This solver performs basic arithmetic operations on polynomials (addition, subtraction, multiplication and division).
The calculator displays complete work and a detailed step-by-step explanation for each operation.
problem
Multiply polynomial $ \, \color{blue}{ 5a+2 } \, $ by $ \, \color{orangered}{ 2a-3 }$.
solution
$$ \left( \color{blue}{5a+2} \right) \cdot \left( \color{orangered}{2a-3} \right) = 10a^2-11a-6 $$explanation
In this example we are multiplying two binomials so FOIL method can be used.
$$ \begin{aligned} \left( \color{blue}{ 5a+2}\right) \cdot \left( \color{orangered}{ 2a-3}\right) &= \underbrace{ \color{blue}{5a} \cdot \color{orangered}{2a} }_{\text{FIRST}} + \underbrace{ \color{blue}{5a} \cdot \left( \color{orangered}{-3} \right) }_{\text{OUTER}} + \underbrace{ \color{blue}{2} \cdot \color{orangered}{2a} }_{\text{INNER}} + \underbrace{ \color{blue}{2} \cdot \left( \color{orangered}{-3} \right) }_{\text{LAST}} = \\ &= 10a^2 + \left( -15a\right) + 4a + \left( -6\right) = \\ &= 10a^2 + \left( -15a\right) + 4a + \left( -6\right) = \\ &= 10a^2-11a-6; \end{aligned} $$
In this short tutorial, you will learn how to perform basic operations on polynomials.
The basic operations are 1. addition 2. subtraction 3. FOIL method for binomial multiplication 4. standard multiplication 5. division by monomial
and 6. long division.
Note that this calculator displays a step-by-step explanation for each of these operations. Nevertheless, let's start with addition.
Example 01: Add $ (2a+5) + (4a-3) $
First we will remove the parenthesis because there are no minus sign in front of the brackets:
$$ (2a+5) + (4a-3) = 2a + 5 + 4a - 3 $$We'll now group the like terms:
$$ 2a + 5 + 4a - 3 = 2a + 4a + 5 - 3$$Finally, we combine like terms:
$$ 2a + 4a + 5 - 3 = 6a + 2 $$Putting all together we have
$$ \begin{aligned} (2a+5) + (4a-3) \overbrace{=}^{\text{remove par.}}& \color{blue}{2a} + 5 + \color{blue}{4a} - 3 = \\ \overbrace{=}^{\text{group like terms}}& \color{blue}{2a + 4a} + 5 - 3 = \\ \overbrace{=}^{\text{combine like terms}}& \color{blue}{6a} + 2 \end{aligned} $$Example 02: Add $ (5x^3 - 3x^2 - 2x + 5) + (-x^3 + 2x^2 - 7) $
To perform vertical addition, we must arrange like terms one above the other.
$$ \begin{array}{rrr} \color{blue}{5x^3} & \color{orangered}{-3x^2} & -2x & \color{purple}{5} \\ \color{blue}{-x^3} & \color{orangered}{2x^2} & & \color{purple}{-7} \\ \hline \end{array} $$It is now quite simple to combine like terms
$$ \begin{array}{rrr} \color{blue}{5x^3} & \color{orangered}{-3x^2} & -2x & \color{purple}{5} \\ \color{blue}{-x^3} & \color{orangered}{2x^2} & & \color{purple}{-7} \\ \hline \color{blue}{4x^3} & \color{orangered}{-x^2} & -2x & \color{purple}{-2} \end{array} $$The answer is
$$ 4x^2 -x^2-2x-2 $$Example 04: Subtract $ (2x^2 + x - 3) - (x^2 - 3x + 5) $
Place like terms one above the other, but in the second polynomial, we must now alter all of the signs.
$$ \begin{array}{rrrr} 2x^2 & x & -2 \\ \color{red}{\bf{-}}x^2 & \color{red}{\bf {+}}3x & \color{red}{\bf{-}}5 \\ \hline \end{array} $$Now we can combine like terms
$$ \begin{array}{rrrr} 2x^2 & x & -2 \\ -x^2 & 3x & -5 \\ \hline x^2 & 4x & -7 \\ \end{array} $$So the answer is $ x^2 + 4x - 7 $
Example 03: Subtract $ (5x - 7) - (3x - 3) $
Here we remove parenthesis by changing the sign of every term in the second bracket.
$$ (5x - 7) \color{blue}{- (3x - 3)} = 5x - 7 \color{blue}{- 3x + 3} $$Now, as in previous example, group the like terms ...
$$ 5x - 7 -3x + 3 = 5x - 3x -7 + 3 $$...and combine them:
$$ 5x - 3x - 7 + 3 = 2x - 4 $$Putting all together we have
$$ \begin{aligned} (5x-7) - (3x-3) \overbrace{=}^{\text{remove par.}}& 5x - 7 - 3x + 3 = \\ \overbrace{=}^{\text{group like terms}}& 5x - 3x - 7 + 3 = \\ \overbrace{=}^{\text{combine like terms}}& 2x - 4 \end{aligned} $$This method is used to multiply two binomials. The best way to explain the FOIL method is to use an example:
Example 05: Use FOIL method to multiply $ (5a + 2) \cdot(2a - 3) $
$$ \begin{array}{lcccc} \text{First} & : & 5a & \cdot & 2a & = & 10a^2 \\ \text{Outer} & : & 5a & \cdot & -3 & = & -15a \\ \text{Inner} & : & 2 & \cdot & 2a & = & 4a \\ \text{Last} & : & 2 & \cdot & -3 & = & -6 \\ \hline \end{array} $$Please tell me how can I make this better.