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Subtext: Factoring Polynomials
Wherein we survey the techniques of factorizing polynomials, covered in the order the
esteemed text presents them.
Let's begin by briefly reviewing what we mean by factoring. The number 6 can be
written as the product of 2 and 3, i.e.
6 = 2 * 3.
We say that 2 and 3 are factors of 6, and we have succeeded in factoring 6, which we
now see is the same as 'multiplication backwards' - we have figured out which numbers 6
is the product of. In this chapter we attempt something a little more ambitious, namely,
factoring polynomials, or breaking them into products of factors. For example, when we
learned the FOIL method, we quickly verified the following statements:
( x  3)( x  2)  x 2  5x  6.
Now we would like to start with the polynomial on the left, and get the product on the
right:
x 2  5x  6  ( x  3)( x  2) .
How? Well, we have to make different decisions on how to factor a given polynomial,
based on certain characteristics of it (for example, the number of terms). Below, we go
over the basic situations.
GCF - Greatest Common Factor
The GCF of two numbers is the greatest number which goes into both of them. For
example, the GCF of 6 and 8 would be 2. It is possible to find the GCF of two numbers
using 'factor trees' and prime factorization of the numbers.
Example 1: Find the greatest common factor of 24 and 36.
24
36
 2
12
(2) 18
(2)
6
(2)
9
(2)
(3)
(3)
(3)
24 = 23*3
36 = 22*32,
the gcf of the two numbers is the product of the lowest powers of the shared factors:
gcf = 22*3 = 12.
When we add variables to the mix, we modify the rule above only slightly
gcf = gcf(numerical coefficients)*(lowest power of shared variable(s)).
Example 2: So the gcf of two terms 24x2 and 36x3 would be
gcf = 12x2. (the gcf of 24 and 36, the lowest power of x, x2).
Exercises:
(1) Find the gcf of the expressions 12x3 and 18x4.
Now, we get down to the business of factoring out the gcf. A rough rule-of-thumb is that
we pull out the gcf by dividing every term of the polynomial by it.
 term1 term2
last  term 
gcf 

 ...
.
gcf
gcf
 gcf

Of course, we need to know how to use exponent rules and do monomial division in
order to do this.
Example 3: Factor the gcf from the polynomial
24x2 + 36x3.
Previously, we found the gcf to be 12x2, so we factor it out:
24x2 + 36x3 = 12x2(2 + 3x)
24x2 36x3
12x2 12x2
Exercises:
(2) Factor out the gcf:
(a) 5 x3  10 x 2  15 x
(b) 4 x 2 y 2  8xy 2  12 xy
Factoring by Grouping
This is another technique that we looked at in section 5.1. Generally, it is used in
situations where you are presented with a polynomial with four terms.
Example 4: Factor the polynomial
xy  4 y  3x  12
Note the four terms (and two variables to boot). The gcf = 1, which is no help. The basic
strategy of factoring by grouping is:
(i) Pair off the first and second, then third and fourth terms, and factor the gcf from these
two pairs.
(ii) Try to combine the result as like terms.
xy  4 y  3x  12 =
gcf = y gcf = 3
y(x + 4) + 3(x + 4) =
(y + 3)(x + 4).
((i) pair the terms and factor the gcf's)
(these are like terms - they share (x + 4). Combine them
by adding the coefficients on top of (x + 4).
One 'trick' you may want to keep in mind here is that to form the final combination of the
like terms, you may have to 'make the signs match' - possibly by picking the second gcf
to be a negative number instead of a positive one.
Example 5: Factor the polynomial
a  3ax  2  6x
(the first two terms are positive, the second two are negative so make the gcf on second two negative, and the signs will
gcf = a gcf = -2
match).
a(1 + 3x) -2(1 + 3x) = (now combine the like terms by adding the coefficients on top of
(a - 2)(1 + 3x).
Exercises:
(3) Factor by grouping:
(a) rx + 2rs + 2x + 4s.
(b) 3xy + 6y -2x - 4.
Difference of Squares (DOS)
The relevant formula is
a 2  b2  (a  b)(a  b).
It can be verfied by using the FOIL method.
We use DOS when we have two terms, both of which are squares or can be re-written as
squares.
Example 6: Factor the polynomial
x2 - 49 =
(x2 is the square of x, 49 is the square of 7)
2
2
(x) - (7) =
(x - 7)(x + 7).
You may have to factor out a GCF from the two terms before you recognize the DOS:
(note - a big principle of factoring is that we always factor out a GCF first if it can be
done).
Example 7: Factor the polynomial
12x2 - 27 =
(the gcf = 3 , so factor it out first:)
2
3(4x - 9) = (there is a DOS inside the pharentheses)
3((2x)2 - 32) =
3(2x - 3)(2x + 3).
You may see other problems where we have variables with even powers other than 2, e.g.
stuff like x4 or x8. It is possible to rewrite these as squares using the exponent rule
(am)n = amn. You may also be able to do a second (or even third) round of factorization,
which goes to another big principle of factoring: always factor as completely as you can.
Example 8: Factor the polynomial
x4 - 16 =
(rewrite x4 as a square: )
(x2)2 - 42 =
(DOS)
(x2 - 4) (x2 + 4) =
(the first polynomial is a difference of squares - factor it, note
(x - 2)(x + 2)(x2 + 4). that the second is a sum of squares - these do not factor!)
Exercises:
(4) Factor:
(a) x2 - 9,
(b) 7x2 - 700,
(c) 2x4 - 2.
Trinomials w/ Lead Coefficient 1.
Trinomial means that the polynomial has 3 terms. The lead coefficient is the coefficient
of the highest-degreed term.
Here, we're going to try and split up polynomials of the form
x 2  bx  c.
The key insight is that if the polynomial factors like this:
x 2  bx  c  ( x  a)( x  b) , then a and b have to multiply to c, and add up to b. We
should look for pairs of factors of c that add up to b.
Example 9: Factor the polynomial
x2  5x  6
1 6
2 +3=5
2
x  5x  6  ( x  2)( x  3).
(check the factors of 6 - find two that add up to 5 they're 2 and 3)
What complicates things here is the presence of the minus signs - in these cases, we have
to be more careful about how the factors of the constant term "add up".
Example 10: Factor the polynomial
x 2  x  12
(here, we need to find the factors of -12 which sum to -1. Start by
listing the factors of 12. In order for the constant term to be
1 -12
negative, and the middle term negative, the bigger of the pair of
2 -6
terms is going to have to be negative. 3 and -4 add up to -1.)
3 -4
2
x  x  12  ( x  4)( x  3).
Example 11: Factor the polynomial
x2  2 x  8
(list the factors of 8 - note that they are going to have to add up to
a positive number (2), so the bigger of the pair will have to be
-1 8
positive this time - 2 and 4 sum to 2, so they're our numbers).
-2 4
x 2  2 x  8  ( x  2)( x  4).
Example 12: Factor the polynomial
x 2  8 x  15
(list the factors of 15 - note that we need two that multiply to 15 and
add up to -8 - so they're going to have to both be negative: -3 and -5
-1 -15 are the choices).
-3 -5
2
x  8x  15  ( x  5)( x  3).
Be aware as well that you may have to factor out a GCF before factoring the trinomial,
and may have to deal with an extra variable.
Example 13: Factor the polynomial
(factor out gcf = 5)
5 x 2  15 xy  140 y 2
2
2
= 5( x  3xy  28 y )
(list the factors of 28 - the larger one will have to be negative)
1 -28
(4 and - 7 add up to -3 - add a y onto them in the factors)
2 -14
4 -7
2
5x  15xy  140 y 2  5( x  4 y)( x  7 y).
Exercises:
(5) Factor the polynomial:
(a) x 2  8 x  7.
(b) x 2  2 x  15 .
(c) x 2  x  20.
(d) x 2  8 x  12.
(e) 7 x 2  14 xy  21y .
Trinomials w/ Lead Coefficient Other than 1
The factoring of polynomials of the form ax2 + bx + c is more involved when a ≠ 1.
There are a couple of methods for factoring these. The first is called the grouping
method. The grouping method is more ‘fail-safe’ than the other method (trial-and-error),
but it may take a bit longer in some situations. Let’s go by example:
Example 14: Factor the polynomial
3x2 – 17x + 10
(first, multiply the constant and lead terms)
3*10 = 30
(find a pair of factors of 30 which sum to -17)
-1 -30
-2 -15
-3 -10
-5 -6
3x2 – 15x – 2x + 10
gcf = 3x
(-2*-15 = 30, -2 + -15 = -17)
(use -15x and -2x to break up -17x)
gcf = -2
(factor by grouping gcf = -2 makes signs match)
3x(x – 5) – 2(x – 5) =
(3x – 2)(x – 5).
(combine like terms – add coefficients on (x – 5))
(answer).
The other method that we saw in this section was the so-called trial-and-error method. In
this approach, we simply begin by writing out factors of the lead coefficients with the
variable in the two sets of pharentheses:
(ax
)(dx
),
Then we try to ‘guess’ the numbers which fill them out by looking at the factors of the
constant term c. The ‘guessing’ is actually done by (hopefully) rapidly calculating the
outer (O) and inner (I) products from the FOIL method, trying to get them to combine
properly into the middle term, bx. The trial-and-error method is easiest when there’s not a
lot of numbers that go into the lead coefficient a and the constant term c.
Example 15: Factor the polynomial
2x2 + 5x – 3
(Then only two factors of a = 2 are 2 and 1 – write:)
(2x )(x )
(The only factors of 3 are 1 and 3: though one of them
1
3
has to be negative. We have to ‘guess’ what goes where
3
1
by Calculating O’s and I’s – I’ll list the possibilities, but this
Would be difficult on, say, a test to do – you want to try
(2x + 1)(x – 3)
and crank out the O’s and I’s in your head.)
O = -6x I = x no
(2x -1)(x+ 3)
O = 6x I = -x yes (add up to 5x – this one works – the others were: )
( 2x + 3)(x – 1)
O = -2x I = 3x no
( 2x - 3)(x + 1)
O = 2x I = -3x no
So, (2x -1)(x+ 3) is the factorization we seek.
As you can see from this fairly simple example, there can possibly be a lot of trial, which
of course increases the chance of error. However, with practice it is possible to get decent
mastery of the trial-and-error method.
Exercises:
(6) Factor the polynomial:
(a) 6x2 – x – 15
(b) 5x2 – 7x – 6
(c) 8x2 + 22x – 21
Difference of Cubes (DOC) & Sum of Cubes (SOC)
The relevant formulas are:
a3 - b3 = (a - b)(a2 + ab + b2)
a3+ b3 = (a+ b)(a2 - ab + b2)
(DOC),
(SOC).
Example 16: Factor the polynomial:
x3  27
(x)3 - (3)3
(we recognize that x3 is the cube of x, 27 is the cube of 3 so this is a DOC. x plays the role of a in the
formula, 3 plays the role of b).
= ( x  3)( x 2  3x  9).
a - b a2 + ab + b2
Example 17: Factor the polynomial:
8 x 3  125 (it's harder to spot, but 8x3 is the square of 2x, and 125 is the square of 5 (2 x)3  (5)3 it's a SOC - 2x plays the role of a, and 5 plays the role of b from the
formula)
2
(2 x  5)(4 x  10 x  25) .
(2x)2 2x*5 52
As with any polynomial, we need to be alert to the possiblity of factoring out a GCF, and
we need to remember to factor completely. Like here:
Example 18: Factor the polynomial
2 x3  128
2( x3  64)
(x)3 -(4)3
2( x  4)( x 2  4 x  16) .
Exercises:
(7) Factor the polynomial:
(gcf = 2:)
(now factor completely - inside the pharentheses is a DOC
so split it up, babe: )
(the complete factorization)
(a) x3  1000
(b) 27 x3  64
(c) 5 x 3  5
General Strategy for Factoring
Now that we have had a look at the various factorization techniques, and had a little
practice using them, it may be in order to summarize the situations in which each should
be used. A general strategy might go as follows:
(1) Always factor out the GCF first, if the GCF is not equal to 1.
(2) Looks at the number of terms of the polynomial. If there are:
(a) two terms - try one of DOS, DOC, SOC.
(b) three terms - we have a trinomial. Look at the lead coefficient a:
(i) if a = 1, then we have to find two numbers which multiply to the constant,
add up to the coefficient on x.
(ii) if a  1, then we need to use the grouping method or trial-and-error.
(c) Four terms - try factoring by grouping.
(3) If none of these works, the polynomial can't be factored and we say it is prime.
Exercises:
(8) Factor the polynomial:
(a) x 2  10 x  25.
(b) 4 x 2  36 x.
(c) 4 x3  36 x.
(d) 3 x 2  11x  4.
(e) x3  216.
(f) 4 x 2  8 x  7 x  14.
(g) 6 x 2  24
(h) 1000 x3  27
(i) x 2  11x  30.
(j) 2 x 2  7 x  15.
Solving Polynomial Equations by Factoring
This is a related question - how do we solve equations formed by setting a polynomial
(usually with degree > 1) = 0. For example:
x 2  5 x  6  0?
In short, we:
(1) Get the left-hand-side (LHS) = 0, if necessary (it's not in the above example),
(2) Factor the LHS,
(3) Set each of the factors = 0, and solve these equations.
(4) Write the aswers in a solution set, enclosed in braces { }.
Example 19: Solve the equation:
x2  5x  6  0
(LHS = 0, so factor LHS)
-1 -6
(trinomial, lead coeff = 1, need two numbers which * to 6
-2 -3
+ to -5 - they're -2 and -3)
( x  3)( x  2)  0
(Set factors = 0)
x - 3 = 0, x - 2 = 0, (Solve.)
+3 +3 +2 +2
x = 3,
x=2
(State answer in solution set: )
x = {2, 3}
Complications may arise - for example, we may need to factor out a GCF first, as in this
example:
Example 20: Solve the equation:
x3  16 x 2  0
(LHS = 0, so factor LHS - the gcf = x)
2
(two terms inside pharentheses, DOS (x)2 - 42 )
x( x  16)  0,
x( x  4)( x  4)  0,
(set all factors = 0)
x  0, x  4  0, x  4  0,
-4 -4 +4 +4,
x = 0, x = -4 , x = 4.
(Form solution set: )
x = {0,4}
(note the super-high-tech )
Another obstacle we may face is being presented with an equation that is not in standard
form:
Example 21: Solve the equation:
x 2  3x  10
(get LHS = 0, by subtracting 10 from both sides)
- 10 -10
x2 - 3x - 10 = 0,
(factor LHS - trinomial, lead coefficient = 1)
( x  5)( x  2)  0,
(set factors = 0, )
x  5  0, x  2  0,
(solve:)
+5 +5 -2 -2
x = 5,
x= - 2
(State answer in solution set: )
x = {-2,5}.
Exercises:
(9) Solve the polynomial equation:
(a) x 2  25  0.
(b) x 2  x  6.
(c) 2 x3  10 x 2  28 x.
Solutions:
(1) the gcf of the coefficients 12 and 13 is 6. The lowest power of x is x3. So the
GCF = 6x3.
(2)
(a) 5 x3  10 x 2  15 x =
5x(x2 + 2x - 3)
5x3 10x2
5x2 5x2
(the gcf = 5x).
-15x
5x2
(b) 4 x 2 y 2  8xy 2  12 xy
4xy(xy - 2y + 3).
(gcf = 4xy)
4x2y2 -8xy2 12xy
4xy 4xy 4xy
(3) (a) rx + 2rs + 2x + 4s.
gcf = r
gcf = 2
r ( x  2s)  2( x  2s)
(r  2)( x  2 s )
(b) 3xy + 6y -2x - 4
gcf = 3y gcf = -2
3 y ( x  2)  2( x  2)
(3 y  2)( x  2) .
(4) (a) x2 - 9,
(x)2 - 32
= (x + 3)(x - 3).
(pair the factors and find the gcf)
(factor out the gcf from the pairs)
(combine like terms on top of (x + 2s) by
adding the coefficients r and 2)
(pair off the terms and find the gcf:)
(Note that I needed to make the signs match by using
gcf = -2 in the second pair. Now pull out the gcf's:)
(Now combine like terms on (x + 2) )
(DOS )
(split it up with the formula)
(b) 7x2 - 700 =
7(x2 - 100) =
(x)2 - 102
7(x - 10)(x + 10).
(gcf = 7, factor it out:)
(DOS in pharentheses).
(c) 2x4 - 2 =
2(x4 - 1) =
(x2)2 - 12,
2( x 2  1)( x 2  1) =
2( x  1)( x  1)( x 2  1)
(gcf = 2, factor it out:)
(rewrite x4 - DOS in pharentheses:)
(split it up using formula:)
(x2 - 1 is a DOS - split it up again)
(the complete factorization).
(5) x 2  8 x  7.
1 7
= (x + 1)(x + 7).
(need two numbers that multiply to 7 add to 8 - there ain't
many choices since 7 is prime - only 1 and 7 are factors)
(1 + 7 = 8, so:)
(b) x 2  2 x  15 (need two numbers which * to -15, add to -2
list the factors of -15. The larger ones will have to be
1 -15 negative.)
3 -5 (3 + -5 = -2, so: )
= (x + 3)(x - 5).
(c) x 2  x  20. (need two numbers which * to -20, add to +1 - the larger factor
has to be positive this time, )
-1 20
-2 10
-4 5 (-4 + 5 = 1)
= (x - 4)(x + 5).
(d) x 2  8 x  12. (need two numbers which * to 12, add to -8 - both factors will have
to be negative to add to a negative, * to a positive)
-1 -12
-2 -6 (-2 + -6 = -8)
-3 -4
= (x - 2)(x - 6).
(e) 7 x 2  14 xy  21y =
7(x2 + 2xy - 3y) =
-1 3
7(x - y)(x + 3y).
(6) (a) 6x2 – x – 15
(6)(-15) = -90
1 -90
2 -45
3 -30
5 -18
6 -15
9 -10
= 6x2 + 9x - 10x - 15
gcf = 3x gcf = -5
= 3x(2x + 3) - 5(2x + 3)
(first, factor out the gcf = 7)
(find the two numbers which * to -3, + to 2)
(add a y to the -1 and 3:)
(multiply the lead and constant terms)
(9 + -10 = -1 - use these to split up -x:)
(pair the terms and get the gcf's)
(picked - 5 to make signs match)
(factor out gcf's, combine terms: )
= (3x - 5)(2x + 3).
(voila. finito. )
(b) 5x2 – 7x – 6
5(-6) = -30,
(Multiply the lead and constant terms:)
1 -30
2 -15
3 -10
2
5x - 10x + 3x - 6
(3 + -10 = -7 - use them to split up -7x)
(pair off, gcf's:)
gcf = 5x gcf = 3
5x(x - 2) + 3(x - 2) =
(5x + 3)(x - 2).
(factor gcf's, combine terms: )
(c) 8x2 + 22x – 21
(8)(-21) = -168
(multiply constant and lead term:)
-1 168
-2 84
-3 56
-4 42
-6 28
(-6 + 28 = 22 : use them to break up 22x)
8x2 + 28x - 6x - 21 =
(pair off, find gcf's, make signs match)
gcf = 4x gcf = -3
4x(2x + 7) - 3(2x + 7) =
(4x - 3)(2x + 7).
(factor out gcf's)
(combine like terms:)
(rock-action)
(7)
(a) x3  1000
(This is a DOC)
3
3
(x) - 10
(use formula with a = x, b = 10).
(x - 10)(x2 + 10x + 100).
a - b a2
ab
b2
(b) 27 x3  64
(This is a SOC)
(3x)3 + (4)3 =
(use formula, with a = 3x, b = 4).
(3x + 4)(9x2 - 12x + 16) (answer)
(3x)2 (3x)(4) (4)2
(c) 5 x 3  5
5(x3 - 1)
(gcf = 5, take it out: )
(DOC inside pharentheses, w/ a = x, b = 1)
5( x  1)( x2  x  1).
(8)(a) x 2  10 x  25.
-1
-25
(yeah)
(no gcf, trinomial, lead coeff = 1)
(need two numbers which * to 25, + to -10:
-5 -5
= (x - 5)(x - 5).
(b) 4 x 2  36 x.
= 4 x( x  9)
(c) 4 x3  36 x. =
4 x( x 2  9) =
4 x( x  3)( x  3).
-5 + -5 = -10 )
4 x2
36 x
 x,
 9 .)
4x
4x
no more factoring can be done)
(gcf = 4x, factor it out:
(gcf = 4x, factor it out:)
(DOS in pharentheses (x)2 - 32)
(complete factorization)
(d) 3 x 2  11x  4.
3*4 = 12
(no gcf, trinomial, lead coeff = 3, grouping)
1 -12
= 3x  x  12 x  4
(use 1 and -12 to break up 11x)
(pair off, gcf's)
2
gcf = x, gcf = -4
(signs have to match)
= x(3x  1)  4(3 x  1) (combine terms)
= ( x  4)(3 x  1).
(e) x3 + 216 =
(x)3 + (6)3
( x  3)( x 2  3x  9).
a + b a2 -ab + b2
(no gcf, two terms, 6 is the cube of 216, so SOC)
( x plays the role of a, 6 plays the role of b)
(f) 4 x 2  8 x  7 x  14.
(no gcf, four terms - grouping, pair off, gcf's)
gcf = 4x gcf = 7
4 x( x  2)  7( x  2) 
(4 x  7)( x  2).
(factor out gcf's)
(combine terms.)
(g) 6 x 2  24
6( x 2  4) 
6( x  2)( x  2).
(gcf = 6, factor it out:)
(two terms, DOS in pharentheses (x)2 - 22 )
(split it up - complete factorization)
(h) 1000 x3  27
(no gcf, two terms, note the cube x3, and 1000 = 103, 27 = 33)
(10x)3 - 33
(DOC, w/ 10x playing a, 3 playing b)
2
(10 x  3)(100 x  30 x  9) .
(10x)2 3*10x 32
(i) x 2  11x  30.
1
2
30
15
(no gcf, trinomial, lead coefficient = 1)
3 10
5
6
= (x + 5)(x + 6).
(j) 2 x 2  7 x  15.
(2)(-15)=-30
( 5 + 6 =11)
(no gcf, trinomial, lead coefficient 1, grouping:)
-1 30
-2 15
-3 10
(-3 + 10 = 7 - use them to break up 7x)
2
= 2 x  3x  10 x  15  (pair off, gcf's)
gcf = x
gcf = 5
(pull out gcf's)
= x(2 x  3)  5(2 x  3)  (combine terms:)
( x  5)(2 x  3).
(9) (a) x 2  25  0.
(x + 5)(x - 5) = 0,
x + 5 = 0, x - 5 = 0,
-5 -5 +5 +5
x = 5, x = -5.
x = {5}
(b) x 2  x  6.
-6 -6
2
x  x  6  0,
( x  3)( x  2)  0,
x  3  0, x  2  0.
+3 +3 -2 -2
x = 3, x = -2,
x = {3, -2}
(c) 2 x3  10 x 2  28 x.
-28x -28x
3
2
2 x  10 x  28 x  0
2 x( x 2  5x  14)  0,
(LHS = 0, so, factor LHS as a DOS)
(set factors = 0,)
(give answer in solution set: )
(get LHS = 0)
(factor LHS)
(set factors = 0)
(get LHS = 0)
(gcf = 2x)
(inside pharentheses - trinomial, lead coefficient = 1)
-1 14
-2 7
(-2 + 7 = 5)
2 x( x  7)( x  2)  0,
(factors = 0)
2 x  0, x  7  0, x  2  0,
x= 0 -7 -7 +2 +2,
x = 0, x = -7, x = 7.
x = {0,7}