Week -3 | System of Equations -3

This is an optional module for the students preparing for engineering entrance examinations

Polynomials

1.1 Introduction

We have already given an introduction to polynomials in our earlier chapters in this module. For any polynomials \(f\) & \(g\), there exists poynomials \(q\) & \(r\) such that \(f(x)=g(x)q(x)+r(x)\) with \(\text{deg}(r)<\text{deg}(g)\).

Note that in this case the degree of \(r\) can be \(0\) as well. Thus, \(q\) is the quotient when \(f\) is divided by \(g\), while \(r\) is the remainder.  Let the degree of \(f\) be \(n\). We must observe that the leading coefficient of RHS cannot come from \(r\) (why?). Thus if the degree of \(g\) be \(k\), then the degree of \(q\) must be \(n-k\).

 

1.2 Multiplicity of Roots

Let’s assume that there exists a natural number n such that \(f(x)=(x-a)^ng(x)\) for some non-zer0 polynomial \(g\). Thus, we call \(a\) to be a root of \(f\) with multiplicity \(n\). In which case we have the following property:

where \(f'(x)\) denotes the derivative of \(f(x)\). We’ll cover details of derivatives in our Calculus module.

Suppose that \(f\) is a polynomial with integer coefficients and constant term \(a_0\). Thus we can write \(f\) as \(f(x)=xg(x)+a_0\). Thus for any \(x_0\) if \(f(x_0)=0\), we must have \(x_0g(x_0)+a_0=0\), which further implies \(x_0|a_0\).

Ex-1

For a polynomial with integer coefficients \(P\), show that if the leading coefficient is \(1\) then all rational roots of the polynomial are integers.

Solution:

We have \(P(x)=a_nx^n+a_{n-1}x^{n-1}+…a_0\). Let \(r=\frac{p}{q}\) be a rational root of this polynomial with \(\text{GCD}(p,q)=1\). Given \(a_n=1\), thus we have

Since RHS is an integer, LHS is also an integer – thus \(q=1\).

 

2.1 Reciprocal Equations

For a polynomial \(P(x)=a_nx^n+a_{n-1}x^{n-1}+…a_0\), we call the polynomial reciprocal if \(a_i=a_{n-i}\) \(\forall\) \(i=0,…n\)

Test Your Concepts:

1. Every polynomial of degree \(n\) with constant term \(a_0 \ne 0\) is reciprocal if and only if \(x^nf\left(\frac{1}{x} \right )=f(x)\)
2. If \(f\) be a reciprocal polynomial of odd degree, show that \(f(-1)=0\). In general show that if \(\alpha\) is a root of any reciprocal equation, then \(\frac{1}{\alpha}\) is also a root of the same equation.

 

2.2 Symmetric Polynomials

A polynomial \(f(x,y)\) is said to be symmetric if \(f(x,y)=f(y,x)\). Some elementary symmetric polynomial examples are \(f(x,y)=x+y\) or \(f(x,y)=xy\).

We can use an interesting substitution technique related to the 2 elementary symmetric polynomials discussed above to solve equations.

Consider \(\sigma_1=x+y\) and \(\sigma_2=xy\). We thus can substitute \(x^2+y^2\) as \(\sigma_1^2-2\sigma_2\). Also \(x^3+y^3=\sigma_1^3-3\sigma_2\sigma_1\).

In general, if we define \(s_i=x^i+y^i\), we can try to recursively derive \(s_n\) in terms of \(\sigma_1\) & \(\sigma_2\). We have \(s_n=x^n+y^n=(x+y)(x^{n-1}+y^{n-1})-xy(x^{n-2}+y^{n-2})=\sigma_1s_{n-1}+\sigma_2 s_{n-2}\), with the base consitions \(s_0=2\) & \(s_1=\sigma_1\).

Ex-2

Solve the system of equations

\(x^5+y^5=33\), \(x+y=3\)

Solution:

Given \(s_1=\sigma_1=3\) & \(s_5=33\)

We have the following sequence of recursive equations:

Puting these values we have the final equation as \(\sigma_2^2-9\sigma_2+14=0\). Solving this gives \(\sigma_2=2,7\). We already have \(\sigma_1=3\).

We thus have 2 sets of equations

We can thus have 4 values of \((x,y)\) that satisfies the above equations.

Ex-3

Solve the equation in \(\mathbb{R}\): \(\sqrt[4]{97-x} +\sqrt[4]{x}=5\)

Solution:

Let \(y=\sqrt[4]{97-x}\) and \(z= \sqrt[4]{x}\). We thus have the following 2 equations to solve

Follwing what we did in our earlier example, we set \(\sigma_1=y+z\) & \(\sigma_2=yz\). We have \(s^4=97\), and using the recursive substitutions we have the final equation as \(\sigma_2^2-50\sigma_2+264=0\). Solving this we have \(\sigma_2=6,44\). We already have \(\sigma_1=5\). And hence we need to solve the following set of equations:

Solving these equations we have \((y,z)=\left \{ (2,3), (3,2) \right \}\). It follows that \(x=(16,81)\).

 

For a polynomial in 3 variables the elementary symmetric polynomials are

 

Test Your concepts:

If the following system of equations has solutions :

\(\begin{Bmatrix}
x+y=a & \\
x^2+y^2=b \\
x^3+y^3=c
\end{Bmatrix}\)

Find the relationship between \(a,b,c\)

 

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