### Video Transcript

In this video, weβll learn how to
prove polynomial identities. After watching this video, you
should be able to identify equivalent polynomials and prove that two polynomials are
equivalent. Before we get started, though,
letβs recap some keywords and notation. The word polynomial comes from
βpolyβ which means many and βnomialβ which here means term, so many terms. A polynomial can have constants,
variables, and exponents or powers, but those exponents will always be integer
values greater than or equal to zero. For example, π₯ cubed minus three
π₯ squared plus π¦ plus four is a polynomial, but one over π₯ is not since the power
of π₯ there would be negative one.

Finally, we need to understand what
we mean by the word identity. An identity is an equation thatβs
true for all values of π₯ or whatever variable weβre using. We sometimes use the three bar sign
to show that we have an identity. To prove identities, we might need
to manipulate expressions by, for example, factoring and simplifying. Letβs see what that might look
like.

Is the equation π₯ cubed minus π¦
cubed equals π₯ plus π¦ times π₯ minus π¦ times π₯ plus π¦ an identity?

Remember, an identity is an
equation thatβs true for all values of the variables involved. Here, our variables are π₯ and
π¦. Itβs simply not enough to
substitute a few values of π₯ and π¦ in and check that it works for these
values. Instead, weβre going to start with
the expression on the right-hand side, π₯ plus π¦ times π₯ minus π¦ times π₯ plus
π¦, and see how we might manipulate it to look more like a standard polynomial. Weβre going to distribute the
parentheses. To distribute three pairs of
parentheses, we begin by distributing any two. So letβs multiply π₯ plus π¦ by π₯
minus π¦.

We multiply the first term in each
expression. π₯ times π₯ gives us π₯
squared. We then multiply the outer
terms. π₯ times negative π¦ is negative
π₯π¦. We multiply the inner terms. π¦ multiplied by π₯ is another
π₯π¦. And finally we multiply the last
terms, giving us negative π¦ squared. And so we see the product of π₯
plus π¦ and π₯ minus π¦ is π₯ squared minus π₯π¦ plus π₯π¦ minus π¦ squared. But negative π₯π¦ plus π₯π¦ is
zero. And so our expression simplifies to
π₯ squared minus π¦ squared times π₯ plus π¦.

Letβs distribute again. Once again, we multiply the first
term in each expression. π₯ squared times π₯ is π₯
cubed. We multiply the outer terms, giving
us π₯ squared π¦. We then multiply the inner terms,
giving us negative π₯π¦ squared. And finally, we multiply the last
terms. Negative π¦ squared times π¦ is
negative π¦ cubed. Now here we need to note that π₯
squared π¦ and π₯π¦ squared are completely different terms. In the first term, the π₯ is being
squared before multiplying by π¦. And in the second, the π¦ is being
squared and then weβre multiplying it by π₯. So we canβt simplify any
further.

And so if we compare this
expression with our original π₯ cubed minus π¦ cubed, we see that these are not
equal. Since the expressions are not
equivalent, we donβt have an identity. This isnβt going to be true for all
values of π₯ and π¦. We can confirm this by finding a
single value of π₯ and π¦ for which the original equation doesnβt hold. Letβs try π₯ is equal to two and π¦
is equal to one. π₯ cubed minus π¦ cubed is then two
cubed minus one cubed, which is equal to seven. Then, π₯ plus π¦ times π₯ minus π¦
times π₯ plus π¦ is two plus one times two minus one times two plus one, which is
equal to nine.

Remember, since identities are true
for all values of the variables, then by showing that there is one set of variable
where this equation isnβt true, we see that we donβt have an identity. The answer then is no, itβs not an
identity.

In our next example, weβll look at
how we work with quotients. Thatβs fractional expressions.

Is the equation π₯ to the fourth
power minus π¦ to the fourth power over π₯ squared minus π¦ squared equals π₯
squared plus π¦ squared an identity?

Remember, an identity is an
equation that holds for all values of π₯ or all values of the variables. In this case, we need to decide
whether this equation holds for all values of π₯ and π¦. Now, to prove an equation is an
identity, itβs simply not enough to substitute a few values of π₯ and π¦ in and
check that it works for those values. Instead, weβre going to start with
the expression π₯ to the fourth power minus π¦ to the fourth power over π₯ squared
minus π¦ squared and see how we might manipulate that to look more like a standard
polynomial. Now, the key to simplifying here is
to spot that we have a quotient of two expressions that are the difference of two
squares.

Remember, π squared minus π
squared can be factored as π plus π times π minus π. Now what this means is we can
factor or factorize both parts, the numerator and the denominator. Now, in fact, weβre just going to
factor the numerator, and weβll see why in a moment. Comparing this expression to the
general form, weβre going to let π be equal to π₯ squared and π be equal to π¦
squared. And this is because π₯ squared
squared gives us π₯ to the fourth power. And we can therefore factor π₯ to
the fourth power minus π¦ to the fourth power as π₯ squared plus π¦ squared times π₯
squared minus π¦ squared.

Now, what do we notice? There is a common factor on both
the numerator and denominator of our fraction. And that common factor is π₯
squared minus π¦ squared. Weβre therefore going to divide
through by π₯ squared minus π¦ squared. On the numerator, that leaves us
with π₯ squared plus π¦ squared. And on the denominator, we get
one. So this simplifies simply to π₯
squared plus π¦ squared. We can therefore say that for all
values of π₯, π₯ to the fourth power minus π¦ to the fourth power over π₯ squared
minus π¦ squared is equal to π₯ squared plus π¦ squared. Since this is true for all values
of π₯, we know we have an identity. And so the answer is yes. This equation is indeed an
identity.

Weβll now consider a similar
example.

Is the equation π₯ squared plus π¦
squared over π₯ plus π¦ equals π₯ plus π¦ an identity?

Remember, an identity is an
equation thatβs true for all values of our variable. So for this equation to be an
identity, π₯ squared plus π¦ squared over π₯ plus π¦ must be equal to π₯ plus π¦ for
all values of π₯ and π¦. One method we have is to manipulate
this fraction and see if we can simplify it. The problem is, we would usually
factor the expressions on the numerator and/or the denominator to do so. And these arenβt factorable. Now, this might be a hint that the
expression on the left is not equal to that on the right for all values of π₯ and
π¦. And since an identity is true for
all values of π₯ and π¦, if we can find just one set of values where this equation
doesnβt hold, then we can show itβs not an identity.

Letβs try letting π₯ be equal to
one and π¦ be equal to two. Then, the expression on the left
becomes one squared plus two squared over one plus two, which is equal to
five-thirds. The expression on the right,
however, is simply one plus two, which is equal to three. Itβs quite clear to us that
five-thirds is not equal to three. And so we found a value of π₯ and
π¦ such that this equation doesnβt hold. It, therefore, cannot be an
identity; it doesnβt hold for all values of π₯ and π¦.

In our next example, weβll look at
how to manipulate expressions to find equivalent expressions.

Determine which of the following
expressions is equivalent to π₯ plus π¦ plus π§ squared. Is it (A) two π₯ squared plus two
π¦ squared plus two π§ squared? Is it (B) two π₯ squared plus two
π¦ squared plus two π§ squared plus π₯π¦ plus π₯π§ plus π¦π§? Is it (C) π₯ squared plus π¦
squared plus π§ squared plus π₯π¦ plus π₯π§ plus π¦π§? (D) π₯ squared plus π¦ squared plus
π§ squared. Or is it (E) π₯ squared plus π¦
squared plus π§ squared plus two π₯π¦ plus two π₯π§ plus two π¦π§?

In order to decide which of the
expressions is equivalent to π₯ plus π¦ plus π§ squared, weβre going to distribute
the parentheses or expand the brackets. Now, in doing so, we need to be
really careful. When we square an expression, we
multiply it by itself. And so π₯ plus π¦ plus π§ squared
is π₯ plus π¦ plus π§ times π₯ plus π¦ plus π§. A common mistake here is to simply
square each of the individual terms. That would give us answer (D),
which is in fact incorrect. So how are we going to distribute
these parentheses? Well, weβll need to be quite
methodical.

Letβs begin by taking the π₯ in the
first set of parentheses and multiplying it by everything in the second. We get π₯ times π₯, which is π₯
squared; π₯ times π¦, which is π₯π¦; then π₯ times π§, which is π₯π§. Weβll now repeat this process with
the π¦. π¦ times π₯ is π₯π¦, π¦ times π¦ is
π¦ squared, and π¦ times π§ is π¦π§. Thereβs just one term left; itβs
this π§. π§ times π₯ is π₯π§. We then multiply π§ by π¦ to get
π¦π§. And finally, π§ times π§ is π§
squared.

So letβs simplify a little
further. We have π₯ squared, π¦ squared, and
π§ squared. Then π₯π¦ plus π₯π¦ is two π₯π¦,
π₯π§ plus π₯π§ is two π₯π§, and finally, π¦π§ plus π¦π§ is two π¦π§. And so when we distribute π₯ plus
π¦ plus π§ squared, we get π₯ squared plus π¦ squared plus π§ squared plus two π₯π¦
plus two π₯π§ plus two π¦π§. And so we see that the expression
that is equivalent to π₯ plus π¦ plus π§ squared is (E).

In fact, what we have is an
identity. We can say that for all values of
π₯, π¦, and π§, π₯ plus π¦ plus π§ squared is equal to π₯ squared plus π¦ squared
plus π§ squared plus two π₯π¦ plus two π₯π§ plus two π¦π§.

In our next example, weβll look at
how to perform a process called equating coefficients.

Given that three times two π₯ plus
five minus four times π₯ minus one is equivalent to ππ₯ plus π for integers π and
π, calculate the value of π and π.

We have three bars equating these
two expressions, and so this is an identity. With identities, the equations hold
for all values of π₯. Our job is to find the values of π
and π, so weβre going to make the expression on the left-hand side look a little
bit more like the one on the right. To do so, weβre going to begin by
distributing the parentheses or expanding the brackets. Letβs begin with the first pair of
parentheses. Weβre going to multiply three by
the two π₯, giving us six π₯. We then multiply three by five,
giving us 15.

We now move on to the second pair
of parentheses, but we need to be careful here. We have a negative four, so weβre
multiplying negative four by π₯ to get negative four π₯. And then weβre multiplying negative
four by negative one, giving us positive four. So this expression is six π₯ plus
15 minus four π₯ plus four, which we can see simplifies to two π₯ plus 19. Now weβre told that this is
equivalent to an expression ππ₯ plus π for all values of π₯. So two π₯ plus 19 is equal to ππ₯
plus π.

Now that the expressions look
similar, weβre actually going to do something called equating coefficients. In other words, weβre going to
compare the numbers in front of the π₯ and the constants. Letβs begin by comparing the
coefficient of π₯. We have two π₯ on the left-hand
side and ππ₯ on the right. The coefficients are two and π,
respectively. So we can say that two must be
equal to π.

Weβll repeat this process comparing
constants. Now if we really think about it,
this is a bit like comparing coefficients of π₯ to the power of zero, since π₯ to
the power of zero is one. On the left-hand side, our constant
is 19, whereas on the right itβs π, so 19 must be equal to π. And so we can say that π is two
and π is 19.

In this video, we learned that an
identity is an equation thatβs true for all values of π₯ or for all values of
whatever variables weβre working with. We also saw that to prove an
equation is an identity, we can manipulate one or both expressions. And we do so by expanding,
factoring, and simplifying, and so on. However, to prove an equation is
not an identity, we simply need to find one value of π₯ or whatever variables weβre
working in for which the equation doesnβt hold.