Factoring is an important skill going into calculus. You will be doing a LOT of factoring in calculus. Master these techniques now and you will do much better in calculus. Many students struggle with calculus because their algebra skills are lacking, including factoring.

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There are lots of factoring techniques and, unfortunately, there is no one technique that works in all cases. So you need to learn them well. You will find several that you will use a lot. Learn those first. Then keep a bookmark for this page and come back here when you need to remind yourself of the other techniques.

The major techniques are labeled as primary. Learn these well. The secondary techniques are usually special cases and you will use them but not as often as the primary techniques.

However, there are a few steps that will help you get started with all equations.

* Steps *

1. Rewrite the polynomial with terms in order, highest term on the left. Of course, it would still work if you wrote them in reverse order but most of the time you will see your instructor and textbook written left to right, highest power on the left. So we suggest that you do the same.

2. Look for obvious common factors in all terms, like \(x\) or constants. You do not need to see all of them at the same time. If you see one term, factor it out and then look for more until you think they are all factored out. Here is an example.

Completely factor this polynomial.

\( 3x^2 + 6x \)

Problem Statement |
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Factor \( 3x^2 + 6x \)

Final Answer |
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\( 3x^2 + 6x = 3x(x + 2) \)

Problem Statement

Factor \( 3x^2 + 6x \)

Solution

First, we will break each factor down into individual terms. |

\( 3x^2 = 3 \cdot x \cdot x\) |

\( 6x = 2 \cdot 3 \cdot x \) |

Now we can see the common factors in each term. |

In both terms we have a 3, so let's factor out a 3 first. |

\( 3 \cdot x \cdot x + 2 \cdot 3 \cdot x = 3( x \cdot x + 2 \cdot x) \) |

Okay, so now let's look at what is left inside the parentheses. Notice that in each of those terms, we have an \(x\). So let's factor that out. |

\( 3( x \cdot x + 2 \cdot x) = 3x(x + 2) \) |

Now, let's again look at what is inside the parentheses. Notice that we have two terms with nothing in common. One has \(x\). The other has \(2\). Since they do not have anything in common, we are done. |

Okay, so that is how we would factor this polynomial. Here is a video showing the GCF method. See which why you think is easier to understand. However, as usual, check with your instructor to see what they require.

video by Freshmen Math Doctor |
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Final Answer

\( 3x^2 + 6x = 3x(x + 2) \)

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Before we get into the details of factoring, let's watch this video clip as an overview. This instructor shows several techniques with a few quick examples. If you don't understand everything in this video, that's okay. We cover most of this techniques on this page and give you a chance to practice them.

video by freeCodeCamp.org |
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Factoring By GCF (Greatest Common Factor)

Before we go on, we want to mention this technique. We have recently watched videos with instructors using this technique (including the last video above). Many instructors just seem to wave their hands and come up with the GCF of each term. We recommend instead to use the step-by-step, one factor at a time technique we demonstrated in the previous example. Essentially, it is the same as GCF but you do not need to come up with the one large factor all at once. Of course, make sure you check with your instructor to see what they expect.

Testing Linear Factors

Before we get into factoring techniques, there is one concept that will help you a lot. Let's say you have a polynomial and you suspect that \((x-1)\) is a factor. What you can do is set this equal to zero and solve for \(x\), i.e. \( x-1=0 \to x=1\) and substitute \(x=1\) into the polynomial. If the result is zero, then \((x-1)\) is a factor of the polynomial. Use long division of polynomials or synthetic division to factor it out. This will reduce the highest power by one and perhaps give you a polynomial that you can then factor using simpler techniques.

So your next question is, why would I think that \((x-1)\) might be a factor? Well, one idea is to plot the polynomial on your calculator (if your instructor allows it) and see where it might cross the *x*-axis, i.e. try to see if you can find any real zeroes/roots. This can reduce the complexity of the polynomial until it is more manageable.

Factoring Quadratics/Trinomials (Primary)

You will see quadratics often in calculus. There are several techniques that you can use on quadratics. Check out the practice problems for examples.

Unless otherwise instructed, factor these polynomials. Give your answers in simplified, completely factored form.

\( x^2 - 4x - 5 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( x^2 - 4x - 5 \)

Solution

video by Freshmen Math Doctor |
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\( x^2 + 8x + 15 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( x^2 + 8x + 15 \)

Solution

video by Freshmen Math Doctor |
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\( x^2 - 14x + 45 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( x^2 - 14x + 45 \)

Solution

video by Freshmen Math Doctor |
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Solve \( x^2 = -11x - 10 \) by factoring.

Problem Statement

Solve \( x^2 = -11x - 10 \) by factoring.

Solution

video by Freshmen Math Doctor |
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Solve \( x^2 - 13x + 36 = 0 \) by factoring.

Problem Statement

Solve \( x^2 - 13x + 36 = 0 \) by factoring.

Solution

video by Freshmen Math Doctor |
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\( 6x^4 - 18x^3 + 12x^2 \)

Problem Statement

Unless otherwise instructed, factor \( 6x^4 - 18x^3 + 12x^2 \)

Solution

video by Freshmen Math Doctor |
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\( x^2+4x - 12 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( x^2+4x - 12 \)

Solution

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\( 3x^2 + 12x - 36 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( 3x^2 + 12x - 36 \)

Solution

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\( 3x^2 + 10x - 8 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( 3x^2 + 10x - 8 \)

Solution

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\( 8x^2 + 35x + 12 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( 8x^2 + 35x + 12 \)

Solution

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\( 6x^2 - 3x - 45 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( 6x^2 - 3x - 45 \)

Solution

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Factor By Grouping (Secondary)

When it appears that you have groups of similar terms, try pairing them up and factoring them together and then seeing if you have the same terms in each group. This technique is best seen by example. Look at the practice problems for examples.

Unless otherwise instructed, factor these polynomials. Give your answers in simplified, completely factored form.

\( x^3 - x^2 - 5x + 5 \)

Problem Statement

Factor \( x^3 - x^2 - 5x + 5 \)

Solution

video by Freshmen Math Doctor |
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\( x^3 - 3x^2 + 4x - 12 \)

Problem Statement

Factor \( x^3 - 3x^2 + 4x - 12 \)

Solution

video by Freshmen Math Doctor |
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\( x^3 + 2x^2 - 5x - 10 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( x^3 + 2x^2 - 5x - 10 \)

Solution

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\( 4x^3 - 8x^2 + 6x - 12 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( 4x^3 - 8x^2 + 6x - 12 \)

Solution

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\( x^3 + 3x^2 - 4x - 12 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( x^3 + 3x^2 - 4x - 12 \)

Solution

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\( x^3 - 4x^2 + x + 6 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( x^3 - 4x^2 + x + 6 \)

Solution

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\( 5v^3 - 2v^2 + 25v - 10 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( 5v^3 - 2v^2 + 25v - 10 \)

Solution

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\( 5r^4 - 7r^2s - 6s^2 \)

Problem Statement

Unless otherwise instructed, factor the polynomial \( 5r^4 - 7r^2s - 6s^2 \)

Solution

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\( 6xy - 9y - 10x -15 \)

Problem Statement

Unless otherwise instructed, factor \( 6xy - 9y - 10x -15 \)

Solution

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\( 15 - 5A^2 - 3B^2 + A^2B^2 \)

Problem Statement

Unless otherwise instructed, factor \( 15 - 5A^2 - 3B^2 + A^2B^2 \)

Solution

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