Polynomials with four terms can sometimes be factored using a method called factoring by grouping. This method involves grouping the terms into two pairs and taking a common factor out of each pair, then taking a common binomial factor out of the two resulting terms.

\displaystyle ax + ay + bx + by	\displaystyle =	\displaystyle a\left(x + y\right) + b\left(x + y\right)
	\displaystyle =	\displaystyle \left(x+y\right)\left(a+b\right)

Sometimes the terms will need to be rearranged before common factors can be taken out of each pair.

It is also possible to factor some polynomials with more than four terms using this method - for instance, a polynomial with six terms might be able to split into three pairs which each leave behind a common binomial.

Note that not every four-term polynomial is factorable in this way. For instance, the polynomial3x + 6y + x + 5zdoesn't have enough common factors between any possible pairing of terms.

Worked examples

Example 1

Factor the expression 10 xy + 4x + 15y + 6.

Approach

We arrange the terms first, grouping those with common factors. We factor out the GCF on each pair and the common binomial factor afterwards.

Solution

\displaystyle 10 xy + 4x + 15y + 6	\displaystyle =	\displaystyle \left(10 xy + 4x\right)+ \left(15y + 6\right)	Split based on common factors
	\displaystyle =	\displaystyle 2x\left(5y + 2\right)+ 3\left(5y + 2\right)	Factor out each GCF \left(2x \text{ and } 3 \right)
	\displaystyle =	\displaystyle \left(5y + 2\right)\left(2x + 3\right)	Factor out the common binomial factor

Since \left(5y + 2\right)\left(2x + 3\right) cannot be factored further, it is the final answer.

Reflection

Alternatively, we can group 10xy \text{ and } 15y and 4x \text{ and } 6 together and get the same answer.

\displaystyle 10 xy + 4x + 15y + 6	\displaystyle =	\displaystyle \left(10 xy + 15y\right)+ \left(4x + 6\right)	Split based on common factors
	\displaystyle =	\displaystyle 5y\left(2x + 3\right)+ 2\left(2x + 3\right)	Factor out each GCF \left(5y \text{ and } 2 \right)
	\displaystyle =	\displaystyle \left(2x + 3\right)\left(5y + 2\right)	Factor out the common binomial factor

We can check the answer by multiplying the factored form \left(5y + 2\right)\left(2x + 3\right).

\displaystyle \left(5y + 2\right)\left(2x + 3\right)	\displaystyle =	\displaystyle 2x\left(5y + 2\right) + 3\left(5y + 2\right)	Distribute \left(5y + 2\right)
	\displaystyle =	\displaystyle 10xy + 4x + 15y + 6	Check

Example 2

Factor the expression 4a^{2} + 5ab - 8a - 10b.

Approach

We arrange the terms first, grouping those with common factor. We factor out the GCF on each pair and the common binomial factor afterwards.

Solution

\displaystyle 4a^{2} + 5ab - 8a - 10b	\displaystyle =	\displaystyle \left(4a^{2} + 5ab \right)+ \left(- 8a - 10b\right)	Split based on common factors
	\displaystyle =	\displaystyle a\left(4a+5b\right) - 2\left(4a+5b\right)	Factor out the GCF \left(a \text{ and } -2 \right)
	\displaystyle =	\displaystyle \left(4a + 5b\right)\left(a - 2\right)	Factor out the common binomial factor

Since \left(4a + 5b\right)\left(a - 2\right) cannot be factored further, it is the final answer.

Reflection

Alternatively, we can group 4a^{2} \text{ and } - 8a and 5ab \text{ and } - 10b together and get the same answer.

\displaystyle 4a^{2} + 5ab - 8a - 10b	\displaystyle =	\displaystyle \left(4a^{2} - 8a \right)+ \left(5ab - 10b\right)	Split based on common factors
	\displaystyle =	\displaystyle 4a\left(a- 2\right) + 5b\left(a-2\right)	Factor out the GCF \left(4a \text{ and } 5b \right)
	\displaystyle =	\displaystyle \left(a - 2\right)\left(4a + 5b\right)	Factor out the common binomial factor

We can check the answer by multiplying the factored form \left(4a + 5b\right)\left(a - 2\right).

\displaystyle \left(4a + 5b\right)\left(a - 2\right)	\displaystyle =	\displaystyle a\left(4a + 5b\right) -2\left(4a + 5b\right)	Distribute \left(4a + 5b\right)
	\displaystyle =	\displaystyle 4a^{2} + 5ab - 8a- 10b	Check

Example 3

Factor the expression 3x^{3} + 3x^{2} - 6x - 6.

Approach

We arrange the terms first, grouping those with common factor. We factor out the GCF on each pair and the common binomial factor afterwards.

Solution

\displaystyle 3x^{3} + 3x^{2} - 6x - 6	\displaystyle =	\displaystyle \left(3x^{3} + 3x^{2} \right)+ \left(- 6x - 6\right)	Split based on common factors
	\displaystyle =	\displaystyle 3x^2\left(x+1\right) - 6\left(x+1\right)	Factor out the GCF \left(3x^{2} \text{ and } -6 \right)
	\displaystyle =	\displaystyle \left(x+1\right)\left(3x^{2} - 6\right)	Factor out the common binomial factor
	\displaystyle =	\displaystyle 3\left(x+1\right)\left(x^{2} - 2\right)	Factor out 3 in 3x^{2} - 6

Since 3\left(x+1\right)\left(x^{2} - 2\right) cannot be factored further, it is the final answer.

Reflection

Alternatively, we can also factor out 3 at the start.

We can check the answer by multiplying the factored form 3\left(x+1\right)\left(x^{2} - 2\right).

\displaystyle 3\left(x+1\right)\left(x^{2} - 2\right)	\displaystyle =	\displaystyle 3\left(x^{2}\left(x+1\right) - 2\left(x+1\right)\right)	Distribute \left(x + 1\right)
	\displaystyle =	\displaystyle 3\left(x^{3}+x^{2} - 2x-2\right)	Multiply
	\displaystyle =	\displaystyle 3x^{3} + 3x^{2} - 6x - 6	Check

Outcomes

A1.A.APR.A.1

Add, subtract, and multiply polynomials. Use these operations to demonstrate that polynomials form a closed system that adhere to the same properties of operations as the integers.

6.06 Factoring by grouping

Concept summary

Worked examples

Example 1

Approach

Solution

Reflection

Example 2

Approach

Solution

Reflection

Example 3

Approach

Solution

Reflection

Outcomes

A1.A.APR.A.1

A1.MP1

A1.MP4

A1.MP7

A1.MP8

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