1. The document provides information about precalculus chapter 2, which covers exponents and radicals, polynomials, factoring, and complex numbers. 2. Key topics include scientific notation, properties of exponents and radicals, adding/subtracting/multiplying polynomials, factoring polynomials, and performing operations with complex numbers. 3. Examples are provided for simplifying expressions with exponents, factoring trinomials and polynomials, using long division and synthetic division to divide polynomials, and solving quadratic equations with complex number solutions.

1. PRECALCULUS 1
(ALGEBRA AND TRIGONOMETRY)
CHAPTER 2: EXPONENTIAL AND RADICALS, POLYNOMIALS
AND FACTORING AND COMPLEX NUMBERS
PREPARED BY:
ENGR. RAYMOND JAY G. SEVERO

2. EXPONENTS AND RADICALS
Repeated multiplication can be written in exponential form.
Repeated Multiplication Exponential form
a * a * a * a * a a5
(-2)(-2)(-2)(-2) (-2)4
(5x) (5x) (5x) (5x) (5x) (5x) (5x)6

4. Scientific Notation
Exponents provide an efficient way of writing and computing with
large (or very small) numbers. For instance, there are about 359 billion
billion gallons of water on Earth—that is, 359 followed by 18 zeros.
359,000,000,000,000,000,000
It is convenient to write such numbers in scientific notation.
3.59 x 100,000,000,000,000,000,000 = 3.59 x 1020
0.000000000009. = 9.0 x 10−12

5. Radicals and Their Properties
A square root of a number is one of its two equal factors. For
5 is a square root of 25 because 5 is one of the two equal factors of
In a similar way, a cube root of a number is one of its three equal
factors, as in 125 = 53.

6. Some numbers have more than one nth root. For example,
both 5 and -5 are square roots of 25. The principal square root
of 25, written as 25 is the positive root, 5. The principal nth
root of a number is defined as follows.

7. A common misunderstanding is that the square root sign
implies both negative and positive roots. This is not correct. The
square root sign implies only a positive root. When a negative
root is needed, you must use the negative sign with the square
root sign.

8. Integers such as 1, 4, 9, 16, 25, and 36 are called perfect squares
because they have integer square roots. Similarly, integers such as 1,
8, 27, 64, and 125 are called perfect cubes because they have
integer cube roots

10. Exercises No. 5:
Use the properties of
exponents to simplify each
expression.
1. (-3ab4)(4ab-3)
2. (2xy2)3
3. 3a(-4a2)0
4. (
5𝑥3
𝑦
)2
5. (
3
8𝑥2
𝑦
)2
Rewrite each expression with
positive exponents
1. x-2y4
2.
𝑥
3𝑦−2
3.
3𝑦𝑥−2
(3)2(𝑥𝑦)−6
4.
12𝑎3 𝑏−4
4𝑎−2 𝑏
5. (
3𝑥2
𝑦
)−2

11. Write each number in scientific notation.
1. 0.0000782 _______________
2. 836,100,000 _______________
3. 148,123,000,000 _______________
Write each number in decimal notation.
1. 9.36 x 10-8 _______________________
2. 1.823 x 109 _______________________

12. Use the properties of radicals to simplify expression.
1. 8 ∗ 2
2.
6
𝑥6
3.
4
48𝑥5
4.
3
−40𝑥6
5. 2 48 − 3 27
6.
3
16𝑥 −
3
54𝑥4
7.
5
2 3
8.
2
3
5
9.
2
3+ 7
10.
5𝑥
4𝑥2+ 12𝑥

13. Change radical to exponential and
exponential to radical form.
1. 2𝑥
4
𝑥3
2.
3
(12𝑥𝑦2)5
3. (𝑥2
+ 𝑦2
)
3
2
4. 2𝑦
3
4 𝑥
1
4

14. Simplify the rational exponent
1. −5𝑥
5
3 3𝑥−
3
4
2.
9
𝑎3
3.
3
125𝑥3
4. (2𝑥 − 1)
4
3(2𝑥 − 1)−
1
3
5.
𝑥−1
(𝑥−1)
−
1
2

15. POLYNOMIALS AND FACTORING
Polynomials the most common algebraic expression. Some
examples are 3x + 9, 4x3 – 2x2 + 24 and x2y2 – 4xy -16. The first
two are polynomials in x and the third is a polynomial in x and y.
The terms of a polynomial in x have the form axk, where a is the
coefficient and k is the degree of the term.

16. Polynomials with one, two, and three terms are called
monomials, binomials, and trinomials, respectively. In
standard form, a polynomial is written with descending
powers of x.

17. Operations with Polynomials
Sum and Difference of
Polynomials
You can add and subtract
polynomials in much the same
way you add and subtract real
numbers. Simply add or
subtract the like terms (terms
having the same variables to
the same powers) by adding
their coefficients.

18. Product of Polynomials
To find the product of two
polynomials, use the left and right
Distributive Properties.
Note in this FOIL Method
(which can only be used to
multiply two binomials) that
the outer (O) and inner (I)
terms are like terms and can
be combined.

19. Special Products
Some binomial products have
special forms that occur
in algebra. You do not need to
memorize these formulas
you can use the Distributive
Property to multiply. However,
becoming familiar with these
formulas will enable you to
manipulate the algebra more
quickly.

20. Special Product Example:
1.5x + 9 and 5x – 9 2. x + y – 2 and x + y + 2
Solution: Solution:
5x + 9 and 5x – 9 x + y – 2 and x + y + 2
= (u + v) * (u – v) = [(x + y) – 2] [(x + y) + 2]
= u2 – v2 = (x + y)2 - 22
= 25x2 – 81 = x2 + 2xy + y2 – 4

21. Exercises No. 6:
Perform the operation and write the
result in standard form.
1. (6x + 5) – (8x + 15)
2. (2x2 + 1) – (x2 – 2x + 1)
9. (9x2 + 3y2 + 5) (2x + 3xy + 2y2)
10. (5x2 + 2x + 4) (3y2 + 4y + 19)
11. ( 𝑥 + 𝑦)( 𝑥 − 𝑦)
12. (𝑥 − 5)2
13. (5 + 𝑥)(5 − 𝑥)

22. Division of Polynomials
There are two procedures in dividing
polynomials. The procedures are especially
valuable in factoring and finding the zeros of
polynomial function.
1. Long Division Polynomials
2.Synthetic Division

23. Example of long division
polynomials
Divide 6x3 – 19x2 +16x – 4 by x
– 2, and use the result to
the polynomial completely.
from the answer on the long division you
can conclude that
6x3 – 19x2 +16x – 4 is the factor of
(x – 2) (6x2 – 7x + 2) and we can factor
(6x2 – 7x + 2) into (2x -1)(3x – 2)
therefore;
6x3 – 19x2 +16x – 4 = (x – 2) (2x -1)(3x – 2)

24. Synthetic Division

25. Example: Use synthetic division to divide
x4 – 10x2 – 2x + 4 by x + 3
so, you have:
𝑥4
− 10𝑥2
− 2𝑥 + 4
𝑥 + 3
= 𝑥3
− 3𝑥2
− 𝑥 + 1 +
1
𝑥 + 3

26. The Remainder and Factor Theorem

27. Sample: Use the remainder theorem to evaluate the
following function at x = -2.
f(x) = 3x3 + 8x2 + 5x – 7
Solution: using synthetic division
Because the remainder is r = -9, you can conclude that f(-
2) = -9 r = f(x)
This means that (-2, -9) is a point on the graph of f(x).

28. Sample: Show that (x – 2) and (x + 3) are factors of
f(x) = 2x4 + 7x3 – 4x2 – 27x – 18.
Solution: using synthetic division (x – 2)
Since the remainder of f(2) = 0, therefore (x – 2) is a
factor. Also perform synthetic division to the result
using factor of (x + 3)

29. Since the remainder of f(-3) = 0,
therefore(x + 3) is a factor.

30. Polynomials with Common Factors
The process of writing a polynomial as a product is
called factoring. It is an important tool for solving
equations and for simplifying rational expressions.
If a polynomial cannot be factored using integer
coefficients, then it is prime or irreducible over the
integers. For instance, the polynomial x2 - 3 is irreducible
over the integers. Over the real numbers, this
polynomial can be factored as (𝑥 + 3 )(𝑥 − 3)

31. The simplest type of factoring involves a polynomial
that can be written as the product of a monomial and
another polynomial. The technique used here is the
Distributive Property, a(b + c) = ab + ac, in the
reverse direction.
Factoring Special Polynomial Forms
Some polynomials have special forms that arise from
the special product forms from the last topic. You should
learn to recognize these forms so that you can factor
such polynomials easily.

33. Exercises No. 7
Division of polynomials.
a.Divide the following using long method.
1. (2x2 + 10x + 12) by (x + 3)
2.(5x2 – 17x – 12) by (x - 4)
b. Divide the following using synthetic
division
1. (3x3 – 17x2 + 15x – 25) by (x – 5)
2.(4x3 – 9x + 8x2 – 18) by (x + 2)
c. Factor each expression.
1. 6x3 – 4x
2.-4x2 + 12x – 16

34. COMPLEX NUMBER
The imaginary unit I there are quadratic equations that
have no real solutions. For instance, the quadratic
of x2 + 1 = 0 has no real solution because there is no real
number x that can be squared to produced -1. To
overcome this efficiency, mathematicians created an
expanded system of numbers using the imaginary unit i.

36. Operation with complex number
Addition and Subtraction of Complex
Number
Sample:
1. (4 + 7i) + (1 – 6i)
(4 + 7i) + (1 – 6i) = (4 + 1) + (7i – 6i)
= 5 + i answer
2. (1 + 2i) – (4 + 2i)
(1 + 2i) – (4 + 2i) = (1 – 4) – (2i – 2i)
= -3 answer
3. (3 + 2i) + (4 – i) – (7 + i)
(3 + 2i) + (4 – i) – (7 + i)
=(3 + 4 – 7)+ (2i – i – i )
= 0 answer

37. Multiplying Complex Number
Many of the properties of real numbers are valid for complex
numbers as well. Here are some examples.
•Associative Properties of Addition and Multiplication
•Commutative Properties of Addition and Multiplication
•Distributive Property of Multiplication Over Addition
Notice below how
these properties are
used when two
complex numbers
are multiplied.

39. Quotient of Complex Number
To write the quotient of a +bi and c +di in standard form,
where c and d are not both zero, multiply the numerator
and denominator by the complex conjugate of the
denominator to obtain

41. Complex Solutions of
Quadratic Equations

42. Exercises No.8:
Perform the addition and subtraction of imaginary numbers.
1. (5 + i) + (6 – 2i)
2. (-2 + −8) + (5 – −50)
Perform the operation and write the result in standard form.
1. (1 + i) ( 3 – 2i)
2. 14 + 10𝑖 14 − 10𝑖
Perform the operation and write the result in standard form.
1.
2
4−5𝑖
2.
6−7𝑖
1−2𝑖

43. Trinomials with Binomial Factors
To factor a trinomial of the form ax2 + bx + c, use
the following pattern.
The goal is to find a
combination of factors of a and
b such that the outer and inner
products add up to the middle
term bx.

44. For instance, in the trinomial 6x2 + 17x + 5, you can write
all possible factorizations and determine which one has
outer and inner products that add up to 17x.
(6x + 5)(x + 1), (6x + 1)(x + 5)
(2x + 1)(3x + 5), (2x + 5)(3x + 1)
You can see that (2x + 5)(3x + 1) is the correct factorization
because outer (O) and inner (I) products add up to 17x.

45. Factoring by Grouping
Sometimes polynomials with more than three terms can be
factored by a method called factoring by grouping. It is not
always obvious which terms to group, and sometimes several
different groupings will work.

46. Exercises No. 9 (Sample)
Factor the following:
1. x2 – 7x + 12
2.2x2 + x – 15
3.x3 – 2x2 – 3x + 6x
4.2x2 + 5x – 3
5.6x3 – 2x + 3x2 – 1

47. The Binomial Theorem
Binomial is a polynomial that has two
terms. In this section, you will study a
formula that gives a quick method of
raising a binomial to a power. To begin,
look at the expansion of (x + y)n for
several values of n.

48. There are several observations you can make about these
expansions.
 In each expansion, there are n + 1 terms.
 In each expansion, x and y have symmetrical roles. The
powers of x decrease by 1 in successive terms, whereas the
powers of y increase by 1.
 The sum of the powers of each term is n. For instance, in the
expansion of (x + y)5, the sum of the powers of each term is
5.

49.  The coefficients increase and then decrease in a symmetric
pattern. The coefficients of a binomial expansion are called
binomial coefficients. To find them, you can use the Binomial
Theorem.

50. Pascal’s Triangle
There is a convenient way to remember the pattern for
coefficients. By arranging the coefficients in a triangular pattern,
you obtain the following array, which is called Pascal’s Triangle.
This triangle is named after the famous French mathematician
Blaise Pascal (1623–1662).

51. The first and last numbers in each row of Pascal’s Triangle are 1. Every
other number in each row is formed by adding the two numbers
immediately above the number. Pascal noticed that numbers in this
triangle are precisely the same numbers that are the coefficients of
binomial expansions, as follows.

52. Example:
Write the expansion of each expression.
1. (2x – 3)4
2. (x – 2y)4
Solution:
1, 4, 6, 4, 1 = is the binomial coefficient (fourth
row)
(2x – 3)4
(1)(2x)4+(4)(2x)3(-3)+(6)(2x)2(-3)2+(4)(2x)(-3)3+(1)(-3)4
= 16x4 – 96x3 + 216x2 – 216x+ 81
(x – 2y)4
(1)(x)4+(4)(x)3(-2y)+(6)(x)2(-2y)2+(4)(x)(-2y)3+(1)(-2y)4
= x4 – 8x3y + 24x2y2 – 32xy3 + 16y4

53. Sometimes you will need to find a specific term, the middle
term, sum of exponents, sum of coefficients and coefficient of
the term containing xn-mym in a binomial expansion. Instead of
writing out the entire expansion, you can use the following
formula.
The rth term of the binomial expansion
𝑛 𝑛 − 1 𝑛 − 2 . . . (𝑛 − 𝑟 + 2)𝑥 𝑛−𝑟+1 𝑦 𝑟−1
𝑟 − 1 !
or
𝑛! 𝑥 𝑛−𝑟+1 𝑦 𝑟−1
(𝑛 − 𝑟 + 1) 𝑟 − 1 !
nCr - 1 (x)n-r+1(y)r-1

54. To find the middle term of a binomial expansion:
𝑟 =
𝑛
2
+ 1
Then solve for the middle term r.
To find the sum of exponents:
𝑆 = 𝑛 𝑛 + 1 (𝑎𝑣𝑒. 𝑜𝑓 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑠)
To find sum of coefficients:
1. (𝑎𝑥 + 𝑏𝑦) 𝑛 = (𝑎 + 𝑏) 𝑛 2. (𝑎𝑥 + 𝑏) 𝑛 = (𝑎 + 𝑏) 𝑛 − (𝑏) 𝑛

55. Example:
1. Find the 8th term, the middle term, sum of exponents, sum
coefficients and coefficient of the term containing x4 in a
binomial expansion of (2x + 3)12.
Solution:
(2x + 3)12 = (ax + b)n
The rth term of the binomial expansion
𝑛! 𝑥 𝑛−𝑟+1
𝑦 𝑟−1
𝑛 − 𝑟 + 1 ! 𝑟 − 1 !
=
12! (2𝑥)12−8+1
(3)8−1
12 − 8 + 1 ! 8 − 1 !
= 55,427,328𝑥5
or
nCr-1 (x)n-r+1(y)r-1 = 12C8 - 1 (2x)12 – 8 +1(3)8 – 1
= 55, 427, 328x5

56. To find the middle term of a binomial
expansion:
𝑟 =
𝑛
2
+ 1 =
12
2
+ 1 = 7
Then solve for the middle term r.
𝑛! 𝑥 𝑛−𝑟+1 𝑦 𝑟−1
𝑛 − 𝑟 + 1 ! 𝑟 − 1 !
=
12! (2𝑥)12−7+1(3)7−1
12 − 7 + 1 ! 7 − 1 !
= 43,110,144𝑥6
Example: (Continuation)
1. Find the 8th term, the middle term, sum of exponents, sum of coefficients
and coefficient of the term containing x4 in a binomial expansion of (2x +
3)12.
or
nCr-1 (x)n-r+1(y)r-1 = 12C7-1 (2x)12 – 8 +1(3)8 – 1
= 43, 110, 144x6

57. To find the sum of exponents:
𝑆 = 𝑛 𝑛 + 1 𝑎𝑣𝑒. 𝑜𝑓 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑠
= 12 12 + 1
1 + 0
2
= 78
To find sum of coefficients:
(2𝑥 + 3)12
= (𝑎𝑥 + 𝑏) 𝑛
(𝑎𝑥 + 𝑏) 𝑛
= (2 + 3)12
− (3)12
= 243, 609, 184

58. Exercises No. 10
Using Pascal Triangle solve the following binomials. Find the 3th
term, the middle term, sum of exponents, sum of coefficients
using the given formula.
1. (2x2 + 2y3)5
4. [
6𝑚2
𝑚−2 + 12𝑛3 − (4𝑚4 +
26𝑛6
2𝑛3 )]4
Find coefficient of the term containing;
1. x8y8 in binomial (2x2 + y2)12
2. x3y3 in binomial (3x + 2)6

59. End of Chapter 2
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