Download Preparation for Chapter 4

Preparation for Chapter 4
1- Synthetic Division
If a polynomial p(x) is divided by (x – c) we can apply synthetic division:
Let’s divide x3 – 4x2 – 5 by x – 3
Step 1: Write the dividend in descending powers of x. Copy the coefficients; write a zero
for any missing power.
1 -4 0 -5
Step 2: Create the following structure:
1!!4!0!!5!Row!1
3 !!!!!!!!!!!!!!!!!!Row!2
!!!!!!!!!!!!!!!!!!!!!Row!3
First row, we place: 1 -4 0 -5
Second row, to the left of the symbol, we place 3 since the divisor is x – 3.
Step 3: Bring the 1 down two rows, and enter it in row 3.
1!!4!!0!!!5
3"
!!!1
Step 4: Multiply the 1, in row 3, by 3, and place the result 3, in row 2 underneath the -4.
1!!4!!0!!!5
3 "!!3
#3!
!! 1
Step 5: Add the entries in row 1 and row 2 (-4 + 3 = -1) and enter the sum in row 3.
1!!!!!4!!0!!!5
3 "!!!!!3
#3!
!!1! ! 1
Step 6: Multiply the -1, in row 3, by 3, and place the result -3, in row 2 underneath the 0.
1!!!!!4!!!0!!!5
3 "!!!!!3!!!3
#3! !!!#3!
!!1! !1
Step 7: Add the entries in row 1 and row 2 (0 + -3 = -3) and enter the sum in row 3.
1!!!!!4!!!!!0!!!5
3 "!!!!!3!!!!!3
#3! !!!#3!
!!1!
!1 ! 3
Step 8: Multiply the -3, in row 3, by 3, and place the result -9, in row 2 underneath the -5.
1!!!!!4!!!!!0!!!5
3 "!!!!!3!!!!!3!!!9
#3! !!!#3! !#3!
!!1! !1 !3
Step 9: add the entries in row 1 and row 2 (-5 + -9 = -14) and enter the sum in row 3.
1!!!!!4!!!!!0!!!!!5
3 "!!!!!3!!!!!3!!!!!9
#3! !!!#3! !#3!
!!1!
!1 !3 !!14
End of the process.
The numbers 1 -1 and -3 are the coefficients of a second degree polynomial that is the
quotient (x2 – x – 3), the number -14 is the remainder, so
x3 - 4x2 – 5 = (x - 3)(x2 – x – 3) + (-14)
Example:
Use synthetic division to divide (2x5 + 5x4 – 2x3 + 2x2 – 2x + 3) by (x + 3)
The divisor is x + 3 = x – (-3), so we place -3 to the left of the symbol.
Remember, we add each entry in row 1 to the corresponding entry in row 2 and the sum
is place in row 3.
2 !!!!!!!!!!5!!!!!!2!!!!!!!!!!2!! ! 2!!!!!!3!Row!1
!3 "!!!!!!!!!6!!!!!!!3!!!!!!!!!!3!!!!!! 3!!!!!!3!Row!2
# ( !3)! # ( !3)! !!!# ( !3)! # ( !3)! # ( !3)!
!!! 2!
!1
1 ! !1 ! 1 !!!0!Row!3
The remainder is 0, the quotient is the polynomial 2x4 – x3 + x2 – x + 1, so
2x5 + 5x4 – 2x3 + 2x2 – 2x + 3) = (x + 3)( 2x4 – x3 + x2 – x + 1)
2- Roots or Zeros of a Polynomial
Given a polynomial P(x), we say that a number c is a root or zero of the polynomial if
P(c) = 0.
The Factor Theorem:
The number c is a root or zero of the polynomial P(x) if and only if (x – c) is a factor of
P(x). (P(x) = (x – c) Q(x) where degree Q(x) equals degree P(x) minus 1)
Example:
Show that x – 1 is a factor of P(x) = 2x3 – x2 + 2x – 3.
Applying the theorem, we have to show that P(1) = 0.
P(1) = 2(1)3 – (1)2 + 2(1) – 3 = 2 – 1 + 2 – 3 = 0
Then P(x) = 2x3 – x2 + 2x – 3 = (x -1)Q(x).
We can find Q(x) using synthetic division,
2 !!!!!1!!!!!2!!!!!3
1 "!!!!!!2!!!!!1!!!!!!!3
#1 ! !!!#1! !!#1!
!!2!
1
3 !!!0
2
3
2
Q(x) = 2x – x + 3 and P(x) = 2x – x + 2x – 3 = (x -1)( 2x2 – x + 3).
3- Quadratic Equation
A quadratic equation has the form: ax2 + bx + c = 0 where a, b, and c are real numbers.
We can use quadratic formula to find the roots or zeros,
x=
!b ± b 2 ! 4ac
2a
b2 – 4ac is called the discriminat.
If b2 – 4ac > 0, there are two unequal real number solutions.
If b2 – 4ac = 0 there is a repeated solution, root of multiplicity 2.
If b2 – 4ac < 0 there are no real solutions, two distinct complex number solutions.
4- The Rational Zeros Theorem
Let P(x) = a n x n + a n!1x n!1 + ... + a1x + a 0 ,!a n " 0,!a 0 " 0 be a polynomial with integer
coefficients.
A rational number p/q in lowest terms, is a rational zero of P(x) if p is a factor of a0 and q
is a factor of an.
Example:
List the potential rational zeros of P(x) = 2x3 + 11x2 – 7x – 6.
Notice that the coefficients of P(x) are integer numbers.
Since a3 = 2 and a0 = -6, then p is a factor of -6 and q is a factor of 2:
p : ±1, ±2, ±3, ±6 (factors of -6)
q: ±1, ±2
(factors of 2)
Now, we form all possible ratios p/q: ±1, ±2, ±3, ±6, ±1/2, ±3/2. There are 12 possible
candidates. So if P(x) has a rational zero, it is one of those listed. It may be the case that
P(x) does not have any rational zero.
By substitution, we can check each potential zero, to determine whether it is indeed a
zero.
Let’s start with 1, P(1) = 2(1)3 + 11(1)2 – 7(1) – 6 = 0. It is a zero, then (x – 1) is a factor
of P(x).
We can use synthetic division to factor P(x)
2 !!!!!11!!!!!!7!!!!!6
1 "!!!!!!2!!!!!!13!!!!!!6
#1 ! !!!#1! !!#1!
!!2! 13 6 !!!!0
Then, P(x) = (x -1)(2x2 + 13x + 6) = (x – 1)(x + 6)(x + ½), using quadratic formula, so 1,
-6, and -1/2 are the zeros of P(x).
5- Determinants
2 by 2 Determinant
If a, b, c, and c are four real numbers, the symbol
a b
c d
is called a 2 by 2 determinant. Its value is the real number ad – bc.
Then
a b
= ad ! bc
c d
Example:
3 !2
= 3(1) ! 6(!2) = 15
6 1
3 by 3 Determinant
A 3 by 3 determinant is symbolize by
a11 a12 a13
a 21 a 22 a 23
a 31 a 32 a 33
aij indicates: it is in row i, column j.
Its value is given by
+
!
a11 a12
a 21 a 22
a 31 a 32
+
a13
a 22
a 23 = a11
a 32
a 33
a 23
a 21 a 23
a 21 a 22
! a12
+ a13
a 33
a 31 a 33
a 31 a 32
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
each 2 by 2 determinant is obtained after removing the row and column that contain a1j
Example:
Evaluate
2 !1 3
5 1
!2 1
!2 5
!2 5 1 = 2
! (!1)
+3
= 2(!45 ! 6) + (18 ! 0) + 3(!12 ! 0) =
6 !9
0 !9
0 6
0 6 !9
2(!51) + 18 ! 36 = !120
6 Complex Numbers
When we discussed set of numbers, we saw that the equation 3x = 1 did not have a
solution in the set of integers since there is no integer number that multiply to 3 is equal
to 1. For that reason the set of rational numbers was created and the number 1/3 is a
solution of the equation.
Now. consider the equation x2 = -1. Since the square of any real number is nonnegative,
then there is no real number that satisfies that equation. To remediate this situation we
will create a new set of numbers such that when we square them, they will produce
negative real numbers. This new set of numbers is called imaginary numbers.
Definition: Imaginary unit, i
The imaginary unit is denoted by the letter i and it is defined as
i = !1
where i2 = -1.
Remark:
i = !1
i2 = !1
i3 = i2 " i = (!1) " i = !i
i4 = i2 " i2 = ( !1) ( !1) = 1
Notice that the fourth power of i is 1. This is telling us that the powers of i are cyclical,
with a period equal 4.
Example: Simplify the Imaginary Unit raised to Power
(a) i7 = i4⋅ i3 = 1(-i) = -i
(b) i17 = i16⋅ i = (i4)4 ⋅i =14 ⋅i = i
(c) i110
since 110 ÷ 4 = (27)⋅4 + 2
( )
i110 = i4
27
! i2 = 127 ! i2 = 1( "1) = "1
Example: Using Imaginary Numbers to Simplify Radicals
(a)! !9 = 9(!1) = 9 !1 = 3i
(b)! !8 = 8(!1) = 8 !1 = 2 2i
Definition: Any number in the form bi, where b is any real number is an imaginary
number.
When a real number and an imaginary number are combined, the result is a complex
number.
Definition: Complex Numbers
A complex number z, in standard form is defined as
z =a + bi
where a and b are real numbers and i is the imaginary unit.
In the above definition, we call a the real part of the complex number z and b is the
imaginary part of the complex number.
Complex numbers in the form z = bi are called pure imaginary numbers.
We dente by ! , the set of all complex numbers.
Examples:
(a) 2 – 3i, 2 is the real part and the imaginary part is -3
(b) -5 + i , -5 is the real part and 1 is the imaginary part.
(c) –πi = 0 – πi, the real part is 0 and the imaginary part is –π.
(d) 7 = 7 + 0i, the real part is 7 and the imaginary part is 0.
(e) 0 = 0 + 0i, the real and the imaginary part are 0.
Remark: The set of real numbers is included in the set of complex numbers.
Definition: Equality of complex numbers
Two complex numbers z1 = a + bi and z2 = c + di are equal if and only if
a = c and b = d
Two complex numbers are equal if they have the same real part and the same imaginary
part.
We can treat complex numbers z = a + bi, similar to binomials a + bx. So we add and
subtract complex numbers as we add and subtract binomials, real parts are combined with
real parts, and imaginary parts are combined to imaginary parts.
Example: Adding and Subtracting Complex Numbers
(a) (3 - 2i) + (-1 + i) = (3 – 1) + (-2 + 1)i = 2 + (-1)i = 2 – i
(b) (2 – i) – (3 – 4i) = 2 – i – 3 + 4i = (2 – 3) + (-1 + 4)i = -1 + 3i
To multiply complex numbers we apply all the methods of multiplying binomials. It is
important to remember that i2 = -1.
Example Multiply the Complex Numbers
(a) (3 – i)(2 + i) = 3(2) + 3i – 2i – i(i) = 6 + i – (-1) = 7 + i
(b) 2i(-5 + i) = (2i)(-5) + (2i)(i) = -10i + 2(i2) = 2(-1) – 10i = -2 -10i
Definition: Complex Conjugates
Given the complex number z = a + bi, the complex number z = a – bi is called the
complex conjugate of z and vice versa.
Remark: z z = (a + bi)(a – bi) = a2 –abi + abi – b2i2 = a2 + b2
When we divide two complex numbers to get the answer in standard form, we multiply
and divide by the complex conjugate of the denominator.
Example: Dividing Complex Numbers
2!i
1 + 3i
2 ! i 1 ! 3i
"
=!!!!!!!!!!!!!!!!!!!!!!!!!!!multiply and divide!by!complex!conjugate
1 + 3i 1 ! 3i
( 2 ! i) (1 ! 3i) =!!!!!!!!!!!!!!!!!!!!!!!!mutliply!numerator!and!deno min ator
(1 + 3i) (1 ! 3i)
2 "1 + 2(!3i) ! i(1) ! i(!3i)
=!!!!!!use!FOIL
1+ 9
2 ! 6i ! i + 3i2
=!!!!!!!!!!!!!!!!!!!!!!!!simplify
10
(2 ! 3) ! 7i
=!!!!!!!!!!!!!!!!!!!!!!!!!!!!!i2 = 1
10
1
7
! ! i!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!s tan dard!notation
10 10
Quadratic Equations with a Negative Discriminant
Given the equation x2 = 9, it has two solutions x = 3 and x = -3.
Now, the equation x2 = -9 is solve
x = ± !9 = ± 9(!1) = ±3 !1 = ±3i
so, it has two solutions -3i and 3i.
Because we know to deal with the square root of a negative number, we can extend the
quadratic formula to all cases of the discriminant..
Example: Solve the Quadratic Equation
x2 + 8 = 4x
Write the equation in standard form
x2 -4x + 8 = 0
a = 1, b = -4, c = 8
Applying quadratic formula
b2 – 4ac = (-4)2 – 4(8)(1) = 16 – 32 = -16 < 0,
(
)
!b ± b 2 ! 4ac !(!4) ± !16 4 ± 16 " ( !1) 4 ± 4 !1 4 1 ± !1
=
=
=
=
= 2 ± 2i
2a
2 "1
2
2
2
The solution set is {2 – 2i, 2 + 2i}, they are complex numbers
So, if the discriminant is negative, the quadratic equation has two unequal solutions that
are conjugate complex numbers.
x=
Character of the Solutions of a Quadratic Equation
In the complex number system consider the quadratic equation with real coefficients
ax2 + bx + c = 0
(1) If b2 – 4ac > 0, the equation has two unequal real solutions
(2) If b2 – 4qc = 0 the equation has a repeated real solution, a double root
(3) If b2 – 4ac < 0, the equation has two complex solutions, they are not real number.
They are the conjugate of each other.
Polar Form of a Complex Number
If P is a point in the complex plane corresponding to the complex number z = a + bi that
can be viewed as the ordered-pair (a, b), then a = r cos ø and b = r sin ø, where r =
a 2 + b 2 is called the modulus or absolute value of z, |z| and ø is called the argument of
z. It follows that z = a + bi = r(cos ø + i sin ø). This is the polar form of z.
The polar form of z is not unique, it is up to a multiple of 2πk, k an integer
z = r(cos(ø + 2πk) + i sin(ø + 2πk) , k any integer
De Moivre’s Theorem
z n = [r(cos ø + i sin ø)]n = r n (cos nø + i sin nø)
"
" ø + 2!k %
" ø + 2!k % %
z = r(cos ø + isin ø) = n r $ cos $
+ i!sin $
'
' ,!k = 0,1, 2,...,(n ( 1)
#
#
#
n &
n & '&
Notice, there are n different n-th roots of the complex number z.
n
Euler’s Formula
By assuming that the infinite series expansion
xn
x2 x3
ex = "
= 1+ x +
+
+!
n!
2!
3!
n!0
that we learned in calculus, holds when x = iø, we can get
( iø)n = (!1)n ø2n + i (!1)n ø2n+1 = cos ø + i!sin ø
e iø = #
#
#
n"0 n!
n"0 ( 2n )!
n"0 ( 2n + 1)!
this is called the Euler’s formula
Therefore, e z = e a+ib = e a e ib = e a (cos b + i!sin b)