1.3 Factoring by grouping

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This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken great care in developing the student's understanding of the factorization process. The technique is consistently illustrated by displaying an empty set of parentheses and describing the thought process used to discover the terms that are to be placed inside the parentheses.The factoring scheme for special products is presented with both verbal and symbolic descriptions, since not all students can interpret symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with leading coefficients different from 1. Objectives of this module: know how to factor a polynomial using the grouping method and when to try the grouping method.

Overview

Using Grouping to Factor a Polynomial
Knowing when to Try the Grouping Method

Using grouping to factor a polynomial

Sometimes a polynomial will not have a particular factor common to every term. However, we may still be able to produce a factored form for the polynomial.

The polynomial $x 3 3 x 2 6 x 18$ has no single factor that is common to every term. However, we notice that if we group together the first two terms and the second two terms, we see that each resulting binomial has a particular factor common to both terms.

The polynomial 'x cubed plus three x squared minus six x minus eighteen'. The first two terms of the polynomial have x square in common, and the last two terms of the polynomial have negative six in common.

Factor $x^{2}$ out of the first two terms, and factor $- 6$ out of the second two terms.

$x^{2} (x + 3) - 6 (x + 3)$

Now look closely at this binomial. Each of the two terms contains the factor $(x + 3)$ .

Factor out $(x + 3)$ .
$(x + 3) (x^{2} - 6)$ is the final factorization.

$x^{3} + 3 x^{2} - 6 x - 18 = (x + 3) (x^{2} - 6)$

Knowing when to try the grouping method

We are alerted to the idea of grouping when the polynomial we are considering has either of these qualities:

no factor common to all terms
an even number of terms

When factoring by grouping, the sign $(+ or -)$ of the factor we are taking out will usually (but not always) be the same as the sign of the first term in that group.

Sample set a

Factor $8 a^{2} b^{4} - 4 b^{4} + 14 a^{2} - 7$ .

We notice there is no factor common to all terms.
We see there are four terms, an even number.
We see that terms 1 and 2 have $+ 4 b^{4}$ in common (since the 1st term in the group is $+ 8 a^{2} b^{4})$ .
We notice that the 3rd and 4th terms have $+ 7$ in common (since the 1st term in the group is $+ 14 a^{2}$ ).
$\begin{array}{l} 8 a^{2} b^{4} - 4 b^{4} + 14 a^{2} - 7 = & {(2a}^{2} {-1)(4b}^{4} +7) \end{array}$

Practice set a

Use the grouping method to factor the following polynomials.

$a x a y b x b y$

$(a + b) (x + y)$

$2 a m + 8 m + 5 a n + 20 n$

$(2 m + 5 n) (a + 4)$

$a^{2} x^{3} + 4 a^{2} y^{3} + 3 b x^{3} + 12 b y^{3}$

$(a^{2} + 3 b) (x^{3} + 4 y^{3})$

$15 m x + 10 n x - 6 m y - 4 n y$

$(5 x - 2 y) (3 m + 2 n)$

$40 a b x - 24 a b x y - 35 c^{2} x + 21 c^{2} x y$

$x (8 a b - 7 c^{2}) (5 - 3 y)$

When factoring the polynomial $8 a^{2} b^{4} 4 b^{4} 14 a^{2} 7$ in Sample Set A, we grouped together terms1 and 2 and 3 and 4. Could we have grouped together terms1 and 3 and 2 and 4? Try this.
$8 a^{2} b^{4} - 4 b^{4} + 14 a^{2} - 7 =$

yes

Do we get the same result? If the results do not look precisely the same, recall the commutative property of multiplication.

Exercises

For the following problems, use the grouping method to factor the polynomials. Some polynomials may not be factorable using the grouping method.

$2 a b + 3 a + 18 b + 27$

$(2 b + 3) (a + 9)$

$x y - 7 x + 4 y - 28$

$x y + x + 3 y + 3$

$(y + 1) (x + 3)$

$m p + 3 m q + n p + 3 n q$

$a r + 4 a s + 5 b r + 20 b s$

$(a + 5 b) (r + 4 s)$

$14 a x - 6 b x + 21 a y - 9 b y$

$12 m x - 6 b x + 21 a y - 9 b y$

$3 (4 m x - 2 b x + 7 a y - 3 b y)$ Not factorable by grouping

$36 a k - 8 a h - 27 b k + 6 b h$

$a^{2} b^{2} + 2 a^{2} + 3 b^{2} + 6$

$(a^{2} + 3) (b^{2} + 2)$

$3 n^{2} + 6 n + 9 m^{3} + 12 m$

$8 y^{4} - 5 y^{3} + 12 z^{2} - 10 z$

Not factorable by grouping

$x^{2} + 4 x - 3 y^{2} + y$

$x^{2} - 3 x + x y - 3 y$

$(x + y) (x - 3)$

$2 n^{2} + 12 n - 5 m n - 30 m$

$4 p q - 7 p + 3 q^{2} - 21$

Not factorable by grouping

$8 x^{2} + 16 x y - 5 x - 10 y$

$12 s^{2} - 27 s - 8 s t + 18 t$

$(4 s - 9) (3 s - 2 t)$

$15 x^{2} - 12 x - 10 x y + 8 y$

$a^{4} b^{4} + 3 a^{5} b^{5} + 2 a^{2} b^{2} + 6 a^{3} b^{3}$

$a^{2} b^{2} (a^{2} b^{2} + 2) (1 + 3 a b)$

$4 a^{3} b c - 14 a^{2} b c^{3} + 10 a b c^{2} - 35 b c^{4}$

$5 x^{2} y^{3} z + 3 x^{3} y w - 10 y^{3} z^{2} - 6 w x y z$

$y (5 y^{2} z + 3 x w) (x^{2} - 2 z)$

$a^{3} b^{2} c d + a b c^{2} d x - a^{2} b x y - c x^{2} y$

$5 m^{10} n^{17} p^{3} - m^{6} n^{7} p^{4} - 40 m^{4} n^{10} q t^{2} + 8 p q t^{2}$

$(m^{6} n^{7} p^{3} - 8 q t^{2}) (5 m^{4} n^{10} - p)$

Exercises for review

( [link] ) Simplify $(x^{5} y^{3}) (x^{2} y)$ .

( [link] ) Use scientific notation to find the product of $(3 \times 10^{- 5}) (2 \times 10^{2})$ .

$6 \times 10^{- 3}$

( [link] ) Find the domain of the equation $y = \frac{6}{x + 5}$ .

( [link] ) Construct the graph of the inequality $y \geq - 2$ .

( [link] ) Factor $8 a^{4} b^{4} + 12 a^{3} b^{5} - 8 a^{2} b^{3}$ .

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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