯ cubed plus three π₯ squared minus 13π₯ minus 15 is
equal to zero. And one technique that we have is
to fully factor that cubic expression. But we canβt do so easily without a
little bit of extra information. And that information is given to us
in the question. Weβre told π of negative one is
equal to zero. This bit of information means we
can use the factor theorem to help us factor our cubic.

The factor theorem says that if π
of π is equal to zero, then π₯ minus π is a factor of π of π₯. Comparing this general form with
our question, and we see we need to let π be equal to negative one. And we can therefore say that π₯
minus negative one, which is of course π₯ plus one, must be a factor of π of
π₯. So how does this help us? Well, this means that we can divide
our cubic π of π₯ by π₯ plus one. Exactly, we wonβt have a
remainder. And so when we do this division, we
should be left with a quadratic, which we can then factor.

There are a number of techniques
that we can use. Iβm going to use polynomial long
division using this bus stop method. I begin by asking myself, what is
π₯ cubed divided by π₯? π₯ cubed divided by π₯ is π₯
squared. Now I multiply π₯ squared by each
term in my divisor. π₯ squared times π₯ is π₯
cubed. And π₯ squared times one is π₯
squared. Next, we subtract the two terms we
just found from the two terms above them. π₯ cubed minus π₯ cubed is
zero. And then three π₯ squared minus π₯
squared is two π₯ squared.

We bring down the next term. And then we ask ourselves, whatβs
two π₯ squared divided by π₯? Two π₯ squared divided by π₯ is two
π₯. So we put two π₯ above three π₯
squared. Next, we multiply two π₯ squared by
each term in our divisor. Two π₯ squared times π₯ is two π₯
squared. And two π₯ times one is two π₯. Once again, we subtract. Now two π₯ squared minus two π₯
squared is zero. So we work out negative 13π₯ minus
two π₯, which is negative 15π₯.

Letβs bring down the last term. We now divide negative 15π₯ by π₯,
which is negative 15. And we then multiply negative 15 by
both terms in our divisor. Thatβs negative 15π₯ minus 15. We subtract once more, but negative
15π₯ minus 15 minus negative 15π₯ minus 15 is zero. And that tells us, as we expected,
that thereβs no remainder when we divide our cubic function by π₯ plus one. And it also means we can now
rewrite our cubic function as π₯ plus one times π₯ squared plus two π₯ minus 15.

And weβre now able to solve this
equation by factoring π₯ squared plus two π₯ minus 15. We know that thereβs going to be an
π₯ in each of our binomials. And then we want two numbers that
multiply to make negative 15 and add to make two. Those numbers are five and negative
three. So π₯ squared plus two π₯ minus 15
is equal to π₯ plus five times π₯ minus three.

Our last job is to solve this
equation. Now for the product of three
numbers to be equal to zero, any one of those numbers must itself be equal to
zero. So either π₯ plus one is equal to
zero, π₯ plus five is equal to zero, or π₯ minus three is equal to zero. If we subtract one from both sides
of our first equation, we get π₯ equals negative one. Solving our second equation, and we
get π₯ equals negative five. And solving our last equation, we
get π₯ equals three. Now we already knew that π₯ equals
negative one was a root to the equation. And so the other roots of π of π₯
are π₯ equals negative five and π₯ equals three.

In our final example, weβll look at
how to find a set of zeros of a quartic function or a polynomial function whose
order is four.

Find the set of zeros of the
function π of π₯ equals π₯ to the fourth power minus 17π₯ squared plus 16.

Remember, the zeros of a polynomial
function are the values of π₯ that make that function equal to zero. And so to find the zeros of our
function, we need to solve the equation π₯ to the fourth power minus 17π₯ squared
plus 16 equals zero. And one technique we have to find
the set of zeros is to factor the expression.

So how do we factor π₯ to the
fourth power minus 17π₯ squared plus 16? Well, itβs a little bit tricky to
spot, but this does look a little bit like a quadratic function. Weβre going to perform a
substitution. And weβre going to let π¦ be equal
to π₯ squared. And then if we consider π₯ to the
fourth power as being equal to π₯ squared squared, π₯ to the fourth power can be
written then as π¦ squared. Similarly, negative 17π₯ squared
can be written as negative 17π¦. So our equation becomes π¦ squared
minus 17π¦ plus 16 equals zero.

We now have a rather nice-looking
quadratic that we can solve by factoring. Weβre going to have two binomials
at the front of which we must have a π¦. We then need to find two numbers
that have a product, they multiply to make 16, and a sum of or they add to make
negative 17. Those numbers are negative one and
negative 16. Remember, a negative multiplied by
a negative is a positive. So our equation is π¦ minus one
times π¦ minus 16 equals zero. And for the product of these
binomials to be equal to zero, we can say that either π¦ minus one must be equal to
zero or π¦ minus 16 must be equal to zero. And if we solve as normal, we see
that π¦ must be equal to one or 16.

But of course, we were trying to
find the zeros of our function. And we said those are the values of
π₯ that make π of π₯ equal to zero. So weβre going to now replace π¦
with our original substitution, π₯ squared. And so π₯ squared is equal to one
or π₯ squared is equal to 16. Weβll solve by taking the square
root of both sides of each equation remembering, of course, to take both the
positive and negative square root of one and 16. And so we get π₯ equals positive or
negative one and π₯ equals positive or negative four. Now, we want to write this using
set notation. And so we use these squiggly
brackets. The set of zeros of the function π
of π₯ is the set containing the elements negative four, negative one, one, and
four.

Weβll now summarize the key points
from this lesson. Firstly, we saw that the zeros of a
function are the values of π₯ that make the function equal to zero. We also call this finding the roots
of an equation. And we also saw that if a function
is factorable, weβre going to factor that polynomial to help us solve π of π₯
equals zero.