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St Pius X ACCP Precalculus
Summer Study Guide
Purpose
Welcome to the ACCP Precalculus Summer Study and Review Guide. This
packet is intended to help review important concepts crucial to success in Precalculus. These topics are supposed to be assumed knowledge and will not be
reviewed at the beginning of the year. You are to complete the portions of the
packet marked Exercise. This is due the first day of school. Away we go....
1
Order of Operations
Everything begins and ends with Order of Operations:
Please Parentheses
Excuse Exponents
My
Multiplication
Dear
Division
Aunt
Addition
Sally
Subtraction
Remember that multiplication and division are taken in order from left to
right, as are addition and subtraction. One operation does not come before the
other in this case.
2
Polynomial Arithmetic
2.1
Polynomials
First start with a monomial. A monomial is a product of a constant (any ol’
number) and any variables. Something like 5x2 or 12abc are monomials. The
degree of a monomial is the sum of all variable powers present. So the degree of
5x2 is 2 and 12abc is 3. The degree of any constant is 0. (Why? There are no
variables, so no powers.) A polynomial is a bunch of monomials strung together
with addition or subtraction signs. So something like 4x4 −x3 +3x2 +9x−30 is a
polynomial. The degree of a polynomial is the highest power that appears. The
example polynomial is of degree 4. We also like to write polynomials in standard
form, which has the highest degree term first, then the rest in descending order.
2.2
Adding and Subtracting
This is easy. All you do are add or subtract the like terms together. Be very
careful when subtracting. Make sure you negate all terms in parentheses before
combining.
Exercise Add or subtract
1. (x5 + 4x3 − 2x2 − 5) + (2x5 + 3x4 − x3 + 5x2 + 3x + 8)
2. (2x3 − 4x + 1) + (5x2 + 3x − 2)
3. (5x4 − 2x3 + x2 − 6x + 2) − (3x4 − 3x2 + 1)
4. (6x3 − 3x2 + x) − (−2x3 + 3x2 + x + 1)
5. 9x3 y − 2x2 y 2 + 5xy 3 − x3 y + 6y 4 + x2 y 2 − 8xy 3
6. (4ab + a2 b2 − 9a2 ) − (−2ab2 + ab − 3a2 + 4b)
2.3
Multiplying
Monomial X Polynomial
All you do here is use the distributive property to multiply.
x2 (5x2 − 2x + 1) = 5x2 ∗ x2 − 2x ∗ x2 + 1 ∗ x2
= 5x4 − 2x3 + x2
Easy, right?
Binomial X Binomial
Everyone remembers FOIL to multiply something like (3x + 5)(2x − 1). But
what about some special products?
Definition The square of a binomial
(a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2
(a − b)2 = (a − b)(a − b) = a2 − ab − ab + b2 = a2 − 2ab + b2
The resulting expression is called a perfect square trinomial.
Exercise Multiply
1. (x + 4)2 = x2 + 8x + 16
2. (x − 4)2
3. (x − 3)2
4. (2x + 5)2
5. (7x − 4y)2
6. (3a2 + 5)2
7. (10a3 − 3b)2
Definition The sum and difference of same terms
(a − b)(a + b) = a2 + ab − ab + b2 = a2 − b2
The resulting expression is called a difference of squares.
Exercise Multiply
1. (x + 4)(x − 4) = x2 − 16
2. (x − 25)(x + 25)
3. (3x − 5y)(3x + 5y)
4. (8x2 + y 3 )(8x2 − y 3 )
Definition The cube of a polynomial
(a + b)3 = (a + b)(a + b)(a + b) = a3 + 3a2 b + 3ab2 + b3
(a − b)3 = (a − b)(a − b)(a − b) = a3 − 3a2 b + 3ab2 − b3
Polynomial X Polynomial
When you have two polynomials of more than two terms multiplied together,
you just take each term of the first polynomial and multiply it to each term of
the second polynomial.
(3x − 1)(2x2 + 4x − 3) = 3x ∗ 2x2 + 3x ∗ 4x + 3x ∗ −3 − 1 ∗ 2x2 − 1 ∗ 4x − 1 ∗ −3
= 6x3 + 12x2 − 9x − 2x2 − 4x + 3
= 6x3 + 10x2 − 13x + 3
3
Exponents and Radicals
3.1
Exponents
Properties of Exponents
1. am an = am+n
2.
am
= am−n
an
−n
3. a
1
= n =
a
( )n
1
a
4. a0 = 1
m
5. (ab) = am bm
( a )m
am
6.
= m
b
b
7. |a2 | = |a|2 = a2
Most of this looks familiar, right?
Exercise Simplify. No negative exponents should remain.
1.
32
34
2. (23 · 32 )2
3. 64(−2)−6
4. (−z)3 (3z 4 )
5.
25y 8
10y 4
6. (z + 2)−3 (z + 2)−1
( −3 4 )−3
x y
7.
5
3.2
Radicals
Square Roots
Definition A number c is a square root of a if c2 = a.
Example
1. 4 and −4 are square roots of 16 since 42 = 16 and (−4)2 = 16
2. Does −9 have a square root? No, since there is no number that we can
square to get −9.
Every positive number has two square roots, a positive and a negative one.
0 is its own square root. Negative numbers do not have square roots.
There can be some confusion when dealing with square roots. Since there
are two, when I say “the square root of 16”, which one am I talking about? We
can clear this up by only using the principal square root, which is defined to be
the positive square root. √The symbol for the principal square root (using the
definition from above) is a.
√
Example
1. 16 = 4
√
16
4
2.
=
9
3
√
3. − 9 = −3
√
25
5
4. −
=−
4
2
Algebraic expressions underneath the radical
sign are a bit tougher. We’re
√
going to look at expressions of the form of x2
√
√
1. 42 = 16 = 4
√
√
2.
(−4)2 = 16 = 4
√
Technically, the general result of x2 = |x|. However, we are going to
assume that all variables underneath the radical sign will represent positive
quantities, so absolute
value notation will not be necessary. Thus, under
√
this assumption x2 = x.
√
√
3. 25t2 = (5t)2 = 5t
√
√
4. 49a2 = (7a)2 = 7a
√
√
5. x2 − 4x + 4 = (x − 2)2 = x − 2
√
So, in general x2 = x when we assume that x only represents non-negative
numbers.
Exercise Simplify these expressions, if possible.
√
1. −25
√
2. 0
√
3. 121
√
4. 36x2
√
5.
81x2 y 2
Cube Roots
Same idea as the square root, except it’s related a power of 3
Definition A number c is a cube root of a if c3 = a.
The cube root of 8 is 2, since 23 = 8. The cube root of -8 is -2, since
(−2)3 = −8. Note right away that negative numbers can have cube roots and
that every number has only one cube root. This is a fact of cube roots: Every
real number has exactly one cube root. A positive number will have a positive
cube root,√a negative number will have a negative cube root. The cube root
symbol is 3 a
Example Simplify
√
1. 3 8 = 2
√
2. 3 −8 = −2
√
3. 3 125 = 5
√
4. 3 −1 = −1
√
3
5. 27x3 = 3x
√
6. 3 −64y 3 = −4y
√
√
√
3
In general, x3 = x, for all x. (Also note that 3 −a = − 3 a.)
Exercise Simplify
√
1. 3 64
√
2. 3 −216
√
3. 3 0
√
4. 3 x3 y 3
√
3
5. 125a3
nth Roots
We can have 4th roots, 5th roots, etc. Instead of listing the characteristics of
each root, we’ll generalize.
Definition
A number c is an nth root of a if cn = a. The symbol of the nth
√
n
root is a. a is the radicand, and n is the index of the radical. (Note that for
a square root, there is no index.)
If n is even,
If n is odd,
√
n
√
n
a “behaves” like
a “behaves” like
√
√
3
a
a
Example Odd nth roots
√
√
7
1. 7 128 = 27 = 2
√
√
2. 7 −128 = 7 (−2)7 = −2
√
5
3. x5 = x
√
19
4.
x19 = x
Example Even nth roots
√
1. 4 81 = 3
√
2. 4 −81 does not exist
√
√
4
3. 16x4 = 4 (2x)4 = 2x
√
√
4. 16 x16 y 16 = 16 (xy)16 = xy
√
√
5. x16 = (x8 )2 = x8
√
√
6. x22 = (x11 )2 = x11
Note how the last two problems worked. You divide the power of the expression by the index to get the simplified form. This only works if it divides
evenly! We will see how to handle situations that aren’t so nice a bit later.
Exercise Simplify if possible
√
1. 5 −32
√
5
2. x10
√
10
3.
x40
√
4. x100
3.3
Rational Exponents
Definition
So far, we have dealt with only integer exponents (like x2 , y −5 ). Fractions can
also be exponents.
x1/2
If we square x1/2 , we get:
(x1/2 )2 = x1/2·2 = x1 = x
This implies that x1/2 is a square root of x. Thus, x1/2 =
√
x.
x1/3
If we cube x1/3 , we get:
(x1/3 )3 = x1/3·3 = x1 = x
This implies that x1/3 is a cube root of x. Thus, x1/3 =
continues for x1/n .
x
1/n
=
√
n
√
3
x. The pattern
x
Exercise Write the following expressions in radical notation
1. x1/7
2. 271/3
3. (xy)1/4
Exercise Write the following expressions in exponential notation
√
1. 3 4xy
√
2. 3x
√
xy
3. 6
3
Okay, now what about something like x2/3 , which doesn’t fit the format
above?
√
3
x2/3 = x2·1/3 = x2
√
x2/3 = x1/3·2 = ( 3 x)2
In general:
x
m/n
=
√
n
xm
√
n
= ( x)m
Exercise Write in radical notation, and simplify when possible
√
1. 274/3 = ( 3 27)4 = (3)4 = 81
2. 813/2
3. b3/2
4. (125a)2/3
5. (9y 6 )3/2
Exercise Write in exponential notation
√
1. a5
√
5
2. n4
√
3. ( 3 7xy)8
To reiterate:
x1/n =
xm/n =
√
n
√
n
x
√
xm = ( n x)m
The Laws of Exponents can be applied to rational exponents as well.
1. am · an = am+n
am
2. n = am−n
a
3. (am )n = am·n
4. (ab)m = am bm
( a )m a m
= n
5.
b
a
1
6. a−m = m
a
1
7. −m = am
a
a−m
bn
=
b−n
am
( a )−m ( b )m
9.
=
b
a
8.
Exercise Simplify. Do not leave negative exponents in your final answers.
1. y −1/4 =
2.
1
y 1/4
1
a−3/5
3. 5 x−2/3 y 4/5 z
4.
2y
5 x−1/3
5. 112/3 · 111/2
6.
87/11
8−4/11
7. (27−2/3 )3/2
8. (x−1/3 y 2/5 )1/4
Rational exponents can be used to simplify difficult looking radicals.
√
√
8
t4 = t4/8 = t1/2 = t
Exercise Simplify. Do not use rational exponents in your final answers.
√
6
1. t4
2.
3.
4.
√
5
a10
√
√
3
6
m
√
4
(xy)12
√
5
5. ( a2 b4 )15
3.4
Multiplying Radical Expressions
The Product Rule
Definition The Product Rule for multiplying two radicals together is:
√
√
√
n
n
n
a· b= a·b
Note that you can multiply radicals like this only when they have the same
index.
√ √
√
1. 5 · 7 = 35
√
√
√
2. 5 8x · 5 3x = 5 24xy
√
√
√
√
3. x − 4 · x + 4 = (x + 4)(x − 4) = x2 − 16
Simplifying Roots by Factoring
√
√ √
The most useful aspect of the product rule is the reverse of it: n ab = n a n b.
This can be used to simplify roots
√ that don’t work out evenly like they did in
10.1. For example to simplify 8, we can do the following:
√
√
√ √
√
8= 4·2= 4· 2=2 2
To simplify a square root, factor out the perfect square, split it off, and take the
square root. Everything else is left as a radical. This generalizes to nth roots
as well. √
Now what about an algebraic expression underneath the radical? Let’s
look at x3 :
√
√
√
√
√
x3 = x2 · x = x2 · x = x x
√
We factor out the perfect square, which happens to be x2 . How about x7 ?
√
√
√
√
√
x7 = x6 · x = x6 · x = x3 x
This time, the perfect square is x6 . It seems that the perfect squares in
these cases are simply even powers of x. In fact, that is true. When simplifying
square roots by factoring, even powers of x (multiples of 2) are factored out,
split off, and “square rooted”. This can be extended to cube roots.
√
√
√
√
√
3
3
3
3
3
x5 = x3 · x2 = x3 · x2 = x x2
This time, the perfect cube is x3 . For cube roots, you factor out powers of x that
are multiples of 3, split them off, and “cube root” them. Again, the “leftover”
factors are left underneath the radical. This idea extends to any nth root. If
you are trying to simplify a 4th root, factor out powers that are multiples of 4,
5th roots, multiples of 5, etc.
√
√
√ √
√
1. 54 = 9 · 6 = 9 · 6 = 3 6
√
√
√
√
√
2. 300 = 100 · 3 = 100 · 3 = 10 3
√
√
√
√ √ √
√
3.
x5 y 8 = x5 · y 8 = x4 x y 8 = x2 y 4 x
√
√
√
√ √
√ √ √
√ √ √
3
3
4. 3 48 x10 y 8 = 3 48· x10 · 3 y 8 = 3 8 3 6 x9 3 x 3 y 6 3 y 2 = 2 x3 y 2 3 6 x y 2
√
√
√
√
√
√ √
4
4
4
4
4
5. 81 x7 = 4 81 · x7 = 4 81 x4 x3 = 3x x3
Exercise Simplify these roots by factoring.
√
1. 27
2.
3.
4.
5.
6.
7.
√
√
80
325
√
√
3
x6 y 9
27 a b7
√
4
16 x5 y 11
√
3
128 a7 b8
√
To sum up, if you have n some expression, you factor out the powers that
are multiples of n. Those are the perfect nth roots that you can simplify, and
all leftover factors are left underneath the radical sign.
Exercise Multiply, then simplify.
√ √
1. 6 3
2.
3.
√
√
3
933
√
√
5a7 15a3
4.
√
√
3
x2 y 4 3 x2 y 6
5.
√
√
4
9x7 y 2 4 9x2 y 9
3.5
Dividing Radical Expressions
The Quotient Rule
Definition The Quotient Rule for dividing two radicals is
√
√
n
a
a
n
√
=
n
b
b
Note that you can only divide radicals as such when they have the same
index.
Example Simplify
√
√
7
49
49
1.
=√ =
25
5
25
√
√
x2
x2
x
2.
=√
=
400
20
400
√
√
√
4x3
4x3
2x x
3.
= 3√
= √
y7
y y
y7
√
√
√
3
3
a3 b
a 3 ab
3 a b
4.
= √
= √
3 8
3
c8
c
c2 c2
Exercise Simplify
√
100
1.
81
√
2.
√
3.
√
4.
4
√
5.
4
121
x2
36a5
b6
81x4
y8 z4
x9 y 12
z6
Example Divide and simplify
√
√
√
√
56x2
56x2
1. √
=
= 8x = 2 2x
7x
7x
√
√
3
7 10
√
128 x7 y 10
3 128x y
√
=
= 3 64x6 y 6 = 4x2 y 2
2.
4
3
4
2xy
2xy
Exercise Divide and simplify
√
28y
1. √
4y
√
3
40
2. √
3
5
√
3
3.
189x5 y 7
√
3
7x2 y 2
√
4.
75ab
√
3 3
Rationalizing Denominators
It is preferred not to have a radical expression in the denominator of a fraction.
We can “get rid” of radicals by rationalizing. What we need to do is multiply
both the numerator and denominator by an expression that will create a perfect
square, cube, or whatever in the denominator. For time considerations, we will
only look at rationalizing square roots.
1
If we want to rationalize √ , we need to create a perfect square in the
2
denominator. Here’s how:
√
√
√
1
1
2
2
2
√ =√ ·√ =√ =
2
2
2
2
4
a square root, multiply by the same square root again, since
√
√ To
√ rationalize
x · x = x2 = x. Be sure to do it to both the numerator and denominator
of the fraction and to simplify the final result.
Exercise Rationalize the denominator and simplify, if possible
√
11
1. √
6
√
2.
√
3.
3x
10
3
8
√
3
4. √
4 6
5
5. √
50
√
5 8
6. √
2 15
3.6
Simplifying Expressions With Several Radicals
Adding and subtracting radicals
We can only add or subtract radicals if they are of√the same kind,
√ meaning they
2
and
−5
2 would be like
must be of the
same
index
and
radicand.
So,
3
√
√
√
√
radicals, but 3 and 2 are not, and 2 and 3 2 are not also. Like radicals are
a similar concept to like√terms √
in polynomials.
We√add or√subtract
√
√ in the same
n
manner as like terms: 3 2 + 5 2 = 8 2. Note: n a + b 6= n a + b
Example Here are a couple of demonstrations
√
√
√
1. 8 5 + 9 5 = 17 5
√
√
√
√
√
√
√
√
√
√
2. 5 7 − 8 4 11 − 7 + 9 4 11 = 5 7 − 7 − 8 4 11 + 9 4 11 = 4 7 + 4 11
Sometimes, though,
√ radicals
√ look different when they really are the same.
For example,
look
at
8
and
2. They look like
√
√
√ different radicals, but remember
that 8 = 2 2. Since now they both have a 2 in them, they are like radicals
and can be added or subtracted. So, to demonstrate:
√
√
√
√
√
8+ 2=2 2+ 2=3 2
√
√
√
√
√
3
3
3
3
3
3 4 − 32 = 3 4 − 2 4 = 4
√
√
√
√
√
75x + 3x3 = 5 3x + x 3x = (5 + x) 3x
Be aware that you may have to simplify the radicals involved before you
actually add or subtract them.
Exercise Add or subtract
√
√
√
√
1. 6 3 5 − 4 7 12 + 6 7 12 − 3 3 5
√
√
2. 9 50 − 4 2
3.
4.
√
3
√
54x −
√
3
2x4
9y + 27 +
√
y+3
Other Multiplications
We know how to
√ multiply
√ two individual radicals together, but how about an
expression like 5(6 + 2)? Notice that it looks a lot like this: x(x2 + 5). We
handled polynomial expressions like that by using the distributive law. So the
radicals would be handled like this:
√
√
√
√ √
√
√
5(6 + 2) = 6 5 + 5 2 = 6 5 + 10
Remember how to multiply
√ (x
√ + 3)(x
√ + 5)
√using FOIL? We can do the same
thing when multiplying (2 6 − 2)( 6 + 4 2).
√
√ √
√
(2 6 − 2)( 6 + 4 2)
√ √
√
√
√ √
√
√
= (2 6)( 6) + (2 6)(4 2) − ( 2)( 6) − ( 2)(4 2)
√
√
= 2(6) + 8 12 − 12 + 4(2)
√
= 12 + 8 + 7 12
√
= 20 + 7 · 2 3
√
= 20 + 14 3
Exercise Multiply out and simplify, if possible
√
√
1. 3(4 + 3)
2.
√
3
√
√
3
3
x( 3x2 − 81x2 )
√
√
√
√
3. (4 5 + 3 3)(3 5 − 4 3)
√
√
4. ( t − 2r)2
5. (2 −
√
√
5)(2 +
5)
√
√
6. (4 + 3 2)(4 − 3 2)
7. (3 −
√
a)(3 +
√
a)
The last three problems resemble difference of squares problems
(A + B)(A − B) = A2 − B 2 .
Notice that all radicals disappear in the final result. We call A + B and A −
B conjugates and they are most useful for rationalizing denominators that
are binomial in nature. You multiply the numerator and denominator by the
conjugate of the denominator.
√
10 + 4
√
2−3
=
=
=
=
=
=
√
√
10 + 4
2+3
√
·√
2−3
2+3
√
√
( 10 + 4)( 2 + 3)
√
√
( 2 − 3)( 2 + 3)
√
√
√
20 + 3 10 + 4 2 + 12
√
( 2)2 − (3)2
√
√
√
2 5 + 3 10 + 4 2 + 12
2−9
√
√
√
12 + 4 2 + 2 5 + 3 10
−7
√
√
√
12 + 4 2 + 2 5 + 3 10
−
7
You should always simplify when possible. In this last case, there was no
way to simplify any further, so we leave it.
Exercise Rationalize the denominator
1.
3
√
4− 7
√
2. √
y
√
x− y
√
√
7+ 5
√
3. √
5+ 2
3.7
Radical Equations
How to solve radical equations
1. Isolate the radical expression onto one side of the equation
2. “Get rid” of a radical by raising both sides of the equation to the same
power (eg., to get rid of a square root, square both sides, to get rid of a
cube root, cube both sides, etc.)
3. Solve the resulting equation through “normal” means (all radicals should
be gone at this point)
4. Check for extraneous solutions (very important!)
x+2=
√
7x + 2
(√
)2
(x + 2)2 =
7x + 2
x2 + 4x + 4 = 7x + 2
x2 − 3x + 2 = 0
(x − 2)(x − 1) = 0
x = 2 or x = 1
Check:
√
7(2) + 2
√
4 = 14 + 2
√
4 = 16 True
2+2=
√
7(1) + 2
√
3= 7+2
√
3 = 9 True
1+2=
√
y =5+ y−3
√
y−5= y−3
(√
)2
(y − 5)2 =
y−3
y 2 − 10y + 25 = y − 3
y 2 − 11y + 28 = 0
(y − 7)(y − 4) = 0
y = 7 or y = 4
Check:
7=5+
7=5+
√
√
7−3
4
7 = 5 + 2 True
4=5+
4=5+
√
√
4−3
1
4 = 5 + 1 False
Therefore, y = 4 is not a solution, and y = 7 is the only solution.
Exercise Solve the following equations
√
1. 3x + 1 = 6
2.
√
x − 7 + 3 = 10
3. x =
√
x−1+3
4
Factoring
Yes, the topic that doesn’t die. It seems like this is the one thing that doesn’t
stick. Factoring is crucial to almost everything that we do.
4.1
Basics
What are we trying to do when we factor? We’re trying to express the result
of a product as its factors. We’re trying to reverse the multiplication, in other
words. It’s real easy to do with numbers, right? 50 = 25 * 2 = 5 * 5 * 2. 84 =
21 * 4 = 7 * 3 * 2 * 2. Why is it so hard with polynomials? Part of the problem
is that nobody remembers the special multiplication formulas.
a2 + 2ab + b2 = (a + b)2
a2 − 2ab + b2 = (a − b)2
a2 − b2 = (a − b)(a + b)
These formulas describe a good percentage of the factoring problems you
will deal with. But first, let’s back up a little.
4.2
Factoring Methods
Common Factors
The first thing you should look for when factoring polynomial is a common
factor. What is a common factor? It’s a factor that appears in all terms of the
polynomial.
Example
1. 10x2 − 4x + 8 = 2(5x2 − 2x + 4)
2. 7x5 + 49x4 − 14x3 + 21x2 = 7x2 (x3 + 7x2 − 2x + 3)
3. 6x2 y + 3x5 y 2 + 18x3 y = 3x2 y(2 + x3 y + 6x)
In each problem, just “divide” out the excess factor and put it in front. It’s like
reversing the distributive property.
Remember, this is always the first thing you look for when factoring.
Binomial Factoring
The next thing you look at is how many terms the polynomial has. If it has
two, then there are three possible formulas to use:
Difference of Squares
Sum of Cubes
Difference of Cubes
a2 − b2 = (a − b)(a + b)
a3 + b3 = (a + b)(a2 − ab + b2 )
a3 − b3 = (a − b)(a2 + ab + b2 )
Example Factor
1. 25x2 − 49 = (5x − 7)(5x + 7)
2. x3 + 27 = (x + 3)(x2 − 3x + 9)
3. 8x3 − 125 = (2x − 5)(4x2 + 10x + 25)
Exercise Factor
1. 16x2 − 36y 2
2. 9x4 − 64
3. x3 + 216
4. x3 + 1
5. x3 − 8
6. x3 − 125
Trinomial Factoring
The vast majority of factor problems are going to be trinomials (3 terms). There
are two formulas:
Perfect Square Trinomial
Perfect Square Trinomial
a2 + 2ab + b2 = (a + b)2
a2 − 2ab + b2 = (a − b)2
Example Factor
1. x2 + 10x + 25 = (x + 5)2
2. 4x2 − 28xy + 49y 2 = (2x − 7y)2
3. 10x3 − 40x2 + 40x = 10x(x2 − 4x + 4) = 10x(x − 2)2
Exercise Factor
1. x2 − 16x + 64
2. 81x2 + 36xy + 16y 2
Unfortunately, there are other trinomials out there that don’t fall into one
of those two categories. You can use a process I call “reverse FOIL” to get your
answer. Here’s how reverse FOIL works.
x2 + bx + c = (x + p)(x + q)
The numbers p and q have to multiply to make c and add together to make
b. Reverse FOIL is easy when there is no leading coefficient to the polynomial.
Exercise Factor
1. x2 + 6x + 5
2. x3 − 9x2 + 14x
3. 10t3 − 70t2 + 100t
4. 3x2 + 15x − 150
5. x2 − 4x − 32
Reverse FOIL is easy when x2 is all by itself out in front. Unfortunately, it ain’t
so easy when there is a lead coefficient attached to x2 . We talked about trial
and error factoring in Algebra II, but here’s another method.
Example Let’s factor 6x2 − 11x + 3 using a new approach. The first thing you
do is set up your parentheses with the lead term in the first spot in both.
(6x +
)(6x +
)
Now, multiply the lead coefficient to the last term: (6)(3) = 18. Find two
numbers that multiply to 18 and add the the middle number, −11. In this case,
−9 and −2 work. So put those numbers in the empty spots.
(6x − 9)(6x − 2)
Now, divide out any extra factors. In the first set, divide out 3, in the second,
divide out 2.
(2x − 3)(3x − 1)
Does this multiply out to the original polynomial? It sure does.
Exercise Factor with this new method
1. 10x2 − 41x + 4
2. 3x2 + 10x − 8
3. 3x2 + 4x − 4
4. 2x2 + 5x − 12
5. 11x2 − 7x − 4
5
Rational Expressions
Rational expressions are just fancier versions of fractions. Formally, a rational
expression is a fraction made up of polynomials.
5.1
Reducing
Reducing a rational expression is just like reducing a fraction. We can “cancel
out” things that factor out of both the numerator and denominator.
Exercise Simplify each fraction
1.
x2 + 5x − 6
(x − 1)(x + 6)
x−1
=
=
x2 + 6x
x(x + 6)
x
2.
x2 + 6x + 5
x2 − x − 2
3.
x2
x2
+ 2x
4.
8xy
12yz
5.
x2 − 5x
5−x
The last problem is an example of “opposite” factors. x − 5 and 5 − x can
cancel out, provided that a −1 is left behind.
5.2
Multiplying and Dividing
Again, just like fractions.
A C
AC
·
=
B D
BD
A
C
A D
AD
÷
=
·
=
B
D
B C
BC
Exercise
1.
2x2 y 2
·
y x3
2.
x2 − x − 6 x3 + x2
·
x2 − 3x
x+2
3.
a a2 − ab
÷
b
ab + b2
4.
x2 + 2x + 1 3x + 3
÷
x2 − 6x + 9
x−3
5.3
Adding and Subtracting
Again, you need an LCD just like regular fractions.
3 5
9
35
44
+ =
+
=
7 3
21 21
21
How do we do this with polynomials? Again, factoring is the key.
3
2
3
2
+
=
+
x2 + x x2 − 1
x(x + 1) (x − 1)(x + 1)
3(x − 1)
2x
=
+
x(x − 1)(x + 1) x(x − 1)(x + 1)
3x − 3
2x
=
+
x(x − 1)(x + 1) x(x − 1)(x + 1)
3x − 3 + 2x
=
x(x − 1)(x + 1)
5x − 3
=
x(x − 1)(x + 1)
It is not necessary to re-multiply out the denominator. We call this factored
form. Most of the time you will be able to simplify the fraction.
Exercise
1.
x 2x
+
5
3
2.
2
1
−
3x2
2x
3.
9
7
+
x − 3 x2 − 9
4.
1
3
2
+
−
a2 − 4 a − 2 a + 2
5.4
Complex Fractions
A complex fraction is a fraction that contains other fractions. There are two
methods to simplify a complex fraction:
Division Method
1. Combine rational expressions in numerator and denominator to form a
single fraction in both places.
2. Your complex fraction should look something like this:
A
B
C
D
3. Multiply what used to be the numerator by the reciprocal of what used
to be the denominator
A D
·
B C
4. Simplify and reduce to lowest terms if possible.
Example
1 1
4
x
−
−
x 4 = 4x 4x
x−4
x−4
4−x
= 4x
x−4
4−x
1
=
·
4x
x−4
1
=−
4x
LCD Method
1. Find the LCD of all mini-fractions within the complex fractions.
2. Multiply the numerator and denominator by that LCD.
3. Use the distributive property to simplify the numerator and denominator.
4. At this point, no mini-fractions should remain.
5. Simplify to lowest terms if possible.
Example
(
)
1
1
1
1
2
4x
−
−
x2
4
x2
4 =
(
)
1 1
1
1
2
−
4x
−
x 2
x 2
=
=
=
=
4x2
4x2
−
x2
4
4x2
4x2
−
x
2
4 − x2
4x − 2x2
(2 − x)(2 + x)
2x(2 − x)
2+x
2x
Either method can be used. Use whatever you are comfortable with.
Exercise
1
1
−
2
x
9
1.
x−3
1
1
−
2
x
16
2.
x+4
1
1
−
x y
3.
1
1
− 2
2
x
y