## The Theory of Equations- Properties of Roots

[latexpage]

In CAT and other MBA entrance exams, there are many questions which fall in the domain of equations (quadratic, cubic, quartic..) and the properties of their roots. Deriving these properties will not only help us remember them but also give us the basic concepts to solve all kinds of equations and their problems. In this chapter we shall learn how the properties of roots of a general polynomial equation are derived. So let’s begin!

Let’s take a polynomial P(x) in a single variable x:

Let $P(x) = a_0(x)^n + a_1(x)^{n-1} + a_2(x)^{n-2} + … + a_{n-1}x + a_n$ where $a_1$, $a_2$ etc. are constants (associated with decreasing powers of x) and n, n-1, etc. are the whole number powers of the variable x. The highest power of the variable x is known as the ‘degree’ of the polynomial P(x).

For example, $P(x) = x^5 – 3x + 1$ is a polynomial with degree 5 with $a_0 = 1, a_1 = 0, a_2 = 0, a_3 = 0, a_4 = -3, a_5 = 1$

### Polynomial Remainder Theorem and Factors

The polynomial remainder theorem states that the remainder of the division of a polynomial P(x) by a linear polynomial $x – a$ is equal to P(a). In short, $x – a$ is a factor of polynomial P(x) if and only if P(a) = 0.

For example we know that the value of $x^2 – 7x + 12$ is 0 at x = 3 and x = 4. Therefore, $x – 3$ and $x – 4$ are factors of the polynomial $x^2 – 7x + 12$. We can write $x^2 – 7x + 12$ as product of its factors. That is, $x^2 – 7x + 12 = (x – 3)(x – 4)$

Therefore, if P(x) = 0 at $\alpha_1$, $\alpha_2$, $\alpha_3$, … $\alpha_n$, i.e. $P(\alpha_1) = 0$, $P(\alpha_2) = 0$, $P(\alpha_3) = 0$, … $P(\alpha_n) = 0$, then $(x – \alpha_1)$, $(x – \alpha_2)$, $(x – \alpha_3)$,… $(x – \alpha_n)$ are factors of polynomial P(x)

Therefore, we can write the polynomial P(x) as a product of its factors. i.e.

$P(x) = (x – \alpha_1) \times (x – \alpha_2) \times (x – \alpha_3) \times … \times (x – \alpha_n)$

We need a small modification in the above equation but, before that, let’s take a digression.

### Multiplying Linear Algebraic Expressions

We can see by multiplying that

$(x + a)(x + b) = x^2 + (a + b)x + ab$ and

$(x – a)(x – b) = x^2 – (a + b)x + ab$

or

$(x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + ac + bc)x + abc$ and

$(x – a)(x – b)(x – c) = x^3 – (a + b + c)x^2 + (ab + ac + bc)x – abc$

Stated simply, the rule would be

$(x \pm a)(x \pm b)(x \pm c)…(x \pm n)$ = $x^n$ + (sum taken one at a time (with sign))$x^{n-1}$ + (sum taken two at a time (with sign))$x^{n-2}$ + … + (product of all with sign)

#### PROBLEM

Let $(x + 1)(x – 2)(x + 3)(x – 4)…. (x + 99)(x – 100) = a_0x^{100} + a_1x^{99} + a_2x^{98} + … + a_{99}x + a_{100}$, then find the values of $a_1$ and $a_{100}$

Now let’s get back to our original polynomial-

### Properties Of Roots

we can write the polynomial P(x) as a product of its factors. i.e.

$P(x) = (x – \alpha_1) \times (x – \alpha_2) \times (x – \alpha_3) \times … \times (x – \alpha_n)$ **OR**

$a_0(x)^n + a_1(x)^{n-1} + a_2(x)^{n-2} + … + a_{n-1}x + a_n = a_0(x – \alpha_1) \times (x – \alpha_2) \times (x – \alpha_3) \times … \times (x – \alpha_n)$

Notice that we have multiplied with $a_0$ in the RHS to keep the coefficient of $x^n$ equal on both sides of the equation. Simplifying the equation on the RHS, we get

$a_0(x)^n + a_1(x)^{n-1} + a_2(x)^{n-2} + … + a_{n-1}x + a_n = a_0[x^n -(\alpha_1 + \alpha_2 + … + \alpha_n)x^{n-1} + (\alpha_1\alpha_2 + \alpha_1\alpha_3 + … \alpha_{n-1}\alpha_n)x^{n-2} – … (-1)^n\alpha_1\alpha_2\alpha_3…\alpha_n$ OR

$a_0(x)^n + a_1(x)^{n-1} + a_2(x)^{n-2} + … + a_{n-1}x + a_n = a_0x^n -a_0(\alpha_1 + \alpha_2 + … + \alpha_n)x^{n-1} + a_0(\alpha_1\alpha_2 + \alpha_1\alpha_3 + … \alpha_{n-1}\alpha_n)x^{n-2} – … (-1)^na_0\alpha_1\alpha_2\alpha_3…\alpha_n$

Comparing the coefficients of equal powers of x on both sides we get,

$(\alpha_1 + \alpha_2 + … + \alpha_n) = -\frac{a_1}{a_0}$

$(\alpha_1\alpha_2 + \alpha_1\alpha_3 + … \alpha_{n-1}\alpha_n) = \frac{a_2}{a_0}$

$(\alpha_1\alpha_2\alpha_3 + \alpha_1\alpha_3\alpha_4 + … \alpha_{n-2}\alpha_{n-1}\alpha_n) = -\frac{a_3}{a_0}$

$…$

$…$

$…$

$(\alpha_1\alpha_2\alpha_3…\alpha_n) = (-1)^n\frac{a_n}{a_0}$

#### PROBLEM

Let the roots of the equation $x^3 -4x^2 + 8x -11 = 0$ be a, b, and c. What is the value of

$1. \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

$2. \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}$

$3. (1 – a)(1 – b)(1 – c)$

$4. (1 + a)(1 + b)(1 + c)$

$5. (a + b)(b + c)(a + c)$

## 0 comments