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polynomial-equation-calculator en Our online expert tutors can answer this problem Get step-by-step solutions from expert tutors as fast as 15-30 minutes. Your first 5 questions are on us! You are being redirected to Course Hero I want to submit the same problem to Course HeroCorrect Answer :) Let's Try Again :( Try to further simplify Number LineGraphHide Plot » Sorry, your browser does not support this application
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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 $. $$ \begin{aligned} 2x^2 - 3 &= 0 \\ x^2 = \frac{3}{2} \\ x_1x_2 = \pm \sqrt{\frac{3}{2}} \end{aligned} $$ 228 282 433 solved problems How do you determine the number of zeros in a polynomial function?Zeros of a polynomial can be found from the graph by observing the points where the graph line cuts the x-axis. The x-coordinates of the points where the graph cuts the x-axis are the zeros of the polynomial.
What are the zeros of the polynomial function?The zeros of a polynomial p(x) are all the x-values that make the polynomial equal to zero. They are interesting to us for many reasons, one of which is that they tell us about the x-intercepts of the polynomial's graph.
How do you find all real zeros of a polynomial function?To find the real zeros of a polynomial, first convert the polynomial to factored form. Once all factors are found, set each individual factor equal to zero to solve for the real zeros.
How do you calculate the zeros?The zeroes of a function are the values of x at which the total equation is equal to zero, so calculating them is as easy as setting the function equal to zero and solving for x. To see a basic example of this, consider the function f(x) = x + 1.
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