We will discuss here about the polynomial equation and its roots.
If f(x) is a polynomial in x of degree ≥ 1 whose coefficients are real or complex numbers then f(x) = 0 is called its corresponding polynomial equation.
Examples of polynomial equation:
(i) 5x\(^{2}\) + 2 x - 7 is a quadratic polynomial and 5x\(^{2}\) + 2 x - 7 = 0 is its corresponding quadratic equation.
(ii) 2x\(^{3}\) + x\(^{2}\) + 5x - 3 is a cubic polynomial and 2x\(^{3}\) + x\(^{2}\) + 5x - 3 = 0 is its corresponding cubic equation.
(iii) x\(^{4}\) + x\(^{2}\) - 2x + 6 is a cubic polynomial and x\(^{4}\) + x\(^{2}\) - 2x + 6 = 0 is its corresponding cubic equation.
(iv) x\(^{5}\) + 2x\(^{4}\) + 2x\(^{3}\) + 4x\(^{2}\) + x + 2 is a cubic polynomial and x\(^{5}\) + 2x\(^{4}\) + 2x\(^{3}\) + 4x\(^{2}\) + x + = 0 is its corresponding equation.
If α be a value of x for which f(x) becomes zero, i.e., f(α) = 0, then α is said to be a root of the equation f(x) n= 0.
In other words,
α is called a root of the polynomial equation f(x) = 0 if f(α) = 0.
Examples of root of the polynomial equation:
(i) Let f(x) = 4x\(^{3}\) + 12x\(^{2}\) - 4x - 12. As 4(1)\(^{3}\) + 12(1)\(^{2}\) - 4(1) - 12 = 4 + 12 - 4 - 12= 0, i.e., f(1) = 0, f(x) = 0 has a root x = 1.
(ii) Let f(x) = x\(^{2}\) - 2x - 3. As (-1)\(^{2}\) - 2(-1) - 3 = 1 + 2 - 3 = 0, i.e., f(-1) = 0, f(x) = 0 has a root x = -1
(iii) Let f(x) = x\(^{4}\) + x\(^{3}\) – 2x\(^{2}\) + 4x - 24. As (2)\(^{4}\) + (2)\(^{3}\) - 2(2)\(^{2}\) + 4(2) - 24 = 16 + 8 – 8 +8 + 8 = 0, i.e., f(2) =0, f(x) has a root x = 2
(iv) Let f(x) = x\(^{3}\) + x\(^{2}\) - x - 1. As (1)\(^{3}\) + (1)\(^{2}\) – (1) – 1 = 1 + 1 - 1 - 1 = 0, i.e., f(1) = 0, f(x) = 0 has a root x = 1.
● Factorization
From Polynomial Equation and its Roots to HOME
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Jul 22, 24 03:27 PM
Jul 22, 24 02:41 PM
Jul 21, 24 02:14 PM
Jul 20, 24 03:45 PM
Jul 20, 24 02:30 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.