Factoring Polynomials

Factoring Polynomials — The Meaning

Let’s start by taking a quick look at something similar to factoring polynomials but simpler — factoring natural (counting) numbers.

To factor a number means to rewrite the number as a product of smaller counting numbers. For example, here are a few ways in which we can factor 45\hspace{0.2em} 45 \hspace{0.2em}.

  • 45=5×9\hspace{0.2em} 45 \hspace{0.25em} = \hspace{0.25em} 5 \times 9 \hspace{0.2em}
  • 45=15×3\hspace{0.2em} 45 \hspace{0.25em} = \hspace{0.25em} 15 \times 3 \hspace{0.2em}
  • 45=3×3×5\hspace{0.2em} 45 \hspace{0.25em} = \hspace{0.25em} 3 \times 3 \times 5 \hspace{0.2em}

Now there’s something special about the last one. There, we have split (or factor) 45\hspace{0.2em} 45 \hspace{0.2em} into the smallest numbers possible. We can’t factor it any further.

Similarly, to factor a polynomial is to rewrite it as a product of two or more smaller polynomials.

Here are a few examples.

  • 2x23x2=(2x+1)(x2)\hspace{0.2em} 2x^2 - 3x - 2 \hspace{0.2em} = \hspace{0.2em} (2x + 1)(x - 2) \hspace{0.2em}
  • a2b2=(a+b)(ab)\hspace{0.2em} a^2 - b^2 \hspace{0.2em} = \hspace{0.2em} (a + b)(a - b) \hspace{0.2em}
  • 2x32x=2x(x21)\hspace{0.2em} 2x^3 - 2x \hspace{0.2em} = \hspace{0.2em} 2x(x^2 - 1) \hspace{0.2em}

And usually, we want to factor it completely (to the level where no more factoring is possible). Of the three examples above, the first two are factored completely, whereas the third can be factored further. Here it is in it completely factored form.

2x32x=2x(x+1)(x1)2x^3 - 2x \hspace{0.2em} = \hspace{0.2em} 2x(x + 1)(x - 1)

So, how to factor polynomials? We’ll start with the most fundamental concept.

Gcf of Monomials

The greatest common factor (or gcf) of a group of numbers refers to the largest number that divides into each of them evenly. For example, the gcf of 12 and 18 would be 6.

Now the idea of gcf can be extended to algebraic monomials (polynomials with a single term), too.

Consider the following monomials.

8x2y3z,12x2y8x^2y^3z, \hspace{0.25em} 12x^2y

Now, let’s split the monomials into their factors so that we can see which factors are common to both.

8x2y3z=222xxyyyz12x2y=223xxy\begin{align*} 8x^2y^3z \hspace{0.2em} &= \hspace{0.2em} {\color{Red} 2} \cdot {\color{Red} 2} \cdot 2 \cdot {\color{Teal} x} \cdot {\color{Teal} x} \cdot {\color{Orchid} y} \cdot y \cdot y \cdot z \\[1em] 12x^2y \hspace{0.2em} &= \hspace{0.2em} {\color{Red} 2} \cdot {\color{Red} 2} \cdot 3 \cdot {\color{Teal} x} \cdot {\color{Teal} x} \cdot {\color{Orchid} y} \end{align*}

Now if we combine all of the common factors, we get the gcf of the monomials.

gcf (8x2y3z,12x2y)=22xxy=4x2y\begin{align*} \text{gcf }(8x^2y^3z, \hspace{0.25em} 12x^2y) \hspace{0.2em} &= \hspace{0.2em} {\color{Red} 2} \cdot {\color{Red} 2} \cdot {\color{Teal} x} \cdot {\color{Teal} x} \cdot {\color{Orchid} y} \\[1em] &= \hspace{0.2em} {\color{Red} 4} {\color{Teal} x^2} {\color{Orchid} y} \end{align*}

Alright, this is a fine way of finding the gcf of monomials, but there is a quicker way. And because you will need to do this very frequently, speed matters.

How to Find the gcf of Monomials Quickly?


Find the gcf for each group of monomials.

(1)6pq4r3,5q2r(2)5ab2,10ab,15b3\begin{align*} &(1) \hspace{1em} 6pq^4r^3 , \hspace{0.3em} 5q^2r \\[1em] &(2) \hspace{1em} 5ab^2 , \hspace{0.3em} 10ab , \hspace{0.3em} 15b^3 \end{align*}

Solution (1\hspace{0.2em} 1 \hspace{0.2em})

Step 1.  Write down the gcf of the constants (numbers before the variables).

In our case here, the constants are 6\hspace{0.2em} 6 \hspace{0.2em} and 5\hspace{0.2em} 5 \hspace{0.2em}, and their gcf would be 1\hspace{0.2em} 1 \hspace{0.2em}. So,

gcf (6pq4r3,5q2r)=1\text{gcf }(6pq^4r^3 , \hspace{0.3em} 5q^2r) \hspace{0.2em} = \hspace{0.2em} 1 \rule{2em}{0.05em}

Step 2.  Write down the variables that are present in each of the monomials. Leave some space for their exponents.

In the present example, only q\hspace{0.2em} q \hspace{0.2em} and r\hspace{0.2em} r \hspace{0.2em} are present in both the monomials. p\hspace{0.2em} p \hspace{0.2em} is not there in the second monomial and so we leave it out.

gcf (6pq4r3,5q2r)=1qr\text{gcf }(6pq^4r^3 , \hspace{0.3em} 5q^2r) \hspace{0.2em} = \hspace{0.2em} 1 q^{\rule{0.5em}{0.05em}} r^{\rule{0.5em}{0.05em}}

Step 3.  For each variable, take the smallest exponent it has in the given monomials.

Here, the smallest exponent for q\hspace{0.2em} q \hspace{0.2em} is 2\hspace{0.2em} 2 \hspace{0.2em} and that for r\hspace{0.2em} r \hspace{0.2em} is 1\hspace{0.2em} 1 \hspace{0.2em}.

gcf (6pq4r3,5q2r)=1q2r1\text{gcf }(6pq^4r^3 , \hspace{0.3em} 5q^2r) \hspace{0.2em} = \hspace{0.2em} 1 q^2 r^1

Note — It's not necessary to explicitly mention a coefficient of 1\hspace{0.2em} 1 \hspace{0.2em}. Same with an exponent of 1.


gcf (6pq4r3,5q2r)=q2r\text{gcf }(6pq^4r^3 , \hspace{0.3em} 5q^2r) \hspace{0.2em} = \hspace{0.2em} q^2 r

Solution (2\hspace{0.2em} 2 \hspace{0.2em})

Again, the same three steps we used for the last example. But this time, let's do it all in one go.

So, the gcf of 5\hspace{0.2em} 5 \hspace{0.2em}, 10\hspace{0.2em} 10 \hspace{0.2em}, and 15\hspace{0.2em} 15 \hspace{0.2em} is 5\hspace{0.2em} {\color{Red} 5} \hspace{0.2em}. The variable common to the three monomials is b\hspace{0.2em} {\color{Teal} b} \hspace{0.2em}. And the smallest exponent of b in the three monomials is 1\hspace{0.2em} 1 \hspace{0.2em}. That means —

gcf (5ab2,10ab,15b3)=5b\text{gcf }(5ab^2 , \hspace{0.3em} 10ab , \hspace{0.3em} 15b^3) \hspace{0.2em} = \hspace{0.2em} {\color{Red} 5} {\color{Teal} b}

How to Factor Polynomials

Depending on the type of polynomial you want to factorize, you may need to use different techniques. Here, we’ll look at the three most common and important.

Factoring Out the Gcf


Factor the polynomials below.

(1)2x2+5xy+x(2)6a3b2+4a2b2+10a2b3\begin{align*} &(1) \hspace{1em} 2x^2 + 5xy + x \\[1em] &(2) \hspace{1em} 6a^3b^2 + 4a^2b^2 + 10a^2b^3 \end{align*}

Solution (1\hspace{0.2em} 1 \hspace{0.2em})

Step 1.  Find the gcf of each term (the monomials) contained in the polynomial.

Our polynomial consists of 3 monomials — 2x2\hspace{0.2em} 2x^2 \hspace{0.2em}, 5xy\hspace{0.2em} 5xy \hspace{0.2em}, and x\hspace{0.2em} x \hspace{0.2em}. Using what we learned in the last section, we can tell their gcf would be x\hspace{0.2em} {\color{Red} x} \hspace{0.2em}.

gcf (2x2,5xy,x)=x\text{gcf }(2x^2, \hspace{0.25em} 5xy, \hspace{0.25em} x) \hspace{0.25em} = \hspace{0.25em} {\color{Red} x}

Step 2.  Write the gcf followed by a pair of parentheses with space for three terms (our polynomial has 3 terms, remember?).

2x2+5xy+x=x(++)2x^2 + 5xy + x \hspace{0.2em} = \hspace{0.2em} {\color{Red} x} \hspace{0.1em} (\rule{1.5em}{0.05em} + \rule{1.5em}{0.05em} + \rule{1.5em}{0.05em})

Step 3.  Divide each term by the gcf and write the result in the respective position.

Here’s what I mean.

2x2+5xy+x=x(2x2x+5xyx+xx)2x^2 + 5xy + x \hspace{0.2em} = \hspace{0.2em} {\color{Red} x} \left (\frac{2x^2}{ {\color{Red} x} } + \frac{5xy}{ {\color{Red} x} } + \frac{x}{ {\color{Red} x} } \right )

Doing the divisions, we get —

2x2+5xy+x=x(2x+5y+1)2x^2 + 5xy + x \hspace{0.2em} = \hspace{0.2em} {\color{Red} x} \hspace{0.1em} (2x + 5y + 1)

And that’s it. We have factored the given polynomial.

How This Works

In case you are wondering what we did in step 2 above, let me explain. TK

Solution (2\hspace{0.2em} 2 \hspace{0.2em})

Again, we'll be repeating the same steps.

Step 1.  This time the gcf of the three terms would be 2a2b2\hspace{0.2em} {\color{Red} 2a^2b^2} \hspace{0.2em}.

gcf (6a3b2,4a2b2,10a2b3)=2a2b2\text{gcf }(6a^3b^2, \hspace{0.25em} 4a^2b^2, \hspace{0.25em} 10a^2b^3) \hspace{0.25em} = \hspace{0.25em} {\color{Red} 2a^2b^2}

Step 2 and 3.  Factoring out the gcf, we have —

6a3b2+4a2b2+10a2b3=2a2b2(6a3b22a2b2+4a2b22a2b2+10a2b32a2b2)=2a2b2(3a+2+5b)\begin{align*} &6a^3b^2 + 4a^2b^2 + 10a^2b^3 \\[1.5em] = \hspace{0.4em} & {\color{Red} 2a^2b^2} \left ( \frac{6a^3b^2}{ {\color{Red} 2a^2b^2} } + \frac{4a^2b^2}{ {\color{Red} 2a^2b^2} } + \frac{10a^2b^3}{ {\color{Red} 2a^2b^2} } \right ) \\[1.5em] = \hspace{0.4em} & {\color{Red} 2a^2b^2} \left ( 3a + 2 + 5b \right ) \end{align*}

And that's it.

Factoring Polynomials by Grouping

Sometimes, there’s no common factor shared by all the terms in a polynomial but you can find groups of terms that share some common factors.

As an example, consider this polynomial.

2x28x+3x122x^2 - 8x + 3x - 12

Here, if we try finding the gcf of all four terms, we won’t get anything other than 1. There are no other common factors.

But if we take the first two terms and the last two terms as separate groups and find their GCFs, it’s a different story.

2x28x+3x12=2x(x4)+3(x4)\begin{align*} & {\color{Red} 2x^2 - 8x} \hspace{0.2em} {\color{Teal} + \hspace{0.2em} 3x - 12} \\[1em] = \hspace{0.4em} & {\color{Red} 2x(x - 4)} \hspace{0.2em} {\color{Teal} + \hspace{0.2em} 3(x - 4)} \end{align*}

So we were able to factor something out from each group — 2x\hspace{0.2em} 2x \hspace{0.2em} from the first group and 3\hspace{0.2em} 3 \hspace{0.2em} from the second.

But we need to go further. If you look at the expression as a whole, it's still a sum of two terms (those in red and green). It's not a product of two or more factors.

Thankfully for us, in the two groups or terms, we have a something common to factor out, (x4)\hspace{0.2em} (x - 4) \hspace{0.2em}. So, let factor it out.

2x(x4)+3(x4)=(x4)(2x+3)\begin{align*} &2x {\color{Orchid} (x - 4)} + 3 {\color{Orchid} (x - 4)} \\[1em] = \hspace{0.4em} & {\color{Orchid} (x - 4)} (2x + 3) \end{align*}

Now our job is complete. We have turned the given expression into a product of two factors.


Factor the polynomials.

(1)2x2xy2+4xy2y3(2)6x3+9x2+10x+15\begin{align*} &(1) \hspace{1em} 2x^2 - xy^2 + 4xy - 2y^3 \\[1em] &(2) \hspace{1em} 6x^3 + 9x^2 + 10x + 15 \end{align*}

Solution (1\hspace{0.2em} 1 \hspace{0.2em})

Step 1.  Identify the groups of terms that have common factors.

Generally, in problems of this kind, you'll have expressions with four terms and so you make two groups with two terms each.

Let's go with the first two terms in the first group and the last two in the second. So,

2x2xy2+4xy2y3 {\color{Red} 2x^2 - xy^2} {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}4xy - 2y^3}

Step 2.  Factorize each group separately.

Here, we can factor our x\hspace{0.2em} x \hspace{0.2em} from the first group and 2y\hspace{0.2em} 2y \hspace{0.2em} from the second.

2x2xy2+4xy2y3=x(2xy2)+2y(2xy2)\begin{align*} & {\color{Red} 2x^2 - xy^2} {\color{Teal} \hspace{0.2em}+ \hspace{0.2em}4xy - 2y^3} \\[1em] = \hspace{0.4em} & {\color{Red} x \hspace{0.05em} (2x - y^2)} {\color{Teal} \hspace{0.2em}+ \hspace{0.2em} 2y \hspace{0.05em} (2x - y^2)} \end{align*}

Step 3.  Check again. Do the new terms have a common factor(s)? If yes, factor it out.

In the present case, we have (2xy2)\hspace{0.2em} (2x - y^2) \hspace{0.2em} present in both the groups/terms. And so, let's factor it out.

x(2xy2)+2y(2xy2)=(2xy2)(x+2y)\begin{align*} &x {\color{Orchid} (2x - y^2)} + 2y {\color{Orchid} (2x - y^2)} \\[1em] = \hspace{0.4em} & {\color{Orchid} (2x - y^2)} (x + 2y) \end{align*}


Solution (2\hspace{0.2em} 2 \hspace{0.2em})

Identify the groups of terms that have common factors.

Factorize each group separately.

Check again. Do the new terms have a common factor(s)? If yes, factorize again.

Factoring Quadratic Trinomials

One common way of solving quadratic equations is by factoring the quadratic polynomial


Factor the polynomials.

(1)4x24x15(2)x2+7x+10\begin{align*} &(1) \hspace{1em} 4x^2 - 4x - 15 \\[1em] &(2) \hspace{1em} x^2 + 7x + 10 \end{align*}

Factoring Polynomials Using Identities

There are certain patterns in algebraic expressions that we encounter so frequently, it makes sense to familarize ourselve with them and even memorize them.

As an example, consider this identity.

a2b2=(a+b)(ab)a^2 - b^2 \hspace{0.2em} = \hspace{0.2em} (a + b)(a - b)

Now, in the light of the above identity, let's examine and factor the expression below.

4x2254x^2 - 25

As a beginner, it's easy to miss the pattern. So let me rewrite it in a way that makes it easier to see.

4x225=(2x)2(5)24x^2 - 25 \hspace{0.2em} = \hspace{0.2em} {\color{Red} (2x)} ^2 - {\color{Teal} (5)} ^2

Can you see it? It's exactly of the form a2b2\hspace{0.2em} {\color{Red} a} ^2 - {\color{Teal} b} ^2 \hspace{0.2em}.

So, using the identity we started with —

(2x)2(5)2=(2x+5)(2x5) {\color{Red} (2x)} ^2 - {\color{Teal} (5)} ^2 \hspace{0.2em} = \hspace{0.2em} ( {\color{Red} 2x} + {\color{Teal} 5} )( {\color{Red} 2x} - {\color{Teal} 5} )

And that's an example of how we can factor an expression using an algebraic identity.

Here are some of the most popular algebraic identities.

  1. a2+2ab+b2=(a+b)2\hspace{0.2em} \hspace{0.25em} a^2 + 2ab + b^2 \hspace{0.2em} = \hspace{0.2em} (a + b)^2 \hspace{0.2em}
  2. a22ab+b2=(ab)2\hspace{0.2em} \hspace{0.25em} a^2 - 2ab + b^2 \hspace{0.2em} = \hspace{0.2em} (a - b)^2 \hspace{0.2em}
  3. a2b2=(a+b)(ab)\hspace{0.2em} \hspace{0.25em} a^2 - b^2 \hspace{3em} = \hspace{0.2em} (a + b)(a - b) \hspace{0.2em}
  4. a3+b3=(a+b)(a2ab+b2)\hspace{0.2em} \hspace{0.25em} a^3 + b^3 \hspace{3em} = \hspace{0.2em} (a + b)(a^2 - ab + b^2) \hspace{0.2em}
  5. a3b3=(ab)(a2+ab+b2)\hspace{0.2em} \hspace{0.25em} a^3 - b^3 \hspace{3em} = \hspace{0.2em} (a - b)(a^2 + ab + b^2) \hspace{0.2em}

And with that, we come to the end of this tutorial on factoring polynomials. Until next time.