Factoring Calculator

Result
\left(x + 2\right)\left(2x - 1\right)\left(x - 3\right)

The factored form is \displaystyle \left(x + 2\right)\left(2x - 1\right)\left(x - 3\right).

  • Type: Polynomial
  • Method: Factor
  • Factors: \displaystyle \left(x + 2\right), \displaystyle \left(2x - 1\right), \displaystyle \left(x - 3\right)
  • Zeros: \displaystyle -2,\ \frac{1}{2},\ 3 (where it equals 0)

Step-by-step solution

The problem

2x^{3} - 3x^{2} - 11x + 6
1

Look for a rational root

2 \cdot \left(-2\right)^{3} - 3 \cdot \left(-2\right)^{2} - 11 \cdot \left(-2\right) + 6 = 2 \cdot \left(-8\right) - 3 \cdot 4 - 11 \cdot \left(-2\right) + 6 = -16 - 12 + 22 + 6 = 0

By the rational-root theorem any rational root is \displaystyle \pm\frac{p}{q} with \displaystyle p a factor of the constant \displaystyle 6 and \displaystyle q a factor of the leading coefficient \displaystyle 2: \displaystyle \pm \frac{1}{2}, \pm 1, \pm \frac{3}{2}, \pm 2, \pm 3, \pm 6. Trying them, \displaystyle x = -2 works (the value comes out as 0).

2

Divide by \displaystyle \left(x + 2\right) (synthetic division)

\begin{array}{r|rrrr} -2 & 2 & -3 & -11 & 6 \\ & & -4 & 14 & -6 \\ \hline & 2 & -7 & 3 & \boxed{0} \end{array}

Bring down the first coefficient, multiply by \displaystyle -2, add to the next coefficient, and repeat. The last number is the remainder 0.

3

Write the factors

2x^{3} - 3x^{2} - 11x + 6 = \left(x + 2\right)\left(2x^{2} - 7x + 3\right)
4

Use the AC method

\begin{aligned} -1 \times \left(-6\right) &= 6 \\[4pt] -1 + \left(-6\right) &= -7 \end{aligned}

Find two numbers that multiply to \displaystyle a \cdot c = 2 \cdot 3 = 6 and add to \displaystyle b = -7: they are \displaystyle -1 and \displaystyle -6.

5

Split the middle term

2x^{2} - 7x + 3 = 2x^{2} - x - 6x + 3

Write \displaystyle -7x as \displaystyle -x - 6x.

6

Factor by grouping

\begin{aligned} &= \left(2x^{2} - x\right) + \left(-6x + 3\right) \\[4pt] &= x\left(2x - 1\right) - 3\left(2x - 1\right) \\[4pt] &= \left(x - 3\right)\left(2x - 1\right) \end{aligned}

Take the common factor out of the first two terms and out of the last two – the same bracket is left in both.

7

Fully factored

2x^{3} - 3x^{2} - 11x + 6 = \left(x + 2\right)\left(2x - 1\right)\left(x - 3\right)

Graph

  • \displaystyle y = 2x^{3} - 3x^{2} - 11x + 6

The graph of \displaystyle y = 2x^{3} - 3x^{2} - 11x + 6. Its zeros, turning points and any holes are marked.

Crosses the x-axis at
x = −2, 1/2, 3
Crosses the y-axis at
(0, 6)
Local maximum
(−0.943, 12.028)
Local minimum
(1.943, −12.028)
−3−2−11234−5050

Type a polynomial like 6x^2 + x - 2 or x^3 - 8. Use ^ for powers.

Try an example

Trinomials x^2 - 5x + 66x^2 + x - 2x^2 + 5xy + 6y^2
Special 4x^2 - 9x^3 - 8x^2 + 6x + 9
Grouping x^3 + x^2 - 4x - 4x^2 + 6x + 9 - y^2
Higher degree 2x^3 - 3x^2 - 11x + 6x^4 - 16x^6 - 64

Type a polynomial and it is factored the way it is taught – common factor first, then the pattern that fits: 6x² + x − 2 = (2x − 1)(3x + 2) by the AC method. It also spots differences of squares, perfect squares and sums or differences of cubes, groups four-term polynomials and uses the rational root theorem on cubics.

The factoring checklist

Work through these in order – it is the order the calculator uses:

  1. Greatest common factor. Always first: 12x^2y - 18xy^2 = 6xy(2x - 3y).
  2. Two terms. Difference of squares a^2 - b^2 = (a - b)(a + b); sum or difference of cubes a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2). A sum of squares a^2 + b^2 does not factor.
  3. Three terms. A perfect square a^2 \pm 2ab + b^2 = (a \pm b)^2? Otherwise find two numbers that multiply to ac and add to b – directly when a = 1, by splitting the middle term and grouping (the AC method) when a ≠ 1. The same works for x^4 + bx^2 + c, treating x^2 as the variable.
  4. Four terms. Group them in pairs; or spot a perfect square trinomial minus a square.
  5. Higher degree. Use the rational-root theorem to find a root r, divide by (x - r) with synthetic division, and factor what is left.
  6. Check every factor again – x^4 - 16 = (x^2 - 4)(x^2 + 4) still has x^2 - 4 = (x - 2)(x + 2) inside.

Special patterns at a glance

PatternFactoredExample
Difference of squaresa^2 - b^2 = (a - b)(a + b)9x^2 - 25 = (3x - 5)(3x + 5)
Perfect squarea^2 + 2ab + b^2 = (a + b)^2x^2 + 10x + 25 = (x + 5)^2
Sum of cubesa^3 + b^3 = (a + b)(a^2 - ab + b^2)8x^3 + 27 = (2x + 3)(4x^2 - 6x + 9)
Difference of cubesa^3 - b^3 = (a - b)(a^2 + ab + b^2)x^3 - 8 = (x - 2)(x^2 + 2x + 4)

The AC method

Factor 6x^2 + x - 2: ac = -12; the numbers −3 and 4 multiply to −12 and add to 1.

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

Factoring by grouping

Four terms often pair up. In x^3 + 3x^2 - 4x - 12 the first two share x^2 and the last two share −4:

x^2(x + 3) - 4(x + 3) = (x + 3)(x^2 - 4) = (x + 3)(x - 2)(x + 2)

The bracket left over, x^2 - 4, is a difference of squares – factoring isn’t finished until no factor splits further. See the grouping in the calculator.

Why factor? Solving equations

A product is zero only when one of its factors is zero. Once x^3 + 3x^2 - 4x - 12 = 0 is written as (x + 3)(x - 2)(x + 2) = 0, the solutions can be read off: x = −3, 2 or −2. The same factors show where a graph crosses the x-axis and simplify algebraic fractions. Check any factoring by multiplying the brackets out again – you must get the original polynomial back.

When a polynomial cannot be factored with whole numbers it is called prime (irreducible over the rationals) – for example x^2 + 1 or x^2 - 2. To find the roots of such a quadratic, use the quadratic equation solver.

Frequently asked questions

What does it mean if a polynomial is prime?

It has no factors other than 1 and itself using whole-number coefficients, like x² + x + 1.