# Why does factoring polynomials by grouping work?

Mar 12, 2015

It works for some polynomials but not for others. Mostly , it works for this polynomial because the teacher, or author, or test-maker, chose a polynomial that could be factored this way.

Example 1

Factor: $3 {x}^{3} + 6 {x}^{3} - 5 x - 10$

I group the first two terms and take out any common factor of those two:
$\left(3 {x}^{3} + 6 {x}^{2}\right) - 5 x - 10 = 3 {x}^{2} \left(x + 2\right) - 5 x - 10$

Now I'll take out any common factors in the other two terms. If i get a monomial times $\left(x + 2\right)$ then factoring by grouping will work. If I get something else, it won't work.

Ther common factor of $\left(- 5 x - 10\right)$ is $- 5$. Taking that factor out leaves $- 5 \left(x + 2\right)$ so we know factoring by grouping will work.

$3 {x}^{3} + 6 {x}^{2} - 5 x - 10 = \left(3 {x}^{3} + 6 {x}^{2}\right) + \left(- 5 x - 10\right)$

$= 3 {x}^{2} \left(x + 2\right) - 5 \left(x + 2\right)$.

Now we have two terms with a common factor $C$ where $C = \left(x - 2\right)$. So we have $3 {x}^{2} C - 5 C = \left(3 x - 5\right) C$

That is: we have $\left(3 {x}^{2} - 5\right) \left(x + 2\right)$

We'll stop there if we're only willing to use integer (or rational) coefficients.

Example 2

Factor: $4 {x}^{3} - 10 {x}^{2} + 3 x + 15$

$4 {x}^{3} - 10 {x}^{2} + 3 x + 15 = \left(4 {x}^{3} - 10 {x}^{2}\right) + 6 x + 15$

$= 2 {x}^{2} \left(2 x - 5\right) + 6 x + 15$

Now if we take a common factor out of $6 x + 15$ and get a monomial times $\left(2 x - 5\right)$, then we can finish factoring by grouping. If we get something else, then factoring by grouping won't work.

In this case we get $6 x + 15 = 3 \left(2 x + 5\right)$. Almost!, But close doesn't work in factoring by grouping. So we can't finish this by grouping.

Example 3 You do the test-maker's job.

I want a problem that CAN be factored by grouping.

I start with $12 {x}^{3} - 28 {x}^{2}$ So, if it CAN be factored by grouping, the the rest of is has to look like what?

It has to be a monomial times $\left(3 x - 7\right)$.

So finishing with $6 x - 14$ would work, or $15 x - 35$, or I could get tricky and use $- 9 x + 21$. In fact ANY number times $\left(3 x - 7\right)$ added to what I already have will give me a polynomial that can be factored by grouping.

$12 {x}^{3} - 28 {x}^{2} + k 3 x - k 7$ for any $k$ can be factored as:

$12 {x}^{3} - 28 {x}^{2} + 3 k x - 7 k = 4 {x}^{2} \left(3 x - 7\right) + k \left(3 x - 7\right) = \left(4 {x}^{2} + k\right) \left(3 x - 7\right)$

Final note: $k = - 1$ or $k = - 9$ would make good choices. Because then the fisrt factor is a difference of 2 squares and can be factored.