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:

https://johnmayhk.wordpress.com/2008/09/24/comparing-coefficients/
https://johnmayhk.wordpress.com/2010/10/21/factorize-quadratic-polynomial/
https://johnmayhk.wordpress.com/2010/12/28/comparing-coefficients-again/

[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.

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