Prove and use polynomial identities.

Assignment

  • All vocabulary copied into notes
  • p152 18–35, 44–69

Polynomial Identities

The entirety of this section focuses on shortcuts for multiplying and factoring polynomials.

Square of a Sum (and Difference)

\[(a+b)^2 = a^2 + 2ab + b^2\] \[(a-b)^2 = a^2 - 2ab + b^2\]

The book doesn’t mention the difference version specifically, but worth keeping in mind.

Difference of Squares

\[a^2 - b^2 = (a+b)(a-b)\]

Keep in mind that perfect squares are not a necessity to move from the left side to the right. The polynomial $x-1$ can be factored with difference of squares with the result $(\sqrt{x}+\sqrt{1})(\sqrt{x}-\sqrt{1})$.

Sum and Difference of Cubes

\[a^3-b^3 = (a-b)(a^2 + ab + b^2)\] \[a^3+b^3 = (a+b)(a^2 - ab + b^2)\]

While the others you have seen before, if not already committed to memory, these will likely be new. If you didn’t know the rule for difference of squares, you could probably still factor it using your other tools like grouping. For the cubed version, that will be tougher with what you know at the moment.

So, write these down often in an effort to start memorizing them.

Pascal’s Triangle and Binomial Expansion

This next shortcut, which is about binomial expansion, requires a little background knowledge first. Here is the binomial $(a+b)$ raised to a number of powers.

\[\begin{align*} (a+b)^0 &= 1\\ (a+b)^1 &= a + b \\ (a+b)^2 &= a^2 + 2ab + b^2 \\ (a+b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3 \\ (a+b)^4 &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \\ \end{align*}\]

The goal here is that given any binomial and any power, you can expand it without having to go through all the tedious multiplication. So, we’re going to look for patterns we can exploit.

Terms and Powers

The first thing is that the number of terms is one more than the power. Pretty straightforward.

Then, note that the terms from the original binomial bookends the result, and their power matches the original. Then, as you move inward, that power decreases. Where you don’t have a $b$, all of sudden you do, then you have 2, and 3, and so on. Meanwhile, the power of $a$ is decreasing until it’s gone $(a^0)$.

So, when dealing with your terms, the first term is raised to the $n$-th power and it’s power goes down as move across. The second term is not present in the first term, but then does in the second with a power of 1 and then increases as you go across.

A good way to check if you’ve done the above correctly is by adding the powers of each term. It should add up to the original.

The Coeffecients

Here are the same binomial expansions from above, but this time I’ve stripped out everything but the coeffecients.

\[\begin{align*} (a+b)^0 &= 1\\ (a+b)^1 &= 1 + 1 \\ (a+b)^2 &= 1 + 2 + 1 \\ (a+b)^3 &= 1 + 3 + 3 + 1 \\ (a+b)^4 &= 1 + 4 + 6 + 4 + 1 \\ \end{align*}\]

OK, let’s strip out some more stuff and realign it to see if we can get a pattern.

\[\begin{array}{c} 1\\ 1\quad 1\\ 1\quad 2\quad 1\\ 1\quad 3\quad 3\quad 1\\ 1\quad 4\quad 6\quad 4\quad 1\\ \end{array}\]

The edges are simple, just ones. But inside, each number is the sum of the two directly above it. This is a famous mathematical pattern called Pascal’s Triangle. It’s one of those things that pops up in a few different places, but here we use it to develop the coeffecients for each term.

To use it, use your power to find your row. So a cubic expansion would need row three. Keep in mind that the top row is really row zero, and count down. You’ll find your four needed coefficients. Put them in from of the terms you created from before.

Example

Probably best that we go through one. Let’s expand $(x-3)^4$. First the terms and powers. I suggest keeping everything in parentheses until you are ready to simplify.

\[(x)^4 + (x)^3(-3) + (x)^2(-3)^2 + (x)(-3)^3 + (-3)^4\]

Next, we look at the Pascal’s Triangle to find our coefficients. Row 4, which is really the fifth row from the top gives us 1, 4, 6, 4, 1.

\[1(x)^4 + 4(x)^3(-3) + 6(x)^2(-3)^2 + 4(x)(-3)^3 + 1(-3)^4\]

Now we can start simplifying.

\[x^4 - 12x^3 + 54x^2 - 108x + 81\]

Example 2

One more, a little more complicated this time. $(2x+y^2)^5$

\[(2x)^5 + (2x)^4(y^2) + (2x)^3(y^2)^2 + (2x)^2(y^2)^3 + (2x)^1(y^2)^4 + (y^2)^5\]

Now the coefficients. Our triangle is missing that row, but we can come up with it pretty quickly.

\[(2x)^5 + 5(2x)^4(y^2) + 10(2x)^3(y^2)^2 + 10(2x)^2(y^2)^3 + 5(2x)^1(y^2)^4 + (y^2)^5\]

And the final result.

\[32x^5 + 80x^4y^2 + 80x^3y^4 + 40x^2y^6 + 10xy^8 + y^{10}\]