Factorization in proof by induction

Question

I have to prove that Sum of series $$\sum_{k=1}^n k^2 = \frac{(n)(n+1)(2n+1)}{6}$$ I tried doing it myself, but was not able to factorise the third grade polynomial. So I watched this youtube video, but there is something a bit odd happening.. https://www.youtube.com/watch?v=Y1GSE8EZle4. At 5:50 it goes from $$2n^2+7n+6$$ to $$(2n+3)(n+2)$$ The question now is.. Is that allowed? I mean, if I proceed with the quadratic formula, I get $$(n+3/2)(n+2)$$ Giving the fact that we are not in an equation, where we can multiply and divide for whatever we want, for sure the two factorisations are not equal.. But from my solution I don't know how to proceed. So for sure I'm wrong here. What am I missing? Thanks.

Answer 1

Given a second order polynomial

$$p(x) = ax^2 + bc + c$$

with real roots $x_1$ and $x_2$, the correct factorization of $p(x)$ is:

$$p(x) = {\color{red}{a}}(x-x_1)(x-x_2).$$

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In your case:

$$p(n) = 2n^2+7n+6.$$

The roots are $n_1 = -\frac{3}{2}$ and $n_2 = -2.$ Therefore, the factorization is:

$$p(n) = {\color{red}{2}}\left(n + \frac{3}{2}\right)(n+2) = (2n + 3)(n+2).$$

Answer 2

Note that $(n+\frac 32)(n+2) = n^2 + \frac 72 n + 3 \ne 2n^2 + 7n + 6$.

The quadratic formula finds the roots but it completely renders the leading coefficient obsolete.

If $p(x) = ax^2 + bx + c$ then the very first thing you do to derive the the quadratic formula is

You set $ax^2 + bx + c = 0$.

Okay..... the very SECOND thing you do it..... you divide by $a$!!!!!!

$x^2 + \frac bax + \frac ca = 0$.

YOU JUST THREW THE LEADING COEFFICIENT AWAY!!!!!!!

So what you get will be roots. $r_1 = \frac {-b+ \sqrt{b^2 -4ac}}{2a},r_2 = \frac {-b- \sqrt{b^2 -4ac}}{2a} $.

And therefor we can conclude that $(x-r_1)(x-r_2)=0$ and that is a factoring of the reduced polynomial

$x^2 + \frac bax + \frac ca =(x-r_1)(x-r_2)$.

But to factor the ORIGINAL polynomial with the leading coefficient of $a$ we must have

$ax^2 + bx + c = a(x^2 + \frac bax + \frac ca) = a(x-r_1)(x-r_2)$.

Now if $r_2 \in \mathbb Z$ and $ar_1 \in \mathbb Z$ we can of course do

$a(x-r_1)(x-r_2) = (ax - ar_1)(x-r_2)$

Now, It might seem like I'm making a fuss about throwing the leading coefficient away. Actually I don't care. Must don't. The leading coefficient $a$ is probably the least important aspect of a polynomial.