# Factor by Grouping, Part One

Reference > Mathematics > Algebra > Factoring Higher Degree PolynomialsSometimes you will look at a polynomial and think, "None of the factoring rules I've been taught will help me factor this thing." And some polynomials can't be factored. But sometimes, a little bit of clever rearranging or grouping will let you do more factoring than you thought you could. Instead of relying on rules you've been taught, you might need to have some intuitive leaps. Let me show you an example.

**Example One**

Factor the cubic polynomial x^{3} + 2x^{2} + 2x + 4

**Solution**

When I look at this, one of the first things I notice is that in the last two terms, the coefficients are twice the first two coefficients. This gives me the idea to split this polynomial up into pairs of terms like this:

(x^{3} + 2x^{2}) + (2x + 4)

Now I'm going to try factoring both groups, and see what happens.

x^{2}(x + 2) + 2(x + 2)

Notice that the part in parentheses matches in both groups: (x + 2). This means that we can factor out (x + 2) to obtain the following:

(x + 2)(x^{2} + 2)

**Example Two**

Factor the cubic polynomial 4x^{3} - x^{2} - 16x + 4

**Solution**

Again, I notice that there is a pattern across the pairs of factors, so I try this:

(4x^{3} - x^{2}) - (16x - 4)

x^{2}(4x - 1) - 4(4x - 1)

Again we factor out the common binomial (4x - 1):

(4x - 1)(x^{2} - 4)

Note that we now have a difference of squares in the second factor, so this becomes:

(4x - 1)(x + 2)(x - 2)

**Example Three**

Factor x^{5} + 3x^{4} + 2x^{3} + x^{2} + 3x + 2

**Solution**

This time I notice that a nice pattern happens if I group the terms into two trinomials:

(x^{5} + 3x^{4} + 2x^{3}) +(x^{2} + 3x + 2)

x^{3}(x^{2} + 3x + 2) + 1(x^{2} + 3x + 2)

(x^{2} + 3x + 2)(x^{3} + 1)

(x + 2)(x + 1)(x + 1)(x^{2} - x + 1)

(x + 1)^{2}(x + 2)(x^{2} -x + 1)

Note that in this problem, we needed to factor a quadratic trinomial and a sum of cubes to finish the problem.

## Questions

^{4}+ 3x

^{3}-27x - 81

^{3}-4x

^{2}- 2x + 8

^{6}+ x

^{5}- x - 1

^{5}+ 2x

^{4}+ x

^{3}+ 8x

^{2}+ 16x + 8

^{12}- 4x

^{8}- 4x

^{4}+ 16

^{4}+ 3x

^{3}+ 3x

^{2}+ 3x + 2 (HINT: Split the 3x

^{2}term into two terms, in such a way that you have two trinomials with identical coefficients)