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# Calculators :: Solving Equations :: Polynomial Equation Solver

The following calculator can be used to solve polynomial equations. These are equations of the form $P(x)=Q(x)$, where $P(x)$ and $Q(x)$ are polynomials. Special cases of such equations are:

1. Linear equation $(2x+1=3)$
2. Quadratic Equation $(2x^2-3x-5=0)$,
3. Cubic equation $(5x^3 + 2x^2 - 3x + 1 = 0)$ . . .

How to use this calculator?

Example 1: to solve (2x + 3)2 - 4(x + 1)2 = 1 type (2x + 3)^2 - 4(x + 1)^2 = 1.

Example 2: to solve $\frac{3x^2-1}{2}+\frac{2x+1}{3} = \frac{x^2-2}{4} + \frac{1}{3}$ type (3x^2 - 1)/2 + (2x + 1)/3 = (x^2 - 2)/4 + 1/3.

## Polynomial Equation Solver

Important: The form will NOT let you enter wrong characters like a, u, %, y, p, ...

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## How to solve polynomial equation?

We use 4 steps to solve a polynomial equation in one variable. The following example uses all four steps but some equations may not require all of them.

Example:

Solve $\frac{3x^2-1}{2}+\frac{2x+1}{3} = \frac{x^2-2}{4} + \frac{1}{3}$

Solution:

Step 1: Eliminate fractions by multiplying each side by the least common denominator.

In this example we multiply both sides by 12.

\begin{aligned} \frac{3x^2-1}{2}+\frac{2x+1}{3} & = \frac{x^2-2}{4} + \frac{1}{3} \\ {\color{red}{12}}\cdot\frac{3x^2-1}{2}+{\color{red}{12}}\cdot\frac{2x+1}{3} & = {\color{red}{12}}\cdot\frac{x^2-2}{4} + {\color{red}{12}}\cdot\frac{1}{3}\\ 6\cdot(3x^2-1)+4\cdot(2x+1) & = 3\cdot(x^2-2) + 4 \end{aligned}

Step 2:Simplify each side by clearing parentheses and combining like terms.

\begin{aligned} 6\cdot(3x^2-1)+4\cdot(2x+1) & = 3\cdot(x^2-2) + 4 \\ 18x^2-6+8x+4 & = 3x^2-6+4\\ 18x^2+8x-2 & = 3x^2-2\\ \end{aligned}

Step 3:Use the addition property to get all terms on one side of the equation.

\begin{aligned} 18x^2+8x-2 & = {\color{red}{3x^2-2}}\\ 18x^2+8x-2{\color{red}{-3x^2+2}} & = 0\\ 15x^2+8x & = 0 \end{aligned}

Step 4:Finally, solve the equation.

Here we have second degree equation so you can use step-by-step quadratic equation solver to find the solutions:

$${\color{blue}{ x_1 = 0, x_2 = -\frac{8}{15} }}$$

If you have an equation of higher degree then you can use polynomial roots calculator.

Unfortunately, this calculator will show you the solution, but without explanation.