Polynomial Factoring Calculator (2024)

Polynomial Calculators

• FactoringPolynomials

• Polynomial Roots
• Synthetic Division
• PolynomialOperations
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• Simplify Polynomials
• Generate From Roots

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• Simplify Expression

• Multiplication / Division
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Radical Expressions

• Rationalize Denominator

• Simplifying

Solving Equations

• Quadratic Equations Solver

• Polynomial Equations
• Solving Equations - WithSteps

Quadratic Equation

• Solving (with steps)

• Quadratic Plotter
• Factoring Trinomials

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Analytic Geometry

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  See Also

  Inches to cm Conversion (Inches To Centimeters) - Inch CalculatorWhat Is Chi Square Test & How To Calculate Formula EquationSynthetic Division CalculatorQuadratischer Formelrechner - Rechner Wow

• Midpoint Calculator
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Calculus Calculators

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Sequences & Series

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• Find nth Term

Trigonometry

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• Fraction Calculator
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• Prime Factorization
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• Dec / Bin / Hex

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Financial Calculators

• Simple Interest

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Other Calculators

• Sets

• Work Problems

factor $x^2 - 10x + 25$

factor $3x^2 - 27$

factor $4x^3 - 21x^2 + 29x -6$

factor $3uv-12u+6v-24 $

factor $a^2 + 2a + 1 - b^2$

Polynomial Factoring Techniques

To find the factored form of a polynomial, this calculator employs thefollowing methods:

1. Factoring GCF, 2 Factoring by grouping, 3 Using thedifference of squares, and 4 Factoring Quadratic Polynomials

Method 1 : Factoring GCF

Example 01: Factor $ 3ab^3 - 6a^2b $

$$\begin{aligned}3ab^3 - 6a^2b &= \color{blue}{3 \cdot a \cdot b} \cdot b \cdot b - 2 \cdot \color{blue}{3 \cdot a } \cdot a\color{blue}{\cdot b} = \\[1 em]&= \color{blue}3ab(b^2-2a)\end{aligned}$$

Method 2 : Factoring By Grouping

The method is very useful for finding the factored form of the four term polynomials.

Example 03: Factor $ 2a - 4b + a^2 - 2ab $

We usually group the first two and the last two terms.

$$ 2a - 4b + a^2 - 2ab = \color{blue}{2a - 4b} + \color{red}{a^2 - 2ab} $$

We now factor $ \color{blue}{2} $ out of the blue terms and $ \color{red}{a}$ out of from red ones.

$$ \color{blue}{2a - 4b} + \color{red}{a^2 - 2ab} = \color{blue}{2 \cdot (a - 2b)} + \color{red}{a \cdot (a- 2b) } $$

At the end we factor out common factor of $ (a - 2b) $

$$ 2 \cdot \color{blue}{(a - 2b)} + a \cdot \color{blue}{(a - 2b)} = (a - 2b) (2 + a) $$

Example 04: Factor $ 5ab + 2b + 5ac + 2c $

This is a rare situation where the first two terms of a polynomial do nothave a common factor, so we have to group the first and third terms together.

$$ \begin{aligned}5ab + 2b + 5ac + 2c &= \color{blue}{5ab + 5ac} + \color{red}{2b + 2c} = \\&=\color{blue}{5a (b + c)} + \color{red}{2 (b + c)} = \\&= (b+c)(\color{blue}{5a} + \color{red}{2})\end{aligned}$$

Method 3 : Special Form - A

The most common special case is the difference of two squares

$$ a^2 - b^2 = (a+b) (a-b) $$

We usually use this method when the polynomial has only two terms.

Example 05: Factor $ 4x^2 - y^2 $.

First, we need to notice that the polynomial can be written as thedifference of two perfect squares.

$$ \color{blue}{4x^2} - \color{red}{y^2} = \color{blue}{ \left(2x\right)}^2 - \color{red}{y}^2 $$

Now we can apply above formula with $ \color{blue}{a = 2x} $ and $\color{red}{b = y} $

$$ \left( \color{blue}{2x} \right)^2 - \color{red}{y}^2 = (2x - b) (2x + b) $$

Example 06: Factor $ 9a^2b^4 - 4c^2 $.

The binomial we have here is the difference of two perfect squares, thus thecalculation will be similar to the last one.

$$ 9a^2b^4 - 4c^2 = \left( 3ab^2 \right)^2 - \left( 2c \right)^2 = (3ab^2-2c)(3ab^2+2c) $$

Method 4 : Special Form - B

The second special case of factoring is the Perfect Square Trinomial

$$ a^2 \pm 2ab + b^2 = (a \pm b)^2 $$

Example 07: Factor $ 100 + 20x + x^2 $

In this case, the first and third terms are perfect squares. So we can usethe above formula.

$$ \color{blue}{100} + 20x + \color{red}{x^2} = \color{blue}{10}^2 + 2 \cdot \color{blue}{10} \cdot\color{red}{x} + \color{red}{x}^2 = (10 + x)^2 $$

Method 5: Factoring Quadratic Polynomials

The best way to explain this method is by using an example.

Example 08: Factor $ x^2 + 3x + 4 $

We know that the factored form has the following pattern

$$ x^2 + 5x + 4 = (x + \_ ) (x + \_ ) $$

All we have to do now is fill in the blanks with the two numbers. To dothis, notice that the product of these two numbers has to be 4 and their sum has to be 5.We conclude, after some trial and error, that the missing numbers are $ \color{blue}{1} $ and $\color{red}{4} $. As a result, the solution is::

$$ x^2 + 5x + 4 = (x + \color{blue}{1} ) (x + \color{red}{4} ) $$

Example 09: Factor $ x^2 - 8x + 15 $

Like in the previous example, we look again for the solution in the form

$$ x^2 - 8x + 15 = (x + \_ ) (x + \_ ) $$

The sum of missing numbers is $-8$ so we need to find two negative numbers such that the productis $15$ and the sum is $-8$.It's not hard to see that two numbers with such properties are $-3$ and $-5$, so the solution is.

$$ x^2 + 5x + 4 = (x + \color{blue}{-3} ) (x + \color{red}{-5} ) = (x-3)(x-5) $$