Question

What is the relationship between the roots and the coefficients of a polynomial?

Answer 1

A polynomial is a mathematical expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a positive integral power (like$a + bx + c{x^2}$)

Complete step-by-step solution:
Let ${\alpha _1},{\alpha _2},{\alpha _3}...{\alpha _n}$ are the roots of any polynomial which is given by an expression,
$f(x) = {a_0}{x^n} + {a_1}{x^{n - `}} + {a_2}{x^{n - 2}} + ... + {a_{n - 1}}x + {a_n} = 0$
Therefore, we can write
$f(x) = {a_0}(x - {\alpha _1})(x - {\alpha _2})(x - {\alpha _3})...(x - {\alpha _n})$
Now, equating the right hand side of both the above equation, we get
${a_0}(x - {\alpha _1})(x - {\alpha _2})(x - {\alpha _3})...(x - {\alpha _n}) = {a_0}{x^n} + {a_1}{x^{n - `}} + {a_2}{x^{n - 2}} + ... + {a_{n - 1}}x + {a_n} = 0$
Now, comparing coefficient of ${x^{n - 1}}$ on both sides we get
${s_1} = {\alpha _1} + {\alpha _2} + {\alpha _3} + ... + {\alpha _n} = \sum {{\alpha _i}} = - \dfrac{{{a_1}}}{{{a_0}}}$
$\Rightarrow {s_1} = - \dfrac{{\text{coeff. of}({x^{n - 1}})}}{{\text{coeff. of} ({x^n})}}$
Comparing coefficient of ${x^{n - 2}}$ on both sides, we get
${s_1} = {\alpha _1}{\alpha _2} + {\alpha _1}{\alpha _3} + ... = \sum\limits_{i \ne j} {{\alpha _i}{\alpha _j}} = {( - 1)^2}\dfrac{{{a_2}}}{{{a_0}}}$
$\Rightarrow {s_1} = {( - 1)^2}\dfrac{{\text{coeff. of}({x^{n - 2}})}}{{\text{coeff. of}({x^n})}}$
Particular case: Quadratic equation $a{x^2} + bx + c = 0$,
$>$The solutions of the quadratic equation, $a{x^2} + bx + c = 0$ is given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$>$The expression ${b^2} - 4ac = D$ is called the discriminant of the quadratic equation.
$>$If α & β are the roots of the quadratic equation $a{x^2} + bx + c = 0$ then;

1. $\alpha + \beta = - \dfrac{b}{a}$
2. $\alpha \beta = \dfrac{c}{a}$
3. $\left| {\alpha - \beta } \right| = \dfrac{{\sqrt D }}{{\left| a \right|}}$
   $>$Quadratic equation whose roots are α & β is $(x - \alpha )(x - \beta ) = 0$i.e.
   ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$ i.e. ${x^2} -$(sum of roots) x$$$$ + product of roots$= 0$
   Nature of Roots: Consider the quadratic equation $a{x^2} + bx + c = 0$ where $a,b,c \in R\& a \ne 0$ then;
   $\to$ $D > 0 \Leftrightarrow$ Roots are real & distinct (unequal).
   $\to$ $D = 0 \Leftrightarrow$ Roots are real & coincident (equal).
   $\to$ $D < 0 \Leftrightarrow$ Roots are imaginary.

Note:
$>$A polynomial is a combination of terms that are only added, subtracted or multiplied.
$>$A quadratic polynomial with real coefficients is of the form $a{x^2} + bx + c$, where a, b, c are real numbers with $a \ne 0$ .
$>$The highest exponent of the variable in the equation is called the degree of polynomial.
$>$Polynomials of degrees $1,2and3$ are called linear, quadratic and cubic polynomials respectively.