Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
270
12.34 Factor the expression: x
6
– 1.
Express the terms of the polynomials as perfect squares: (x
3
)
2
= x
6
and 1
2
= 1.
Apply the difference of perfect squares formula.
The resulting factors are a sum of perfect cubes (x
3
+ 1) and a difference of
perfect cubes (x
3
– 1). Apply the factoring formulas for each, setting a = x and
b = 1.
Factoring Quadratic Trinomials
Turn one trinomial into two binomials
12.35 Factor the expression: x
2
+ 3x + 2.
The quadratic binomial x
2
+ ax + b, once factored, has the form below.
Each of the empty boxes represents a number. Those missing values have sum a
and product b. In this problem, a = 3 and b = 2, so the two missing values must
have a sum of 3 and a product of 2. The correct values are 1 and 2, because
1 + 2 = 3 and 1 · 2 = 2. Substitute 1 and 2 into the boxes of the formula.
x
2
+ 3x + 2 = (x + 1)(x + 2)
The order of the factors is irrelevant, so the answer (x + 2)(x + 1) is equally
correct.
12.36 Factor the expression: x
2
+ 8x + 12.
To factor this quadratic binomial, you must identify two numbers with sum 8
and product 12. Begin by subtracting 1 from the required sum (8 – 1 = 7) to
identify two numbers with the correct sum: 1 + 7 = 8. Unfortunately, 1 and 7 do
not have the required product (1 · 7 ≠ 12).
You could
factor x
6
– 1
using the difference
of perfect cubes
formula, because
(x
2
)
3
= x
6
and 1
3
= 1.
You’ll end up with
(x + 1)(x – 1)(x
4
+ x
2
+ 1).
There’s no easy way
to factor the trinomial
in there, though. The
difference of perfect
squares formula gives
a better answer with
more factors.
In other
words, when
you add the
numbers, you get
the coefcient of
x in the trinomial,
and when you
multiply them, you
get the constant.
This works only when
the coefcient
of x
2
is 1 (like in
this problem).
Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
271
To continue the search for the correct values, subtract two from the required
sum (8 – 2 = 6) to identify another possible pair of numbers: 2 + 6 = 8.
The numbers 2 and 6 satisfy both requirements, the correct sum and the
required product: 2 + 6 = 8 and 2(6) = 12. Substitute 2 and 6 into the boxes of
the factoring formula.
12.37 Factor the expression: x
2
– 9x + 20.
The two numbers needed to factor this expression have a sum of –9 and a
product of 20. Those values are only possible if both numbers are negative; two
negative numbers have a negative sum and a positive product. Test a series of
negative numbers that add up to –9 until you find a pair of values that equals 20
when multiplied: –8 – 1, –7 – 2, –6 – 3, and –5 – 4.
Notice that –5 and –4 have the required sum and product: –5 + (–4) = –9 and
–5(–4) = 20. Use those values to factor the polynomial.
x
2
– 9x + 20 = (x – 5)(x – 4)
12.38 Factor the expression: x
2
– 16x + 64.
Like Problem 12.37, the numbers required to factor the expression have a
negative sum and a positive product, so both of the numbers must be negative.
Test a series of sums that equal –16 until you find values that have a product of
64: –15 – 1, –14 – 2, –13 – 3, –12 – 4, –11 – 5, –10 – 6, –9 – 7, and –8 – 8.
Because –8 + (–8) = –16 and –8(–8) = 64, –8 and –8 are the necessary values.
Factor the expression.
x
2
– 16x + 64 = (x – 8)(x – 8)
Because the factor (x – 8) appears twice, you should use exponential notation.
x
2
– 16x + 64 = (x – 8)
2
12.39 Factor the expression: x
2
+ 3x – 18.
To factor, you need to identify two numbers with a sum of 3 and a product of
–18. Because the product is negative, the numbers must have different signs; the
products of two positive numbers and the products of two negative numbers are
positive. Because the sum is positive, the positive value must be larger than the
negative value.
Generate pairs of numbers with opposite signs that have a sum of 3 until you
identify a pair with a product of –18: 4 – 1, 5 – 2, and 6 – 3. Because 6 – 3 = 3
and 6(–3) = –18, the values required to factor the expression are –3 and 6.
x
2
+ 3x – 18 = (x – 3)(x + 6)
If that
didn’t work, you’d
subtract 3, then 4,
and so on. From start
to nish, you’d try 7 +
1, 6 + 2, 5 + 3, and 4
+ 4 until you nd the
pair of numbers that
multiplies to give
you 12.
You don’t
HAVE to write out
this list, but it gives
you a place to start
if you can’t gure
out what the two
numbers are
right away.
In other
words, if you
ignore the sign for a
moment, the positive
number will be greater
than the negative.
For example, +9 and
–6 have a positive sum
because the positive
value 9 is greater than
the negative value 6.
However, –9 and +6
have a negative sum
because 6 < 9.
Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
272
12.40 Factor the expression: x
2
– 5xy – 36y
2
.
Much like the polynomial x
2
+ ax + b has factors , the polynomial
x
2
+ axy + by
2
has factors . To factor, you must identify two
numbers that have a sum of –5 and a product of –36. Because the product is
negative, the numbers have different signs. The sum is negative, so the larger of
the two numbers is negative.
Generate a list of number pairs that have a sum of –5 until you encounter
numbers with a product of –36: –6 + 1, –7 + 2, –8 + 3, and –9 + 4. Notice that
–9 + 4 = –5 and –9(4) = –36, so factor the polynomial by placing –9 and 4 into
the boxes of the formula .
x
2
– 5xy – 36y
2
= (x – 9y)(x + 4y)
12.41 Factor the expression: x
4
+ 6x
2
+ 9.
Identify two numbers with sum 6 and product 9 and place them into the boxes
of the formula to factor the polynomial x
4
+ 6x
2
+ 9. The sum
and the product of the numbers are positive, so the numbers themselves are
also positive.
Generate a list of number pairs that sum to 6 until you identify a pair that has a
product of 9: 5 + 1 = 6, 4 + 2 = 6, and 3 + 3 = 6. Notice that 3 + 3 = 6 and 3(3) = 9,
so substitute 3 and 3 into the boxes of the factoring formula stated above.
x
4
+ 6x
2
+ 9 = (x
2
+ 3)(x
2
+ 3)
Use an exponent to indicate the presence of a repeated factor.
x
4
+ 6x
2
+ 9 = (x
2
+ 3)
2
12.42 Factor the expression: 2x
2
+ 17x + 36.
When the coefficient of x
2
is not 1, the factoring technique described
in Problems 12.35–12.41 will not work. Instead, you should factor by
decomposition. Like the method presented in the preceding problems, you
must identify two numbers with specific characteristics. Given the polynomial
ax
2
+ bx + c, the two required numbers have sum b and product a · c.
In this problem, the two required numbers have a sum of 17 and a product of
2 · 36 = 72. Both the sum and product are positive, so the individual numbers
are positive as well. List pairs of numbers that have a sum of 17 until one pair
has a product of 72 as well: 16 + 1, 15 + 2, 14 + 3, 13 + 4, 12 + 5, 11 + 6, 10 + 7, and
9 + 8. Notice that 9 + 8 = 17 and 9(8) = 72.
So when you
nd the numbers,
write them as
coefcients of y in
the factors.
The numbers
still add up to
the x-coefcient
like they did before,
but now they don’t
multiply to give you the
constant; their product
is equal to the product
of the constant and
the coefcient of
x
2
.
Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
273
When factoring by decomposition, you do not immediately place the required
numbers into factors. Instead, rewrite the x-coefficient as a sum.
According to the commutative property, the order in which you multiply does
not affect the product. Therefore, (8 + 9)x and x(8 + 9) are equivalent. Apply
the distributive property: x(8 + 9) = 8x + 9x.
2x
2
+ 8x + 9x + 36
Factor by grouping, as explained in Problems 12.20–12.25.
Therefore, the factored form of 2x
2
+ 17x + 36 is (x + 4)(2x + 9).
12.43 Factor the expression: 8x
2
– 2xy – 3y
2
.
Identify two numbers with a sum of –2 and a product of –24. List pairs of
numbers with a sum of –2 until one pair also has a product of –24: –3 + 1,
–4 + 2, –5 + 3, and –6 + 4. Notice that –6 + 4 = –2 and –6(4) = –24. Replace the
x-coefficient with the sum of –6 and 4.
Apply the distributive property: (–6 + 4)xy = –6xy + 4xy.
8x
2
– 6xy + 4xy – 3y
2
Factor by grouping.
Therefore, the factored form of 8x
2
– 2xy – 3y
2
is (4x – 3y)(2x + y).
RULE
OF THUMB:
When the x
2
-
coefcient is 1,
nding the two
“mystery” numbers
means you’re done
factoring. However,
when the x
2
-
coefceint is NOT 1
(in this problem, it’s
2), you replace the x-
coefcient with the
sum of the mystery
numbers and then
factor by
grouping.
When you multiply
the “mystery”
numbers, you should
get the same result
as multiplying the x
2
-
coefcient (8) and the
constant (–3).
This sign is
always positive, no
matter what’s in
parentheses.
Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
274
12.44 Factor the expression: 6x
2
– 41x + 30.
To factor by decomposition, list pairs of numbers that have a sum of –41 until
one pair also has a product of 6 · 30 = 180: –40 – 1, –39 – 2, –38 – 3, –37 – 4, and
–36 – 5. Notice that –36(–5) = 180, so replace the x-coefficient with the sum of
–36 and –5.
Apply the distributive property and factor by grouping.
The
product
is positive,
so both
numbers
have to have
the same sign.
You also know
that the sum is
negative, which
means the
numbers you’re
looking for
must both be
negative.
In order to
get matching (x – 6)
binomials, you have to
factor –5 out of
(–5x + 30) instead
of +5. If you don’t
understand why, look
at Problem 12.24.
You can check this
(or any other factoring answer) by
multiplying the factors to make sure you
get the original polynomial back:
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