Question Video: Proving Polynomial Identities Mathematics

Determine which of the following expressions is equivalent to (π‘₯ + 𝑦 + 𝑧)Β². [A] 2π‘₯Β² + 2𝑦² + 2𝑧² [B] 2π‘₯Β² + 2𝑦² + 2𝑧² + π‘₯𝑦 + π‘₯𝑧 + 𝑦𝑧 [C] π‘₯Β² + 𝑦² + 𝑧² + π‘₯𝑦 + π‘₯𝑧 + 𝑦𝑧 [D] π‘₯Β² + 𝑦² + 𝑧² [E] π‘₯Β² + 𝑦² + 𝑧² + 2π‘₯𝑦 + 2π‘₯𝑧 + 2𝑦𝑧

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Video Transcript

Determine which of the following expressions is equivalent to π‘₯ plus 𝑦 plus 𝑧 squared. Is it (A) two π‘₯ squared plus two 𝑦 squared plus two 𝑧 squared? Is it (B) two π‘₯ squared plus two 𝑦 squared plus two 𝑧 squared plus π‘₯𝑦 plus π‘₯𝑧 plus 𝑦𝑧? Is it (C) π‘₯ squared plus 𝑦 squared plus 𝑧 squared plus π‘₯𝑦 plus π‘₯𝑧 plus 𝑦𝑧? (D) π‘₯ squared plus 𝑦 squared plus 𝑧 squared. Or is it (E) π‘₯ squared plus 𝑦 squared plus 𝑧 squared plus two π‘₯𝑦 plus two π‘₯𝑧 plus two 𝑦𝑧?

In order to decide which of the expressions is equivalent to π‘₯ 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 π‘₯ plus 𝑦 plus 𝑧 squared is π‘₯ plus 𝑦 plus 𝑧 times π‘₯ 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 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 𝑦, which is π‘₯𝑦; then π‘₯ times 𝑧, which is π‘₯𝑧. We’ll now repeat this process with the 𝑦. 𝑦 times π‘₯ is π‘₯𝑦, 𝑦 times 𝑦 is 𝑦 squared, and 𝑦 times 𝑧 is 𝑦𝑧. There’s just one term left; it’s this 𝑧. 𝑧 times π‘₯ is π‘₯𝑧. 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 π‘₯ squared, 𝑦 squared, and 𝑧 squared. Then π‘₯𝑦 plus π‘₯𝑦 is two π‘₯𝑦, π‘₯𝑧 plus π‘₯𝑧 is two π‘₯𝑧, and finally, 𝑦𝑧 plus 𝑦𝑧 is two 𝑦𝑧. And so when we distribute π‘₯ plus 𝑦 plus 𝑧 squared, we get π‘₯ squared plus 𝑦 squared plus 𝑧 squared plus two π‘₯𝑦 plus two π‘₯𝑧 plus two 𝑦𝑧. And so we see that the expression that is equivalent to π‘₯ plus 𝑦 plus 𝑧 squared is (E).

In fact, what we have is an identity. We can say that for all values of π‘₯, 𝑦, and 𝑧, π‘₯ plus 𝑦 plus 𝑧 squared is equal to π‘₯ squared plus 𝑦 squared plus 𝑧 squared plus two π‘₯𝑦 plus two π‘₯𝑧 plus two 𝑦𝑧.

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