If an algebraic expression is equal to zero then it is called an algenraic equation.
Example: \(f(x) = 4x^2 + 2x + 1\), the given expression is an algebraic function and if we put this equation equal to zero then it is called an algebraic equation. $$ 4x^2 + 2x + 1 = 0 $$ this is an algebraic equation.
Note: The certain values of variable \(x\), which can satisfy the equation are called the roots of that equation.
The algebraic equation and identity are same, the only difference is an equation can be satisfied by some certain values of variable \(x\) whereas an identity can be satisfied by all values of variable \(x\).
Hence an equation which satisfy by all the values of variable \(x\) is called an identity.
According to the factor theorem if "\(\alpha\)" is a root of an equation \(f(x) = 0\) then the polynomial \(f(x)\) will be completely divisible by \((x - \alpha)\) or we can say \((x - \alpha)\) will be a factor of polynomial equation \(f(x)\).
If \(\alpha\) is a root of \(f(x) = 0\) then it will satify the equation \(f(x) = 0\). $$ f(\alpha) = 0.....(1) $$ Let by dividing \(f(x)\) from \((x - \alpha)\) we get remainder \(R\) and quotient \(k(x)\) then
dividend = divisor \(\times\) quotient + remainder $$ f(x) = (x - \alpha) \times k(x) + R.....(2) $$ By putting \(x = \alpha\) in equation (2) we get $$ f(\alpha) = (\alpha - \alpha) \times k(\alpha) + R $$ $$ f(\alpha) = R $$ from equation (1) $$ f(\alpha) = 0 $$ So $$ f(\alpha) = R = 0 $$ Here, it is clear that by dividing \(f(x)\) from \((x - \alpha)\) we get remainder zero hence \(x - \alpha\) is a facror of polynomial equation \(f(x)\).
The roots of two degree equation can be find out by the given formula. $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Here \((b^2 - 4ac)\) is called discriminant because the roots of the equation are dependent of discriminant.
Notes: (1): For an \(n\) degree equation, the number of roots will also be \(n\).
(2): For algebraic equation with real factors, if one root is \((\alpha + i \beta)\) then second root must be \((\alpha - i \beta)\).
(3): For algebraic equation with rational factors, if one root is \((\alpha + \sqrt{\beta})\) then the second root must be \((\alpha - \sqrt{\beta})\).