Skip to 0 minutes and 11 secondsHello. Welcome to week 3. Let us start with the first exercise of the "Polynomial and identities in practice" step. The first exercise, we have to check two notable identities, precisely. We have to verify that for any natural number n greater or equal than 3, and any pair a, b of real numbers, we have this equality. And then from the first equality, we want to deduce what is equal this product. OK, let us see in a closer way these two products. You see, in the first one you have two factors. The first factor is very easy, a minus b.

Skip to 1 minute and 9 secondsThe second factor is more complicated, but if you look with attention, you immediately realize that you have a, which appears with exponents, which are decreasing from n minus 1 to 0. a to the 0 is equal to 1. And b is appearing with increasing exponents. b to the 0, which is 1, b to the 1, b to the 2, until b to n minus 1. And this product is to be equal to a to the n minus b to the n. Then, the exercise asks us to consider this other product, which is a small variation, if you would, of the first one. We will see why. Let us start with our first product.

Skip to 2 minutes and 10 secondsAnd you immediately get that a minus b times a to n minus 1 plus a n minus to b plus a, n minus 3 b square plus blah, blah, blah, blah. Plus a b n minus 2 plus b to n minus 1 is equal to what? Let us start considering this product, and this product. You get a to the n minus b to a n minus 1. And now consider this product, and this product, and what you get. Plus a n minus 1 b minus a n minus 2 b square.

Skip to 3 minutes and 17 secondsAnd then again, we consider this product, and this product, and you get plus a n minus 2 b square minus a n minus 3 b cube. And so on until when you arrive at the end and you get plus a b n minus 1 minus b to the n. And now attention, look, this summand and this summand can be erased, but the same happens for this summand, and this summand, and then continuing for this summand, and the next one. And at the end, for this summand and the previous one, at the end you remain only with a to the n minus b to the n.

Skip to 4 minutes and 42 secondsCheck this equality for n equal 3, n equal 4, n equal 5 if you want, just to be sure to have understood how things go on. And now let us try to deduce the second identity. You see, this second one is obtained by the first one. Oh, let us see. Consider the real number c equal minus b. Then we have that b is equal minus c. OK, substitute b with minus c in this expression. What do you get?

Skip to 5 minutes and 29 secondsYou get a minus c times a n minus 1 minus a n minus 2 minus c plus blah, blah, blah, plus minus 1 n minus 2, a minus c to n minus 2 plus minus 1 to n minus 1 times minus c to n minus 1.

Skip to 6 minutes and 7 secondsOK, it's very easy to check that you get a minus c times-- OK, a n minus 1. And here, you have minus a n minus 2 minus c, which is a n minus 2 c plus blah, blah, blah, plus. And you see here, you have minus 1 to n minus 2 times a times minus 1 to n minus 2 times c to n minus 2. Therefore, you get minus 1 to 2 times n minus 2 a c to n minus 2. And finally, plus-- and here, again, you have minus 1 to n minus 1 times minus 1 at the n minus 1 times c n minus 1.

Skip to 7 minutes and 13 secondsTherefore, you get minus 1 to 2 times n minus 1 c n minus 1. And you'll see this is minus 1 to even power. Also this one, all the signs-- therefore now-- becomes plus. And we get a minus c times an minus 1 plus an means 2 c plus blah, blah, blah, blah, plus ac n minus 2 plus c to n minus 1. But this is exactly the first identity where there is only c, where before we had b. Therefore, this product is equal to a to the n minus c to the n. And now, let us remember what c is. c was equal to minus b. Therefore, this is a to the n minus minus b to the n.

Skip to 8 minutes and 33 secondsAnd now what do we get? We get that if n is even, then minus b to the n is exactly as b to the n, and then we get a to the n minus b to the n. If n is odd, minus b to the n is minus b to the n. And with this other minus before, we get a to the n plus b to the n. Thank you for your attention.

# Polynomial and identities in practice

The following exercises are solved in this step.

We invite you to try to solve them **before** watching the video.

In any case, you will find below a PDF file with the solutions.

### Exercise 1.

Verify the following identity for \(n\geq 3\):

\[(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})=a^n-b^n.\]

Deduce that if \(n\) is odd then

\[(a+b)(a^{n-1}-a^{n-2}b+\cdots+(-1)^{n-2}ab^{n-2}+(-1)^{n-1}b^{n-1})=\] \[=a^n+b^n,\]

and if \(n\) is even then

\[(a+b)(a^{n-1}-a^{n-2}b+\cdots+(-1)^{n-2}ab^{n-2}+(-1)^{n-1}b^{n-1})=\] \[=a^n-b^n.\]

### Exercise 2. [Solved only in the PDF file]

Decompose the following polynomials using the notable identities:

i) \(9x^2-12x+4\)

ii) \(x^4+2x^2y+y^2\)

iii) \(9x^2-1\)

iv) \(3x^2-y^4\)

v) \(8x^3+27y^3\)

vi) \(\frac{1}{8}x^6-1\)

### Exercise 3. [Solved only in the PDF file]

Simplify the following expressions:

i) \((a^m+a^n)^2 - (a^m +a^n)(a^m-a^n)-2a^n(a^m+a^n)\)

ii) \((x-y+\frac{1}{2})(x-y-\frac{1}{2})-(x-y)^2+(\frac{1}{3}x^2-\frac{3}{2})^3+\)

\(-x^2(\frac{1}{27}x^4-\frac{1}{2}x^2+\frac{9}{4})\)

iii) \(\frac{1}{2}(2x+\frac{1}{2})^2-2(2x-\frac{1}{2})^2\)

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