Lesson Video: Proving Polynomial Identities Mathematics

In this video, we will learn how to prove polynomial identities.

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Video Transcript

In this video, we’ll 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, let’s recap some keywords and notation. The word polynomial comes from “poly” which means many and “nomial” 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, 𝑥 cubed minus three 𝑥 squared plus 𝑦 plus four is a polynomial, but one over 𝑥 is not since the power of 𝑥 there would be negative one.

Finally, we need to understand what we mean by the word identity. An identity is an equation that’s true for all values of 𝑥 or whatever variable we’re 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. Let’s see what that might look like.

Is the equation 𝑥 cubed minus 𝑦 cubed equals 𝑥 plus 𝑦 times 𝑥 minus 𝑦 times 𝑥 plus 𝑦 an identity?

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

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

Let’s distribute again. Once again, we multiply the first term in each expression. 𝑥 squared times 𝑥 is 𝑥 cubed. We multiply the outer terms, giving us 𝑥 squared 𝑦. We then multiply the inner terms, giving us negative 𝑥𝑦 squared. And finally, we multiply the last terms. Negative 𝑦 squared times 𝑦 is negative 𝑦 cubed. Now here we need to note that 𝑥 squared 𝑦 and 𝑥𝑦 squared are completely different terms. In the first term, the 𝑥 is being squared before multiplying by 𝑦. And in the second, the 𝑦 is being squared and then we’re multiplying it by 𝑥. So we can’t simplify any further.

And so if we compare this expression with our original 𝑥 cubed minus 𝑦 cubed, we see that these are not equal. Since the expressions are not equivalent, we don’t have an identity. This isn’t going to be true for all values of 𝑥 and 𝑦. We can confirm this by finding a single value of 𝑥 and 𝑦 for which the original equation doesn’t hold. Let’s try 𝑥 is equal to two and 𝑦 is equal to one. 𝑥 cubed minus 𝑦 cubed is then two cubed minus one cubed, which is equal to seven. Then, 𝑥 plus 𝑦 times 𝑥 minus 𝑦 times 𝑥 plus 𝑦 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 isn’t true, we see that we don’t have an identity. The answer then is no, it’s not an identity.

In our next example, we’ll look at how we work with quotients. That’s fractional expressions.

Is the equation 𝑥 to the fourth power minus 𝑦 to the fourth power over 𝑥 squared minus 𝑦 squared equals 𝑥 squared plus 𝑦 squared an identity?

Remember, an identity is an equation that holds for all values of 𝑥 or all values of the variables. In this case, we need to decide whether this equation holds for all values of 𝑥 and 𝑦. Now, to prove an equation is an identity, it’s simply not enough to substitute a few values of 𝑥 and 𝑦 in and check that it works for those values. Instead, we’re going to start with the expression 𝑥 to the fourth power minus 𝑦 to the fourth power over 𝑥 squared minus 𝑦 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, 𝑎 squared minus 𝑏 squared can be factored as 𝑎 plus 𝑏 times 𝑎 minus 𝑏. Now what this means is we can factor or factorize both parts, the numerator and the denominator. Now, in fact, we’re just going to factor the numerator, and we’ll see why in a moment. Comparing this expression to the general form, we’re going to let 𝑎 be equal to 𝑥 squared and 𝑏 be equal to 𝑦 squared. And this is because 𝑥 squared squared gives us 𝑥 to the fourth power. And we can therefore factor 𝑥 to the fourth power minus 𝑦 to the fourth power as 𝑥 squared plus 𝑦 squared times 𝑥 squared minus 𝑦 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 𝑥 squared minus 𝑦 squared. We’re therefore going to divide through by 𝑥 squared minus 𝑦 squared. On the numerator, that leaves us with 𝑥 squared plus 𝑦 squared. And on the denominator, we get one. So this simplifies simply to 𝑥 squared plus 𝑦 squared. We can therefore say that for all values of 𝑥, 𝑥 to the fourth power minus 𝑦 to the fourth power over 𝑥 squared minus 𝑦 squared is equal to 𝑥 squared plus 𝑦 squared. Since this is true for all values of 𝑥, we know we have an identity. And so the answer is yes. This equation is indeed an identity.

We’ll now consider a similar example.

Is the equation 𝑥 squared plus 𝑦 squared over 𝑥 plus 𝑦 equals 𝑥 plus 𝑦 an identity?

Remember, an identity is an equation that’s true for all values of our variable. So for this equation to be an identity, 𝑥 squared plus 𝑦 squared over 𝑥 plus 𝑦 must be equal to 𝑥 plus 𝑦 for all values of 𝑥 and 𝑦. 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 aren’t 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 𝑥 and 𝑦. And since an identity is true for all values of 𝑥 and 𝑦, if we can find just one set of values where this equation doesn’t hold, then we can show it’s not an identity.

Let’s try letting 𝑥 be equal to one and 𝑦 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. It’s quite clear to us that five-thirds is not equal to three. And so we found a value of 𝑥 and 𝑦 such that this equation doesn’t hold. It, therefore, cannot be an identity; it doesn’t hold for all values of 𝑥 and 𝑦.

In our next example, we’ll look at how to manipulate expressions to find equivalent expressions.

Determine which of the following expressions is equivalent to 𝑥 plus 𝑦 plus 𝑧 squared. Is it (A) two 𝑥 squared plus two 𝑦 squared plus two 𝑧 squared? Is it (B) two 𝑥 squared plus two 𝑦 squared plus two 𝑧 squared plus 𝑥𝑦 plus 𝑥𝑧 plus 𝑦𝑧? Is it (C) 𝑥 squared plus 𝑦 squared plus 𝑧 squared plus 𝑥𝑦 plus 𝑥𝑧 plus 𝑦𝑧? (D) 𝑥 squared plus 𝑦 squared plus 𝑧 squared. Or is it (E) 𝑥 squared plus 𝑦 squared plus 𝑧 squared plus two 𝑥𝑦 plus two 𝑥𝑧 plus two 𝑦𝑧?

In order to decide which of the expressions is equivalent to 𝑥 plus 𝑦 plus 𝑧 squared, we’re 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 𝑥 plus 𝑦 plus 𝑧 squared is 𝑥 plus 𝑦 plus 𝑧 times 𝑥 plus 𝑦 plus 𝑧. 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, we’ll need to be quite methodical.

Let’s begin by taking the 𝑥 in the first set of parentheses and multiplying it by everything in the second. We get 𝑥 times 𝑥, which is 𝑥 squared; 𝑥 times 𝑦, which is 𝑥𝑦; then 𝑥 times 𝑧, which is 𝑥𝑧. We’ll now repeat this process with the 𝑦. 𝑦 times 𝑥 is 𝑥𝑦, 𝑦 times 𝑦 is 𝑦 squared, and 𝑦 times 𝑧 is 𝑦𝑧. There’s just one term left; it’s this 𝑧. 𝑧 times 𝑥 is 𝑥𝑧. We then multiply 𝑧 by 𝑦 to get 𝑦𝑧. And finally, 𝑧 times 𝑧 is 𝑧 squared.

So let’s simplify a little further. We have 𝑥 squared, 𝑦 squared, and 𝑧 squared. Then 𝑥𝑦 plus 𝑥𝑦 is two 𝑥𝑦, 𝑥𝑧 plus 𝑥𝑧 is two 𝑥𝑧, and finally, 𝑦𝑧 plus 𝑦𝑧 is two 𝑦𝑧. And so when we distribute 𝑥 plus 𝑦 plus 𝑧 squared, we get 𝑥 squared plus 𝑦 squared plus 𝑧 squared plus two 𝑥𝑦 plus two 𝑥𝑧 plus two 𝑦𝑧. And so we see that the expression that is equivalent to 𝑥 plus 𝑦 plus 𝑧 squared is (E).

In fact, what we have is an identity. We can say that for all values of 𝑥, 𝑦, and 𝑧, 𝑥 plus 𝑦 plus 𝑧 squared is equal to 𝑥 squared plus 𝑦 squared plus 𝑧 squared plus two 𝑥𝑦 plus two 𝑥𝑧 plus two 𝑦𝑧.

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

Given that three times two 𝑥 plus five minus four times 𝑥 minus one is equivalent to 𝑎𝑥 plus 𝑏 for integers 𝑎 and 𝑏, calculate the value of 𝑎 and 𝑏.

We have three bars equating these two expressions, and so this is an identity. With identities, the equations hold for all values of 𝑥. Our job is to find the values of 𝑎 and 𝑏, so we’re going to make the expression on the left-hand side look a little bit more like the one on the right. To do so, we’re going to begin by distributing the parentheses or expanding the brackets. Let’s begin with the first pair of parentheses. We’re going to multiply three by the two 𝑥, giving us six 𝑥. 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 we’re multiplying negative four by 𝑥 to get negative four 𝑥. And then we’re multiplying negative four by negative one, giving us positive four. So this expression is six 𝑥 plus 15 minus four 𝑥 plus four, which we can see simplifies to two 𝑥 plus 19. Now we’re told that this is equivalent to an expression 𝑎𝑥 plus 𝑏 for all values of 𝑥. So two 𝑥 plus 19 is equal to 𝑎𝑥 plus 𝑏.

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

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

In this video, we learned that an identity is an equation that’s true for all values of 𝑥 or for all values of whatever variables we’re 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 𝑥 or whatever variables we’re working in for which the equation doesn’t hold.

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