ORCCAOpen Resources for Community College Algebra

Factoring is the opposite process from multiplying polynomials together. We can gain some insight into how to factor complicated polynomials by taking a closer look at what happens when two generic polynomials are multiplied together:

When you encounter a trinomial like \(4x^2+5x-6\) and you wish to factor it, the leading coefficient, \(4\text{,}\) is the \((pr)\) from Equation (7.4.1). Similarly, the \(-6\) is the \(qs\text{,}\) and the \(5\) is the \((qr+ps)\text{.}\)

Now, if you multiply the leading coefficient and constant term from Equation (7.4.1), you have \((pr)(qs)\text{,}\) which equals \(pqrs\text{.}\) Notice that if we factor this number in just the right way, \((qr)(ps)\text{,}\) then we have two factors that add to the middle coefficient from Equation (7.4.1), \((qr+ps)\text{.}\)

Can we do all this with the example \(4x^2+5x-6\text{?}\) Multiplying \(4\) and \(-6\) makes \(-24\text{.}\) Is there some way to factor \(-24\) into two factors which add to \(5\text{?}\) We make a table of factor pairs for \(-24\) to see:

So that \(5\) in \(4x^2+5x-6\text{,}\) which is equal to the abstract \((qr+ps)\) from Equation (7.4.1), breaks down as \(-3+8\text{.}\) We can take \(-3\) to be the \(qr\) and \(8\) to be the \(ps\text{.}\) Once we intentionally break up the \(5\) this way, factoring by grouping (see Section 7.2) can take over and is guaranteed to give us a factorization.

And this is the factorization of \(4x^2+5x-6\text{.}\) This whole process is known as the “AC method,” since it begins by multiplying \(a\) and \(c\) from the generic \(ax^2+bx+c\text{.}\)

Always start the factoring process by examining if there is a greatest common factor. Here there is not one. Next, note that this is a trinomial with a leading coefficient that is not \(1\text{.}\) So the AC Method may be of help.

1. Multiply \(2\cdot(-3)=-6\text{.}\)

2. Examine factor pairs that multiply to \(-6\text{,}\) looking for a pair that sums to \(-5\text{:}\)

   Factor Pair	Sum of the Pair
   \(1\cdot-6\)	\(-5\) (what we wanted)
   \(2\cdot-3\)	(no need to go this far)

   Factor Pair	Sum of the Pair
   \(-1\cdot6\)	(no need to go this far)
   \(-2\cdot3\)	(no need to go this far)

3. Intentionally break up the \(-5\) as \(1+(-6)\text{:}\)

   \begin{align*} 2x^2\overbrace{{}-5x}-3\amp=2x^2\overbrace{{}+x-6x}-3\\ \amp=\left(2x^2+x\right)+(-6x-3)\\ \amp=x\highlight{(2x+1)}-3\highlight{(2x+1)}\\ \amp=\highlight{(2x+1)}(x-3) \end{align*}

So we believe that \(2x^2-5x-3\) factors as \((2x+1)(x-3)\text{,}\) and we should check by multiplying out the factored form:

Our factorization passes the tests.

There is no greatest common factor. Since this is a trinomial, we try the AC Method.

1. Multiply \(6\cdot(-6)=-36\text{.}\)

2. Examine factor pairs that multiply to \(-36\text{,}\) looking for a pair that sums to \(5\text{:}\)

   Factor Pair	Sum of the Pair
   \(1\cdot-36\)	\(-35\)
   \(2\cdot-18\)	\(-16\)
   \(3\cdot-12\)	\(-9\)
   \(4\cdot-9\)	\(-5\) (close; wrong sign)
   \(6\cdot-6\)	\(0\)

   Factor Pair	Sum of the Pair
   \(-1\cdot36\)	\(35\)
   \(-2\cdot18\)	\(16\)
   \(-3\cdot12\)	\(9\)
   \(-4\cdot9\)	\(5\) (what we wanted)

3. Intentionally break up the \(5\) as \(-4+9\text{:}\)

   \begin{align*} 6p^2\overbrace{{}+5pq}-6q^2\amp=6p^2\overbrace{{}-4pq+9pq}-6q^2\\ \amp=\left(6p^2-4pq\right)+\left(9pq-6q^2\right)\\ \amp=2p\highlight{(3p-2q)}+3q\highlight{(3p-2q)}\\ \amp=\highlight{(3p-2q)}(2p+3q) \end{align*}

So we believe that \(6p^2+5pq-6q^2\) factors as \((3p-2q)(2p+3q)\text{,}\) and we should check by multiplying out the factored form:

Our factorization passes the tests.

Notice that \(3\) is a common factor in this trinomial. We should factor it out first:

Now we are left with two factors, one of which is \(6n^2-7n-20\text{,}\) which might factor further. Using the AC Method:

1. \(6\cdot-20=-120\)

2. Examine factor pairs that multiply to \(-120\text{,}\) looking for a pair that sums to \(-7\text{:}\)

   Factor Pair	Sum of the Pair
   \(1\cdot-120\)	\(-119\)
   \(2\cdot-60\)	\(-58\)
   \(3\cdot-40\)	\(-37\)
   \(4\cdot-30\)	\(-26\)
   \(5\cdot-24\)	\(-19\)
   \(6\cdot-20\)	\(-14\)
   \(8\cdot-15\)	\(-7\) (what we wanted)
   \(10\cdot-12\)	(no need to go this far)

   Factor Pair	Sum of the Pair
   \(-1\cdot120\)	(no need to go this far)
   \(-2\cdot60\)	(no need to go this far)
   \(-3\cdot40\)	(no need to go this far)
   \(-4\cdot30\)	(no need to go this far)
   \(-5\cdot24\)	(no need to go this far)
   \(-6\cdot20\)	(no need to go this far)
   \(-8\cdot15\)	(no need to go this far)
   \(-10\cdot12\)	(no need to go this far)

3. Intentionally break up the \(-7\) as \(8+(-15)\text{:}\)

   \begin{align*} 18n^2-21n-60 \amp=3\left(6n^2\overbrace{{}-7n}-20\right)\\ \amp=3\left(6n^2\overbrace{{}+8n-15n}-20\right)\\ \amp=3\left((6n^2+8n)+(-15n-20)\right)\\ \amp=3\left(2n\highlight{(3n+4)}-5\highlight{(3n+4)}\right)\\ \amp=3\highlight{(3n+4)}(2n-5) \end{align*}

So we believe that \(18n^2-21n-60\) factors as \(3(3n+4)(2n-5)\text{,}\) and you should check by multiplying out the factored form.

Notice that \(2xy\) is a common factor in this trinomial. Also the leading coefficient is negative, and as discussed in Section 7.1, it is wise to factor that out as well. So we find:

Now we are left with one factor being \(8x^2+6x-9\text{,}\) which might factor further. Using the AC Method:

1. \(8\cdot-9=-72\)

2. Examine factor pairs that multiply to \(-72\text{,}\) looking for a pair that sums to \(6\text{:}\)

   Factor Pair	Sum of the Pair
   \(1\cdot-72\)	\(-71\)
   \(2\cdot-36\)	\(-34\)
   \(3\cdot-24\)	\(-21\)
   \(4\cdot-18\)	\(-14\)
   \(6\cdot-12\)	\(-6\) (close; wrong sign)
   \(8\cdot-9\)	\(-1\)

   Factor Pair	Sum of the Pair
   \(-1\cdot72\)	\(71\)
   \(-2\cdot36\)	\(34\)
   \(-3\cdot24\)	\(21\)
   \(-4\cdot18\)	\(14\)
   \(-6\cdot12\)	\(6\) (what we wanted)
   \(-8\cdot9\)	(no need to go this far)

3. Intentionally break up the \(6\) as \(-6+12\text{:}\)

   \begin{align*} -16x^3y-12x^2y+18xy\amp=-2xy\left(8x^2\overbrace{{}+6x}-9\right)\\ \amp=-2xy\left(8x^2\overbrace{{}-6x+12x}-9\right)\\ \amp=-2xy\left((8x^2-6x)+(12x-9)\right)\\ \amp=-2xy\left(2x\highlight{(4x-3)}+3\highlight{(4x-3)}\right)\\ \amp=-2xy\highlight{(4x-3)}(2x+3) \end{align*}

So we believe that \(-16x^3y-12x^2y+18xy\) factors as \(-2xy(4x-3)(2x+3)\text{,}\) and you should check by multiplying out the factored form.