Factoring by Grouping

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Factoring by grouping is a way to factor a polynomial with four terms. Group the terms in pairs, factor the GCF out of each pair, and then factor out the binomial they share. For example, x^3 + 3x^2 + 2x + 6 = x^2(x + 3) + 2(x + 3) = (x + 3)(x^2 + 2). It also finishes the AC method for trinomials.

Before this lesson: You should know factoring out the GCF, including a binomial GCF, and factoring trinomials when a is not 1.

How Grouping Works

A GCF can be a whole binomial. In Factoring Out the GCF, you saw 3x(x + 2) + 5(x + 2) = (x + 2)(3x + 5). Grouping turns four terms into that shape.

First, split the four terms into two pairs. Then pull the GCF out of each pair. If the two leftover binomials match, that binomial is a common factor.

Three rows. Top: x³ + 3x² + 2x + 6, with the first two terms marked group 1 by a blue bracket and the last two marked group 2 by a green dashed bracket. Middle: factor each group to get x²(x + 3) + 2(x + 3), with both copies of (x + 3) in dashed orange boxes. Bottom: factor out the common binomial to get (x + 3)(x² + 2).
Each group leaves the same binomial, x + 3. Factor it out to finish.

Steps for Factoring by Grouping

  1. Factor out the GCF of all four terms, if there is one.
  2. Group the first two terms and the last two terms.
  3. Factor the GCF out of each group. If the third term is negative, factor out a negative.
  4. Check that the two binomials match. Then factor out the common binomial.
  5. If they don’t match, try putting the terms in a different order.
  6. Check by multiplying.
ax + ay + bx + by = a(x + y) + b(x + y) = (x + y)(a + b)

Worked Examples

Example 1: Four positive terms

Factor x^3 + 3x^2 + 2x + 6.

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

Check: (x + 3)(x^2 + 2) = x^3 + 2x + 3x^2 + 6. That is the original polynomial in a different order.

Example 2: Coefficients in each group

Factor 2x^3 - 8x^2 + 3x - 12.

The first group has GCF 2x^2. The second group has GCF 3.

\begin{aligned} 2x^3 - 8x^2 + 3x - 12 &= 2x^2(x - 4) + 3(x - 4) \\ &= (x - 4)(2x^2 + 3) \end{aligned}

Example 3: A negative third term

Factor x^3 - 2x^2 - 5x + 10.

The third term is negative, so factor -5 out of the second group. Watch the sign of the last term: 10 \div (-5) = -2.

\begin{aligned} x^3 - 2x^2 - 5x + 10 &= x^2(x - 2) - 5(x - 2) \\ &= (x - 2)(x^2 - 5) \end{aligned}

If you factored out +5 instead, you would get 5(-x + 2), which does not match x - 2.

Example 4: Finishing the AC method

Factor 3x^2 + 10x + 8 by grouping.

From the AC method, ac = 24, and 6 and 4 multiply to 24 and add to 10. Split 10x into 6x + 4x. Now there are four terms to group.

\begin{aligned} 3x^2 + 10x + 8 &= 3x^2 + 6x + 4x + 8 \\ &= 3x(x + 2) + 4(x + 2) \\ &= (x + 2)(3x + 4) \end{aligned}

This matches the box method answer, (3x + 4)(x + 2). Order does not matter when you multiply.

Example 5: Word problem

A rectangle has an area of xy + 2x + 5y + 10 square feet. Find binomials for its length and width. Then find the area when x = 3 and y = 4.

\begin{aligned} xy + 2x + 5y + 10 &= x(y + 2) + 5(y + 2) \\ &= (y + 2)(x + 5) \end{aligned}

The sides are x + 5 feet and y + 2 feet. When x = 3 and y = 4, they are 8 feet and 6 feet, so the area is 48 square feet. Check: 12 + 6 + 20 + 10 = 48.

Common Mistakes

MistakeExample of the errorFix
Wrong sign in the second groupx^3 - 2x^2 - 5x + 10 = x^2(x - 2) - 5(x + 2)-5(x + 2) = -5x - 10, not -5x + 10. Use -5(x - 2).
Stopping after groupingx^2(x + 3) + 2(x + 3)This is still a sum. Factor out (x + 3): (x + 3)(x^2 + 2).
Forgetting the 1x^3 + 4x^2 + x + 4 = (x + 4)(x^2)The second group is 1(x + 4). The answer is (x + 4)(x^2 + 1).
Giving up when the binomials don’t matchx^2 + 3y + xy + 3x: the first pair has no common factorReorder the terms: x^2 + 3x + xy + 3y = x(x + 3) + y(x + 3) = (x + 3)(x + y).

Grouping Tips

SituationWhat to do
All four terms share a factorTake it out first. 2x^3 + 6x^2 + 4x + 12 = 2(x^3 + 3x^2 + 2x + 6)
Third term is negativeFactor a negative out of the second group.
Binomials don’t matchSwap the middle two terms and try again.
Trinomial with a not 1Split the middle term with the AC method, then group.

Grouping only works when the two binomials match. If no order of the terms gives a match, the polynomial may not factor by grouping. Sometimes a factor like x^2 - 9 can be factored again. You will learn that in the next lesson.

You can always check your answer by multiplying the factors back together with the distributive property.

Practice Problems

  1. Factor x^3 + 2x^2 + 4x + 8.
  2. Factor x^3 - 6x^2 + 2x - 12.
  3. Factor 3x^3 + 12x^2 + 2x + 8.
  4. Factor x^3 + 7x^2 - 3x - 21.
  5. Factor 5x^3 - 15x^2 - x + 3.
  6. Factor ab + 4a + 2b + 8.
  7. Factor 10xy - 15x + 4y - 6.
  8. Factor 2x^2 + 11x + 12 by splitting the middle term and grouping.
  9. Factor 2x^3 + 6x^2 + 4x + 12 completely.
  10. A garden has an area of xy + 3x + 4y + 12 square meters. Find binomials for its length and width. Then find the area when x = 2 and y = 5.
Show answers
  1. x^2(x + 2) + 4(x + 2) = (x + 2)(x^2 + 4)
  2. x^2(x - 6) + 2(x - 6) = (x - 6)(x^2 + 2)
  3. 3x^2(x + 4) + 2(x + 4) = (x + 4)(3x^2 + 2)
  4. x^2(x + 7) - 3(x + 7) = (x + 7)(x^2 - 3)
  5. 5x^2(x - 3) - 1(x - 3) = (x - 3)(5x^2 - 1). Factor out -1 from the second group.
  6. a(b + 4) + 2(b + 4) = (b + 4)(a + 2)
  7. 5x(2y - 3) + 2(2y - 3) = (2y - 3)(5x + 2)
  8. ac = 24, and 3 and 8 add to 11. 2x^2 + 3x + 8x + 12 = x(2x + 3) + 4(2x + 3) = (2x + 3)(x + 4)
  9. Take out 2 first: 2(x^3 + 3x^2 + 2x + 6) = 2(x + 3)(x^2 + 2)
  10. x(y + 3) + 4(y + 3) = (y + 3)(x + 4). When x = 2 and y = 5: 6 \cdot 8 = 48 square meters.

FAQ

When do you use factoring by grouping?

Use it when a polynomial has four terms, or when you split the middle term of a trinomial like ax^2 + bx + c with the AC method.

What if the two binomials don’t match?

Check the signs first, especially in the second group. If they still don’t match, reorder the terms and try again. Some polynomials can’t be factored by grouping.

When should I factor out a negative?

When the third term is negative, factor a negative out of the second group. That keeps the binomials the same, as in -5(x - 2).

Does factoring by grouping work for trinomials?

Yes, after you split the middle term into two terms. For 3x^2 + 10x + 8, write 3x^2 + 6x + 4x + 8 and then group.

Summary

  • Factoring by grouping works on four terms.
  • Group the terms in pairs and factor the GCF out of each pair.
  • If the binomials match, factor out the common binomial.
  • Factor out a negative when the third term is negative, and don’t forget a leftover 1.
  • Use grouping to finish the AC method for trinomials. Check by multiplying.

Previous lesson: Factoring Trinomials When a Is Not 1

Next lesson: Factoring the Difference of Squares (coming soon)

Back to the Algebra 1 Index

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