Factor by grouping is an essential method used when factoring trinomials and polynomials. This method applies fundamental concepts such as the greatest common factor (GCF) and the distributive property. Factor by grouping is an important building block in factoring and solving quadratic expressions as well as higher degree polynomials.
Through factor by grouping, you’ll have a systematic approach when dealing with trinomials where the leading coefficient is not equal to 1. In addition, more complex polynomials (with four terms normally) can now easily be factored by grouping the terms into pairs of binomials.
This article breaks down the steps when factoring expressions by grouping. You’ll also have enough examples to work on so you can understand and master this lesson by heart. When you’re ready, head over to the section below and begin by understanding the definition and conditions of this method!
What Is Factor by Grouping?
Factor by grouping is a factoring method where the expression being factored is split into simpler expressions so that they can be grouped as pairs of two terms that share common factors. We use factor by grouping to write the factored form of polynomials with four terms or trinomials that have leading coefficients that are not equal to 1.
\begin{aligned}4x^2 - 4x + 8x-8 &\rightarrow ({\color{DarkOrange}4x^2 - 4x}) + ({\color{DarkGreen}8x -8})\\5mn -n +10m-2&\rightarrow({\color{DarkOrange}5mn - n}) + ({\color{DarkGreen}10m -2})\\6w^3 + 16w^2 -15w -40&\rightarrow({\color{DarkOrange}6w^3 + 16w^2}) + ({\color{DarkGreen}-15w-40})\end{aligned}Take a look at the three examples of polynomials shown above. These three polynomials contain different pairs of terms that share common factors each. Breaking them down into pairs of terms will help identify common factors shared between these terms. To appreciate this method, take a look at how 5mn -n +10m-2 is factored by grouping.
\begin{aligned}5mn -n +10m-2&= ({\color{DarkOrange}5mn - n}) + ({\color{DarkGreen}10m -2})\\&= n(\underline{5m -1}) + 2(\underline{5m -1})\\&= (5m - 1)(n + 2)\end{aligned}By grouping the terms into pairs, it’s much simpler to find common terms shared between two terms and it makes the process of factoring complex expressions less intimidating. Don’t worry, we’ll break down the key steps for you and give you the chance to work on different examples as well!
How to Factor by Grouping?
To factor by grouping, divide the polynomial into pairs of terms. For each pair, look out for the greatest common factor (or GCF) that the terms share. Factor out the GCF from each pair of terms then observe if the resulting expression share common factors from the binomials. This shows that the distributive property of multiplication plays a key part when factoring by grouping.
\begin{aligned}\textbf{Distributive Prop} &\textbf{erty of Multiplication}\\\\\color{DarkOrange}a(b \pm c) = ab \pm ac &\Leftrightarrow \color{DarkGreen}ab \pm ac =a(b \pm c)\end{aligned}The distributive property is often applied twice when factoring by grouping – 1) when we factor out the common factors shared between pair of terms and 2) when we factor the expression completely by looking for other common factors. Learn the specific steps when applying the factor by grouping method on polynomials with four terms or on trinomials.
How To Use Factor by Grouping on Polynomials?
Group the expression into pairs of binomials (expression with two terms) when factoring polynomials by groupings. This opens for an opportunity to look for common factors shared between the paired terms first. We’ve summarized the steps for you as shown below while demonstrating it to factor the polynomial, 6w^3 + 16w^2 -15w -40 .
1. Factor out any common factor shared by the polynomial’s term if any.
2. Group the polynomial’s terms in pairs.
\begin{aligned}6w^3 + 16w^2 -15w -40&= ({\color{DarkOrange}6w^3 + 16w^2}) + ({\color{DarkGreen}-15w-40})\end{aligned}3. Factor out common factors shared between each pair of terms.
\begin{aligned}6w^3 + 16w^2 &= {\color{DarkOrange}2w^2} (3w + 8)\\-15w-40 &= {\color{DarkGreen}-5} (3w + 8)\\&\,\,\,\downarrow\\(6w^3 + 16w^2) + (-15w-40) &= 2w^2(3w+ 8) -5(3w + 8)\end{aligned}4. Observe if the binomials are identical.
\begin{aligned}(6w^3 + 16w^2) + (-15w-40) &= 2w^2(\underline{3w+ 8}) -5(\underline{3w +8})\end{aligned}5. When you see a common binomial shared between the polynomial, reverse the distributive property to factor the polynomial completely.
\begin{aligned}2w^2(\underline{3w+ 8}) -5(\underline{3w +8}) &= {\color{DarkOrange}2w^2}(\underline{3w+ 8}) -{\color{DarkGreen}5}(\underline{3w +8})\\&= (3w + 8)({\color{DarkOrange}2w^2} - {\color{DarkGreen}5})\\&= (3w + 8)(2w^2 -5) \end{aligned}Use these steps when factoring polynomials with four terms. Try out these sample problems below to help you master this method.
Problem 1
Write down the factored forms of the following polynomials.
a. 2v^3 +5v^2 +6v +15
When given a polynomial with four terms, factor out any common factor shared by all four terms. For 2v^3 +5v^2 +6v +15, the terms don’t share any common terms. Group the terms into two pairs of terms.
\begin{aligned}2v^3 +5v^2 +6v +15 &= ({\color{DarkOrange}2v^3 +5v^2}) + ({\color{DarkGreen}6v +15})\end{aligned}Observe each resulting binomial for common factors shared between them then factor them out.
\begin{aligned}(2v^3 +5v^2) + (6v +15) &= {\color{DarkOrange}v^2}(2v + 5) +{\color{DarkGreen}3}(2v + 5)\end{aligned}
Now, notice that 2v +5 is a common binomial factor for each group of terms, so factor it out for each term using the distributive property once more.
\begin{aligned}v^2(2v + 5) +3(2v + 5) &= v^2(\underline{2v + 5}) +3(\underline{2v + 5})\\&= (2v +5)(v^2 + 3)\end{aligned}This means that 2v^3 +5v^2 +6v +15 is equivalent to (2v +5)(v^2 +3) in factored form.
b. 7ab -3m – a + 21mb
Now, there are instances when rearranging the terms will lead to a better result for this method. When factoring 7ab -3m – a + 21mb by grouping, it’s best to rearrange the terms so that terms with a and m are each grouped together.
\begin{aligned}7ab -3m – a + 21mb&= 7ab – a-3m +21mb\\&= ({\color{DarkOrange}7ab – a}) + ({\color{DarkGreen}-3m +21mb})\end{aligned}Now factor out a and 3m from the binomials respectively. See if the resulting binomials inside each of the parentheses are identical.
\begin{aligned}(7ab – a) + (-3m +21mb) &= {\color{DarkOrange}a}(7b - 1) +{\color{DarkGreen}3m}(-1 + 7b)\\&= a(\underline{7b - 1}) + 3m(\underline{7b - 1})\\&= (7b -1)(a + 3m)\end{aligned}Hence, the factored form of 7ab -3m – a + 21mb is (7b -1)(a + 3m).
We’ve now established the process of factoring polynomials, so it’s time to learn how to use this method when factoring trinomials by grouping.
How To Use Factor by Grouping on Trinomials?
When factoring trinomials by grouping, write the trinomial in standard form, ax^2 +bx +c, where a is not equal to one or zero. Multiply the first and last coefficients – a and c. This is why this method is also called the ac-method.
Use the guide below as you learn how to factor the trinomial, ax^2 + bx + c, by grouping. We’ve included an example to help you understand each step by heart.
1. Multiply the leading and last coefficient of the trinomial. Always remember: we’re using the ac-method.
| General Form | Example |
| \begin{aligned}{\color{DarkOrange}a}x^2 +&\,\,bx \,\,{\color{DarkGreen} +x}\\&\Downarrow\\{\color{DarkOrange}a}({\color{DarkGreen}c}) &= ac\end{aligned} | \begin{aligned}{\color{DarkOrange}6}x^2 +&7x \,\,{\color{DarkGreen} +2}\\&\Downarrow\\{\color{DarkOrange}6}({\color{DarkGreen}2}) &= 12\end{aligned} |
2. Now, think of two numbers that have a product of ac and a sum of the middle coefficient, b. For our case, let’s establish that a + c = b. Rewrite the middle term as sum of these factors.
\begin{aligned}ax^2 + {\color{Purple}b}x + c &=ax^2 +{\color{Purple}(a +c)}x+c\\&=ax^2 +ax + cx + c\\&\Downarrow\\6x^2 + {\color{Purple}7}x + 2 &=6x^2 +{\color{Purple}(3 + 4)}x+2\\&=6x^2 +3x + 4x + 2\end{aligned}3. The trinomial is now a polynomial with four terms, so apply the same process to factor the expression. Begin by grouping the terms as pairs.
\begin{aligned}ax^2 +ax + cx + c&= ({\color{DarkOrange}ax^2+ax})+({\color{DarkGreen}cx+ c})\\&\Downarrow\\6x^2 +3x + 4x + 2 &= ({\color{DarkOrange}6x^2 +3x})+({\color{DarkGreen}4x+2})\end{aligned}4. Factor out common factors for each binomial then factor out any common binomial from the resulting expression.
\begin{aligned}(ax^2+ax)+(cx+ c)&= ax(\underline{x + 1}) + c(\underline{x + 1})\\&=(ax +c)(x +1)\\&\Downarrow\\(6x^2 + 3x) + (4x +2)&= 3x(\underline{2x + 1}) + 2(\underline{2x + 1})\\&=(3x + 2)(2x + 1)\end{aligned}You can double check by applying the FOIL method and expand the factored form. The result should return the original trinomial.
\begin{aligned}(3x + 2)(2x + 1) &= \overbrace{(3x)(2x)}^{F}+ \overbrace{(3x)(1)}^{O} +\overbrace{(2)(2x)}^{I}+\overbrace{(2)(1)}^{L}\\&= 6x^2 + 3x + 4x + 2\\&= 6x^2 + 7x+2\end{aligned}Now that we’ve laid out the steps for factoring trinomials by grouping, it’s time to apply what you’ve learned to factor different trinomials.
Problem 2
Factor the following trinomials completely.
a. 3x^2 -14x-5
Since the leading coefficient of the trinomial is 3, we can use factor by grouping to find the factored of 3x^2 -14x-5. The trinomial is already in its standard form, so it’s time to multiply the leading coefficient and the constant.
\begin{aligned}{\color{DarkOrange}3}x^2 -&14x \,\,{\color{DarkGreen} -5}\\&\Downarrow\\{\color{DarkOrange}3}({\color{DarkGreen}-5}) &= -15\end{aligned}Now think of two numbers that have a product of -15 and a sum of -14. For this to be possible, the two numbers will have opposite sides, so we have -15 and 1. Split the middle term using these factors.
\begin{aligned}3x^2 {\color{Purple}-14x} -5 &= 3x^2 + ({\color{Purple}-15 + 1})x - 5\\&=3x^2 </p> <p> -15x + x -5\end{aligned}Group the polynomials’ terms into pairs of binomials then factor the resulting expression completely as shown below. Keep in mind that when the binomial has no common factor, consider 1 as its common factor and see if the resulting expression has common binomials.
\begin{aligned}3x^2 -15x + x -5 &= ({\color{DarkOrange}3x^2 -15x}) + ({\color{DarkGreen}x - 5})\\&= 3x(\underline{x - 5}) + 1(\underline{x - 5})\\&= (3x + 1)(x - 5)\end{aligned}This means that the factored form of 3x^2 -14x-5 is (3x + 1)(x -5).
b. 12x^2+30x -42
By inspection, we can see that the coefficients and constant are multiples of 6, so factor it out first.
\begin{aligned}12x^2+30x -42 &= 6(2x^2 -5x -7)\end{aligned}Factor 2x^2-5x -7 now by applying the factor by grouping method. Begin by multiplying the leading coefficient by the constant.
\begin{aligned}{\color{DarkOrange}2}x^2 -&5x \,\,{\color{DarkGreen} -7}\\&\Downarrow\\{\color{DarkOrange}2}({\color{DarkGreen}-7}) &= -14\end{aligned}For two numbers to have a product of -14 and a sum of -5, they must be equal to -7 and 2. Use this these two factors as the coefficients when breaking the middle term into two smaller terms. Group the terms into binomials then factor the expression completely.
\begin{aligned}12x^2+30x -42&= 2(2x^2 {\color{Purple}-5x} -7)\\&= 2[2x^2 +({\color{Purple}-7 + 2})x - 5]\\&= 2(2x^2 -7x + 2x-7)\\&= 2[({\color{DarkOrange}2x^2 -7x}) +({\color{DarkGreen}2x -7})]\\&= 2[x(\underline{2x -7}) + 1(\underline{2x - 7})]\\&=2[(x + 1)(2x - 7)]\end{aligned}This means that when factored completely, 12x^2+30x -42 is equivalent to 2[(x + 1)(2x – 7)].