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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 $. $$ \begin{aligned} 2x^2 - 3 &= 0 \\ x^2 = \frac{3}{2} \\ x_1x_2 = \pm \sqrt{\frac{3}{2}} \end{aligned} $$ 228 282 618 solved problems How do you algebraically find the zeros of a function?In general, given the function, f(x), its zeros can be found by setting the function to zero. The values of x that represent the set equation are the zeroes of the function. To find the zeros of a function, find the values of x where f(x) = 0.
How do you find all real zeros of a function?Generally, for a given function f (x), the zero point can be found by setting the function to zero. The x value that indicates the set of the given equation is the zeros of the function. To find the zero of the function, find the x value where f (x) = 0.
How do you find the zeros of a function on a TI 84?How do I find the zero of a function using the TI-84 Plus C Silver Edition?. Press [Y=].. Enter the function.. Press [GRAPH].. Press [2nd] [CALC] [2].. When prompted for "Left Bound?", specify the left bound for x using the arrow keys or enter a value and press [ENTER].. |