The zeros of a polynomial equation are the solutions of the function f(x) = 0. A value of x that makes the equation equal to 0 is termed as zeros. It can also be said as the roots of the polynomial equation. Find the zeros of an equation using this calculator. Show Find the Roots of a Polynomial EquationThe zeros of a polynomial equation are the solutions of the function f(x) = 0. A value of x that makes the equation equal to 0 is termed as zeros. It can also be said as the roots of the polynomial equation. Find the zeros of an equation using this calculator. Code to add this calci to your website  Enter the equation into the text box and you will get the zeros values. Also, the roots will displayed in a complex plane. EMBEDMake your selections below, then copy and paste the code below into your HTML source. ThemeOutput TypeLightbox Widget controls
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This free math tool finds the roots (zeros) of a given polynomial. The calculator computes exact solutions for quadratic, cubic, and quartic equations. Enter polynomial: = 0 Examples: x^2 - 4x + 3 2x^2 - 3x + 1 x^3 – 2x^2 – x + 2 EXAMPLES find roots of the polynomial $4x^2 - 10x + 4$ find polynomial roots $-2x^4 - x^3 + 189$ solve equation $6x^3 - 25x^2 + 2x + 8 = 0$ find polynomial roots $2x^3-x^2-x-3$ find roots $2x^5-x^4-14x^3-6x^2+24x+40$ Search our database of more than 200 calculators TUTORIAL How to find polynomial roots ?The process of finding polynomial roots depends on its degree. The degree is the largest exponent in the polynomial. For example, the degree of polynomial $ p(x) = 8x^\color{red}{2} + 3x -1 $ is $\color{red}{2}$. We name polynomials according to their degree. For us, the most interesting ones are: quadratic - degree 2, Cubic - degree 3, and Quartic - degree 4. Roots of quadratic polynomialThis is the standard form of a quadratic equation $$ a\,x^2 + b\,x + c = 0 $$ The formula for the roots is $$ x_1, x_2 = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} $$ Example 01: Solve the equation $ 2x^2 + 3x - 14 = 0 $ In this case we have $ a = 2, b = 3 , c = -14 $, so the roots are: $$ \begin{aligned} x_1, x_2 &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{3^2-4 \cdot 2 \cdot (-14)}}{2\cdot2} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{9 + 4 \cdot 2 \cdot 14}}{4} \\ x_1, x_2 &= \dfrac{-3 \pm \sqrt{121}}{4} \\ x_1, x_2 &= \dfrac{-3 \pm 11}{4} \\ x_1 &= \dfrac{-3 + 11}{4} = \dfrac{8}{4} = 2 \\ x_2 &= \dfrac{-3 - 11}{4} = \dfrac{-14}{4} = -\dfrac{7}{2} \end{aligned} $$ Quadratic equation - special casesSometimes, it is much easier not to use a formula for finding the roots of a quadratic equation. Example 02: Solve the equation $ 2x^2 + 3x = 0 $ Because our equation now only has two terms, we can apply factoring. Using factoring we can reduce an original equation to two simple equations. $$ \begin{aligned} 2x^2 + 3x &= 0 \\ \color{red}{x} \cdot \left( \color{blue}{2x + 3} \right) &= 0 \\ \color{red}{x = 0} \,\,\, \color{blue}{2x + 3} & \color{blue}{= 0} \\ \color{blue}{2x } & \color{blue}{= -3} \\ \color{blue}{x} &\color{blue}{= -\frac{3}{2}} \end{aligned} $$ Example 03: Solve equation $ 2x^2 - 10 = 0 $ This is also a quadratic equation that can be solved without using a quadratic formula. . $$ \begin{aligned} 2x^2 - 18 &= 0 \\ 2x^2 &= 18 \\ x^2 &= 9 \\ \end{aligned} $$ The last equation actually has two solutions. The first one is obvious $$ \color{blue}{x_1 = \sqrt{9} = 3} $$ and the second one is $$ \color{blue}{x_2 = -\sqrt{9} = -3 }$$ Roots of cubic polynomialTo solve a cubic equation, the best strategy is to guess one of three roots. Example 04: Solve the equation $ 2x^3 - 4x^2 - 3x + 6 = 0 $. Step 1: Guess one root. The good candidates for solutions are factors of the last coefficient in the equation. In this example, the last number is -6 so our guesses are 1, 2, 3, 6, -1, -2, -3 and -6 if we plug in $ \color{blue}{x = 2} $ into the equation we get, $$ 2 \cdot \color{blue}{2}^3 - 4 \cdot \color{blue}{2}^2 - 3 \cdot \color{blue}{2} + 6 = 2 \cdot 8 - 4 \cdot 4 - 6 - 6 = 0$$ So, $ \color{blue}{x = 2} $ is the root of the equation. Now we have to divide polynomial with $ \color{red}{x - \text{ROOT}} $ In this case we divide $ 2x^3 - x^2 - 3x - 6 $ by $ \color{red}{x - 2}$. $$ ( 2x^3 - 4x^2 - 3x + 6 ) \div (x - 2) = 2x^2 - 3 $$ Now we use $ 2x^2 - 3 $ to find remaining roots $$ \begin{aligned} 2x^2 - 3 &= 0 \\ 2x^2 &= 3 \\ x^2 &= \frac{3}{2} \\ x_1 & = \sqrt{ \frac{3}{2} } = \frac{\sqrt{6}}{2}\\ x_2 & = -\sqrt{ \frac{3}{2} } = - \frac{\sqrt{6}}{2} \end{aligned} $$ Cubic polynomial - factoring methodTo solve cubic equations, we usually use the factoting method: Example 05: Solve equation $ 2x^3 - 4x^2 - 3x + 6 = 0 $. Notice that a cubic polynomial has four terms, and the most common factoring method for such polynomials is factoring by grouping. $$ \begin{aligned} 2x^3 - 4x^2 - 3x + 6 &= \color{blue}{2x^3-4x^2} \color{red}{-3x + 6} = \\ &= \color{blue}{2x^2(x-2)} \color{red}{-3(x-2)} = \\ &= (x-2)(2x^2 - 3) \end{aligned} $$ Now we can split our equation into two, which are much easier to solve. The first one is $ x - 2 = 0 $ with a solution $ x = 2 $, and the second one is $ 2x^2 - 3 = 0 $. $$ \begin{aligned} 2x^2 - 3 &= 0 \\ x^2 = \frac{3}{2} \\ x_1x_2 = \pm \sqrt{\frac{3}{2}} \end{aligned} $$ 228 042 888 solved problems |
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