# Quod Erat Demonstrandum

## 2011/03/03

### Polynomial identity

Is the following an identity? Prove or disprove your claim:

$(x+1)(x-3)+1=(x+2)(x-1)$

This is a trivial question in junior form mathematics.

Many students may expand and re-group the polynomials so as to draw their conclusion, like

$L.H.S.=(x+1)(x-3)+1=x^2-2x-3+1=x^2-2x-2$

$R.H.S.=(x+2)(x-1)=x^2+x-2$

By their eyes, it is easy to draw the conclusion that

$(x+1)(x-3)+1=(x+2)(x-1)$ is NOT an identity since $x^2-2x-2$ and $x^2+x-2$ are not identical.

Then, normally, I will add the comment:

for disproving identity, all we need to do is putting value(s) of x

and hope to see that $L.H.S.\neq R.H.S.$

For example, put $x=0$

$L.H.S.=(0+1)(0-3)+1=-3$ and

$R.H.S.=(0+2)(0-1)=-2$,

since $L.H.S.\neq R.H.S.$, $(x+1)(x-3)+1=(x+2)(x-1)$ is NOT an identity for sure.

Then a student asked, in a lesson couple months ago, ‘How about if the given one is really an identity? Can we use some values of x to check?’

‘Yes, of course.’I replied quickly.

Let say

$x^2-1=(3+x)(x-2)-x+5$

is it an identity?

We, of course, can expand the $R.H.S.$ and then judge ‘by naked eyes’ that whether it is an identity or not.

However, we can tackle the problem by the following way:

Put $x=0$, $L.H.S.=-1$, $R.H.S.=(3)(-2)+5=-1$, so $L.H.S.=R.H.S.$
Put $x=1$, $L.H.S.=0$, $R.H.S.=(1+3)(1-2)-1+5=0$, so $L.H.S.=R.H.S.$
Put $x=-1$, $L.H.S.=0$, $R.H.S.=(-1+3)(-1-2)+1+5=0$, so $L.H.S.=R.H.S.$

Then, we can draw the conclusion that

$x^2-1=(3+x)(x-2)-x+5$

is an identity and we may write $x^2-1\equiv(3+x)(x-2)-x+5$.

Why?

Urm, just consider that

$x^2-1=(3+x)(x-2)-x+5$

$x^2-1-((3+x)(x-2)-x+5)=0$

Observe that, the above a polynomial equation of degree at most 2 (well, in this case, the degree is 1 actually),

(i.e. it may be a quadratic equation or a linear equation etc.)

and, from above, we see that

$x=0,1,-1$ are roots of that polynomial equation.

That is, a quadratic equation (i.e. a polynomial equation of degree TWO) has THREE DISTINCT roots (0,1 and -1), strange?

Not really.

Be more abstract,

if the following equation has THREE DISTINCT roots

$ax^2+bx+c=0$

then, the equation should not be a quadratic equation (since a quadratic equation has at most TWO DISTINCT roots), so what is it?

Actually, it is not just an equation, it is an identity! It is the zero polynomial, that is

$0x^2+0x+0=0$　(see, it has infinitely many ‘roots’)

or we may write

$ax^2+bx+c\equiv 0$

For example, it is not necessary to expand the following

$x^4-10x^3+35x^2-50x+29=(x-1)(x-2)(x-3)(x-4)+5$

for showing it is an identity, just check that

there are more than 4 distinct roots to the equation. (Student, check it out.)

In general, if a degree n polynomial equation has more than n DISTINCT roots, then the polynomial is the zero polynomial.

Up to now, junior form students should laugh at my stupidity of using lengthy presentation for that trivial stuff, I guess.

But, this trick will be found in Pure Mathematics exercises, let me post some in random.

e.g.1

Suppose $a,b,c$ are distinct numbers, is the following an identity? Prove or disprove your assertion.

$\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}=1$

At a glance, it is a polynomial equation of degree at most TWO.

Putting $x=a,b,c$, $L.H.S.=1=R.H.S.$

Hence a polynomial equation of degree at most TWO has THREE distinct roots $a,b,c$,

the equation is actually an identity!

That is

$\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}\equiv 1$

Since we can substitute any value of $x$ for identity, say $x=d$, we can create a question like

find the value of

$\frac{(d-b)(d-c)}{(a-b)(a-c)}+\frac{(d-c)(d-a)}{(b-c)(b-a)}+\frac{(d-a)(d-b)}{(c-a)(c-b)}$

(ANS: 1)

e.g.2

For distinct numbers $a,b,c$, prove that

$\frac{a^2(x-b)(x-c)}{(a-b)(a-c)}+\frac{b^2(x-c)(x-a)}{(b-c)(b-a)}+\frac{c^2(x-a)(x-b)}{(c-a)(c-b)}\equiv x^2$

Easy to know that $x=a,b,c$ are the roots of it, hence it is an identity.

Then we can create a not-that-trivial equation like

suppose $(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)\ne 0$ and

$\frac{a^2(d-b)(d-c)}{(a-b)(a-c)}+\frac{b^2(d-c)(d-a)}{(b-c)(b-a)}+\frac{c^2(d-a)(d-b)}{(c-a)(c-b)}=2025$

find the value(s) of $d$

(ANS: $\pm 45$)

Well, follow similar way of thinking, we may create something involve $a^3,b^3,c^3,...$, boring…

e.g.3

Suppose $a_1,a_2,\dots a_n,b_1,b_2,\dots b_n$ are $2n$ distinct numbers such that

$\displaystyle \prod_{k=1}^{n}(a_k+b_1)=\displaystyle \prod_{k=1}^{n}(a_k+b_2)=\dots \displaystyle \prod_{k=1}^{n}(a_k+b_n)$.

Prove that

$\displaystyle \prod_{k=1}^{n}(a_1+b_k)=\displaystyle \prod_{k=1}^{n}(a_2+b_k)=\dots \displaystyle \prod_{k=1}^{n}(a_n+b_k)$.

This is an old question, the trick is considering

$P(x)=\displaystyle \prod_{k=1}^{n}(x+a_k)-\displaystyle \prod_{k=1}^{n}(x-b_k)$

Note that, this is a polynomial of degree at most $n-1$, but there are $n$ distinct zeros*.

Done.

The following are very boring related posts, just for your debugging:

[Off topic]

Many days ago, I saw a very beautiful picture on web:

it is something involves roots of polynomials, read the following for detailed description if you want to:

http://math.ucr.edu/home/baez/week285.html

Enjoy!

*Zeros of a polynomial $P(x)$ means the roots of $P(x)=0$.