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Question 1

Can we multiply polynomials? You bet!

Here’s the polynomial 2x2x+12x^2-x+1.
I6SS-image208

If we want to multiply this polynomial by 33 we just have to replace each dot and each antidot with three copies of it. (We want to triple all the quantities we see.)
I6SS-image210

We literally see that 3(2x2x+1)3\left(2x^2-x+1\right) is 6x23x+36x^2-3x+3.

Suppose we wish to multiply 2x2x+12x^2-x+1 by 3-3 instead. This means we want the anti-version of tripling all the quantities we see. So each dot in the picture of 2x2x+12x^2-x+1 is to be replaced with three antidots and each antidot with three dots.
I6SS-image215

We have 3(2x2x+1)=6x2+3x3-3\left(2x2-x+1\right)=-6x2+3x-3. We could also say that 3(2x2x+1)-3\left(2x^2-x+1\right) is the anti-version of 3(2x2x+1)3\left(2x^2-x+1\right).

Now suppose we wish to multiply 2x2x+12x^2-x+1 by x+1x+1. Since x+1x+1 looks like this
I6SS-image221

we need to replace each dot in the picture of 2x2x+12x^2-x+1 with one-dot-and-one-dot, and each antidot with the anti-version of this, which is one-antidot-and-one-antidot. (This is now getting fun!)
I6SS-image222

After some annihilations we see that (x+1)×(2x2x+1)\left(x+1\right) \times \left(2x^2-x+1\right) equals 2x3+x2+12x3+x2+1.

Now let’s multiply 2x2x+12x^2-x+1 with x2x-2, which looks like this.
I6SS-image227

Each dot is to be replaced by one-dot-and-two-antidots, and each antidot with the opposite of this.
I6SS-image228

We see (x2)(2x2x+1)=2x35x2+3x2\left(x-2\right) \left(2x^2-x+1\right) = 2x^3 - 5x^2 +3x - 2.

Okay, your turn. Try 2x2x+12x^2-x+1 times 2x2+3x12x^2 +3x -1. Do you get this picture?
I6SS-image232

Do you see the answer 4x4+4x33x2+4x14x^4 + 4x^3 - 3x^2 +4x -1?

See how Goldfish & Robin multiply polynomials in this video where kids explain math to kids.

You can either play with some of the optional stations below or go to the next island!

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