## Tag Archives: Algebra for IITJEE Foundation

### Polynomials — commuting and cubics

Let us start delving deeper in Algebra. But, I will be providing only an outline to you in the present article. I encourage you to fill in the  details. This is a well-known way to develop mathematical aptitude/thinking. (This method of  learning works even in Physics and esoteric/hardcore programming).

Definition. Commuting polynomials. Two polynomials are said to commute under composition if and only if $(p \circ q)(t)=(q \circ p)(t)$

(i.e., $p(q(t)=q(p(t)))$). We define the composition powers of a polynomial as follows

$p^{(2)}(t)=p(p(t))$

$p^{(3)}(t)=p(p(p(t)))$

and in general, $p^{(k)}(t)=p(p^{(k-1)}(t)$ for $k=2,3 \ldots$

Show that any two composition powers of the same polynomial commute with each other.

One might ask whether two commuting polynomials must be composition powers of the same polynomial. The  answer is no. Show that any pair of  polynomials in the  following  two sets commute

I. ${t^{n}: n=1,2 \ldots}$

II. ${T^{n}(t):n=1,2 \ldots}$

Let a and b be any constants with a not equal to zero. Show that, if p and q are two polynomials which commute under composition, then the polynomials

$(t/a-b/a) \circ p \circ (at+b)$ and $(t/a-b/a) \circ q \circ (at+b)$ also  commute under the composition. Use this fact to find from sets I and II other  families which commute under composition.

Can you find pairs of polynomials not  comprised in the foregoing discussion which commute under composition? Find families of polynomials which commute under composition and within which  there is exactly one polynomial of each positive degree.

The Cubic Equation. Cardan’s Method. An elegant way to solve the general cubic is due to Cardan. The strategy is to replace an equation in one variable by one in two variables. This provides an extra degree of  freedom by which we can impose a convenient second constraint, allowing us to  reduce the problem to that of solving a quadratic.

(a) Suppose  the given equation is $t^{3}+pt+q=0$. Set $t=u+v$ andn obtain the equation $u^{3}+v^{3}+(3uv+p)(u+v)+q=0$.

Impose the second condition $3uv+p=0$ (why do we do  this?) and argue that we can obtain solutions for  the cubic by solving the system

$u^{3}+v^{3}=-q$

$uv = -p/3$

(b) Show that $u^{3}$ and $v^{3}$ are roots of  the quadratic equation

$x^{2}+qx-p^{3}/27=0$

(c) Let $D=27q^{2}+4p^{3}$. Suppose that p and q are both real and that $D>0$. Show that the quadratic in (b) has real solutions, and that if

$u_{0}$ and $v_{0}$ are the real cubic roots of  these solutions, then the system in (a) is satisfied by

$(u,v)=(u_{0},v_{0}), (u_{0}\omega, v_{0}\omega^{2}), (u_{0}\omega^{2}, v_{0}\omega)$

where $\omega$ is the imaginary cube root $(0.5)(-1+\sqrt{-3})$ of  unity. Deduce that the cubic polynomial $t^{3}+pt+q$ has one real and two nonreal zeros.

(d) Suppose that p and q are both real and that $D=0$. Let $u_{0}$ be the real cube root of the solution of  the quadratic in (b). Show  that, in this case, the cubic has all its zeros real, and in fact can be written in the form

$sy^{2}$ where $y=(t+u_{0})$ and $s=t-2u_{0}$

(e) Suppose that p and q are both real and that $D<0$. Show that the solutions  of  the quadratic equation in (b) are nonreal complex conjugates, and that it is possible to choose cube roots u and v of  these solutions which are complex conjugates and satisfy the system in (a). If

$u=r(cos \theta + isin \theta)$ and $v=r(cos \theta - isin \theta)$, show that the three roots of  the cubic equation are the reals

$2r cos \theta$

$2r cos (\theta + (2/3)\pi)$

$2r cos(\theta + (4/3)\pi)$.

(f) Prove that every cubic equation with real coefficients has at least one real root.

Use Cardan’s Method to solve the cubic equation.

(a) $x^{3}-6x+9=0$

(b) $x^{3}-7x+6=0$.

Part (b) above will require  the use of a pocket calculator and some trigonometry. You will also need De Moivre’s Theorem and give a solution to an accuracy of 3 decimal places.

More later…

-Nalin

### Factorization uses and tricks

One of the first concepts of elementary algebra is factorization. Apart from its intrinsic use in Math itself, it is applied in several areas like pole-placement method in Control System Design and design of digital filters for use in digital communications, the latter being the foundation stone of IT. Another use is in error control coding which is used in wireless/mobile/cellular communications. One of the trickiest uses of factorization is in Trigonometry, which of course, has countless applications. Let me present to you some tricks, which are not so obvious in factorizing complicated algebraic expressions.

1) Factorization by guessing! Solve the equation $x^{3}-3x^{2}-6x+16=0$.

Solution: The product of the coefficient of highest degree term and constant is 16. So, think of the factors of 16. You will see that the LHS vanishes when $x=2$. (We have used the factor theorem here).

Hence, $x=2$ is one root of the equation and corresponding to this root, we have a factor $x-2$.

The equation may now be written $x^{2}(x-2)-x(x-2)-8(x-2)=0$ or $(x^{2}-x-8)(x-2)=0$

Removing the factor $x-2$, we have $x^{2}-x-8=0$

hence, the three roots of the above cubic $x = 0.5(1+\sqrt{33}$ or $x = 0.5(1-\sqrt{33})$ or $x=2$.

Moral of the story: Guessing the factor reduces the degree of the equation. There will be some trials and errors though.

2) The Method of Undetermined Coefficients:

Resolve into two factors: $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy$

Solution: This equation might seem not factorizable, at first sight, as the *middle term* involves no *c*. (Note: a, b, and c are parameters, and the variables are x and y). Also, the brute force technique for using the formula method for quadratics is way too cumbersome. So, let’s give a shot to it using the method of undetermined coefficients.

Let $2a^{2}x^{2} - 2(3b-4c)(b-c)y^{2}+abxy=(Px+Qy)(Mx+Ny)$

Hence, equating like coefficients, $PM=2a^{2}$, $NQ= -2(3b-4c)(b-c)$, $MQ+NP=ab$.

Again, you have to guess!

Let’s try: let $N=2(b-c)$, $Q=-(3b-4c)$, $M=a$, $P=2a$

We need to check: $MQ+NP=-a(3b-4c)+2(b-c)(2a)=ab$ Bingo:-)

So, the factors are $(ax+2(b-c)y)(2ax-y(3b-4c))$

3) An important factorization identity is:

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Would you like to try proving this?(It is not so easy!)

More later, — Nalin