6
Chapter
Complex numbers
and polynomials
Syllabus reference: 1.5, 1.8, 2.5, 2.6
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Real quadratics with ¢ < 0
Complex numbers
Real polynomials
Zeros, roots, and factors
Polynomial theorems
Graphing real polynomials
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Contents:
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174
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
A
REAL QUADRATICS WITH ¢ < 0
In Chapter 1, we determined that:
If ax2 + bx + c = 0, a 6= 0 and a, b, c 2 R , then the solutions or roots are found using the
p
¡b § ¢
where ¢ = b2 ¡ 4ac is known as the discriminant.
formula x =
2a
² ¢ > 0 we have two real distinct solutions
² ¢ = 0 we have two real identical solutions
² ¢ < 0 we have no real solutions.
We also observed that if:
However, it is in fact possible to write down two solutions for the case where ¢ < 0. To do this we
need imaginary numbers.
p
In 1572, Rafael Bombelli defined the imaginary number i = ¡1. It is called ‘imaginary’ because we
cannot place it on a number line. With i defined, we can write down solutions for quadratic equations
with ¢ < 0. They are called complex solutions because they include a real and an imaginary part.
Any number of the form a + bi where a and b are real and i =
is called a complex number.
Self Tutor
Example 1
2
a x = ¡4
Solve the quadratic equations:
a
x2 = ¡4
p
) x = § ¡4
p p
) x = § 4 ¡1
) x = §2i
2
b z +z+2= 0
p
¡1,
If the coefficient of i
is a square root, we
write the i first.
b z 2 + z + 2 has a = 1, b = 1, c = 2
p
¡1 § 12 ¡ 4(1)(2)
) z=
2(1)
p
¡1 § ¡7
) z=
2
p
p
¡1 § i 7
) z=
= ¡ 12 § 27 i
2
In Example 1 above, notice that ¢ < 0 in both cases. In each case we have found two complex
solutions of the form a + bi, where a and b are real.
Self Tutor
Example 2
Write as a product of linear factors:
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b x2 + 11
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a x2 + 4
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
175
Self Tutor
Example 3
Solve for x:
a x2 + 9 = 0
b x3 + 2x = 0
Self Tutor
Example 4
Solve for x:
a x2 ¡ 4x + 13 = 0
b x4 + x2 = 6
EXERCISE 6A
1 Write in terms of i:
p
p
a
¡9
b
¡64
c
q
¡ 14
d
p
¡5
e
p
¡8
2 Write as a product of linear factors:
a x2 ¡ 9
e 4x2 ¡ 1
b x2 + 9
f 4x2 + 1
c x2 ¡ 7
g 2x2 ¡ 9
d x2 + 7
h 2x2 + 9
i x3 ¡ x
j x3 + x
k x4 ¡ 1
l x4 ¡ 16
3 Solve for x:
a x2 ¡ 25 = 0
e 4x2 ¡ 9 = 0
b x2 + 25 = 0
f 4x2 + 9 = 0
c x2 ¡ 5 = 0
g x3 ¡ 4x = 0
i x3 ¡ 3x = 0
j x3 + 3x = 0
k x4 ¡ 1 = 0
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c x4 + 5x2 = 36
f x4 + 2x2 + 1 = 0
25
b x4 = x2 + 6
e x4 + 1 = 2x2
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5 Solve for x:
a x4 + 2x2 = 3
d x4 + 9x2 + 14 = 0
c x2 + 14x + 50 = 0
5
d 2x2 + 5 = 6x
l x4 = 81
b x2 + 6x + 25 = 0
p
e x2 ¡ 2 3x + 4 = 0
100
4 Solve for x:
a x2 ¡ 10x + 29 = 0
d x2 + 5 = 0
h x3 + 4x = 0
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=1
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176
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
B
COMPLEX NUMBERS
Any number of the form a + bi where a, b 2 R and i =
p
¡1, is called a complex number.
Notice that all real numbers are complex numbers in the special case where b = 0.
A complex number of the form bi where b 2 R , b 6= 0, is called an imaginary number or
purely imaginary.
THE ‘SUM OF TWO SQUARES’
a2 + b2 = a2 ¡ b2 i2
fas i2 = ¡1g
= (a + bi)(a ¡ bi)
Notice that
a2 ¡ b2 = (a + b)(a ¡ b)
a2 + b2 = (a + bi)(a ¡ bi)
Compare:
fthe difference of two squares factorisationg
fthe sum of two squares factorisationg
If we write z = a + bi where a, b 2 R , then:
For example:
² a is the real part of z
and we write a = Re (z)
² b is the imaginary part of z
and we write b = Im (z).
then Re (z) = 2 and Im (z) = 3.
p
then Re (z) = 0 and Im (z) = ¡ 2.
If z = 2 + 3i,
p
If z = ¡ 2i,
OPERATIONS WITH COMPLEX NUMBERS
Operations with complex numbers are identical to those with radicals, but with i2 = ¡1 rather than
p
p
( 2)2 = 2 or ( 3)2 = 3.
For example:
² addition:
² multiplication:
p
p
p
p
(2 + 3) + (4 + 2 3) = (2 + 4) + (1 + 2) 3 = 6 + 3 3
(2 + i) + (4 + 2i) = (2 + 4) + (1 + 2)i = 6 + 3i
p
p
p
p
p
p
(2 + 3)(4 + 2 3) = 8 + 4 3 + 4 3 + 2( 3)2 = 8 + 8 3 + 6
(2 + i)(4 + 2i) = 8 + 4i + 4i + 2i2 = 8 + 8i ¡ 2
So, we can add, subtract, multiply, and divide complex numbers in the same way we perform these
operations with radicals:
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi) ¡ (c + di) = (a ¡ c) + (b ¡ d)i
(a + bi)(c + di) = ac + adi + bci + bdi2
µ
¶µ
¶
ac ¡ adi + bci ¡ bdi2
a + bi
c ¡ di
a + bi
=
=
c + di
c + di
c ¡ di
c2 + d2
addition
subtraction
multiplication
division
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Notice that in the division process, we use a multiplication technique to obtain a real number in the
denominator.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
177
Self Tutor
Example 5
If z = 3 + 2i and w = 4 ¡ i find:
b z¡w
a z+w
c zw
d
z
w
You can use your calculator to perform operations with complex numbers.
After solving the questions in the following exercise by hand, check your
answers using technology.
GRAPHICS
CALCUL ATOR
INSTRUCTIONS
EXERCISE 6B.1
1 Copy and complete:
Re (z)
z
Im (z)
z
3 + 2i
¡3 + 4i
5¡i
¡7 ¡ 2i
3
¡11i
p
i 3
0
Re (z)
Im (z)
2 If z = 5 ¡ 2i and w = 2 + i, find in simplest form:
a z+w
e 2z ¡ 3w
b 2z
f zw
c iw
g w2
d z¡w
h z2
3 For z = 1 + i and w = ¡2 + 3i, find in simplest form:
b z2
f zw
a z + 2w
e w2
c z3
g z2w
d iz
h izw
a Simplify in for n = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and also for n = ¡1, ¡2, ¡3, ¡4, ¡5.
4
b Hence, simplify i4n+3 where n is any integer.
5 Write (1 + i)4 in simplest form. Hence, find (1 + i)101 in simplest form.
6 Suppose z = 2 ¡ i and w = 1 + 3i. Write in exact form a + bi where a, b 2 R :
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178
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
7 Simplify:
i
1 ¡ 2i
a
i(2 ¡ i)
3 ¡ 2i
b
1
2
¡
2¡i
2+i
c
8 If z = 2 + i and w = ¡1 + 2i, find:
a Im (4z ¡ 3w)
c Im (iz 2 )
b Re (zw)
d Re
³ ´
z
w
EQUALITY OF COMPLEX NUMBERS
Two complex numbers are equal when their real parts are equal and their imaginary parts are equal.
a + bi = c + di , a = c and b = d.
Suppose b 6= d. Now if a + bi = c + di where a, b, c, and d are real,
then bi ¡ di = c ¡ a
) i(b ¡ d) = c ¡ a
Proof:
) i=
c¡a
b¡d
fas b 6= dg
This is false as the RHS is real but the LHS is imaginary.
Thus, the supposition is false. Hence b = d and furthermore a = c.
For the complex number a + bi, where a and b are real, a + bi = 0 , a = 0 and b = 0.
Self Tutor
Example 6
Find real numbers x and y such that:
a (x + yi)(2 ¡ i) = ¡i
b (x + 2i)(1 ¡ i) = 5 + yi
EXERCISE 6B.2
1 Find exact real numbers x and y such that:
a 2x + 3yi = ¡x ¡ 6i
b x2 + xi = 4 ¡ 2i
c (x + yi)(2 ¡ i) = 8 + i
d (3 + 2i)(x + yi) = ¡i
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d (x + yi)(2 + i) = 2x ¡ (y + 1)i
0
c (x + i)(3 ¡ iy) = 1 + 13i
5
b (x + 2i)(y ¡ i) = ¡4 ¡ 7i
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a 2(x + yi) = x ¡ yi
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2 Find exact x, y 2 R such that:
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
179
3 The complex number z satisfies the equation 3z + 17i = iz + 11. Write z in the form a + bi
p
where a, b 2 R and i = ¡1.
³
´2
4
4 Express z in the form a + bi where a, b 2 Z , if z =
+ 7 ¡ 2i .
1+i
5 Find the real values of m and n for which 3(m + ni) = n ¡ 2mi ¡ (1 ¡ 2i).
6 Express z = p
3i
2¡i
+ 1 in the form a + bi where a, b 2 R are given exactly.
7 Suppose (a + bi)2 = ¡16 ¡ 30i where a, b 2 R and a > 0.
Find the possible values of a and b.
HISTORICAL NOTE
18th century mathematicians enjoyed playing with these new ‘imaginary’
numbers, but they were regarded as little more than interesting curiosities
until the work of Gauss (1777 - 1855), the German mathematician,
astronomer, and physicist.
For centuries mathematicians had attempted to find a method of trisecting
an angle using a compass and straight edge. Gauss put an end to this
when he used complex numbers to prove the impossibility of such a
construction. By his systematic use of complex numbers, he was able to
convince mathematicians of their usefulness.
Carl Friedrich Gauss
Early last century, the American engineer Steinmetz used complex numbers to solve electrical
problems, illustrating that complex numbers did have a practical application.
Complex numbers are now used extensively in electronics, engineering, and physics.
COMPLEX CONJUGATES
Complex numbers a + bi and a ¡ bi are called complex conjugates.
If z = a + bi we write its conjugate as z ¤ = a ¡ bi.
We saw on page 176 that the complex conjugate is important for division:
z
z w¤
zw¤
=
=
¤
w
w w
ww¤
which makes the denominator real.
Quadratics with real coefficients are called real quadratics. This does not necessarily
mean that their zeros are real.
² If a quadratic equation has rational coefficients and an irrational root of the form
p
p
c + d n, then the conjugate c ¡ d n is also a root of the quadratic equation.
² If a real quadratic equation has ¢ < 0 and c + di is a complex root, then the
complex conjugate c ¡ di is also a root.
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¢ = (¡2)2 ¡ 4(1)(5) = ¡16
the solutions are x = 1 + 2i and 1 ¡ 2i
¢ = 02 ¡ 4(1)(4) = ¡16
the solutions are x = 2i and ¡2i
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² x2 + 4 = 0
has
and
has
and
5
² x2 ¡ 2x + 5 = 0
For example:
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180
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
If c + di and c ¡ di are roots of a quadratic equation, then the quadratic equation
is a(x2 ¡ 2cx + (c2 + d2 )) = 0 for some constant a 6= 0.
Theorem:
The sum of the roots = 2c and the product = (c + di)(c ¡ di) = c2 + d2
Proof:
) x2 ¡ (sum)x + (product) = 0
) x2 ¡ 2cx + (c2 + d2 ) = 0
In general, a(x2 ¡ 2cx + c2 + d2 ) = 0 for some constant a 6= 0.
² the sum of complex conjugates c + di and c ¡ di is 2c which is real
Notice that:
² the product is (c + di)(c ¡ di) = c2 + d2
which is also real.
Self Tutor
Example 7
Find all real quadratic equations having 1 ¡ 2i as a root.
Example 8
p
Find exact values for a and b if 2 + i is a root of x2 + ax + b = 0, a, b 2 R .
Self Tutor
EXERCISE 6B.3
1 Find all quadratic equations with real coefficients and roots of:
a 3§i
p
e 2§ 3
b 1 § 3i
c ¡2 § 5i
p
g §i 2
f 0 and ¡ 23
d
p
2§i
h ¡6 § i
2 Find exact values for a and b if:
a 3 + i is a root of x2 + ax + b = 0, where a and b are real
p
b 1 ¡ 2 is a root of x2 + ax + b = 0, where a and b are rational
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c a + ai is a root of x2 + 4x + b = 0, where a and b are real. [Careful!]
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
181
PROPERTIES OF CONJUGATES
INVESTIGATION 1
PROPERTIES OF CONJUGATES
The purpose of this investigation is to discover any properties that complex conjugates might have.
What to do:
1 Given z1 = 1 ¡ i and z2 = 2 + i find:
b z2¤
a z1¤
c (z1¤ )¤
d (z2¤ )¤
h z1¤ ¡ z2¤
p (z2¤ )3
e (z1 + z2 )¤
f z1¤ + z2¤
i (z1 z2 )¤
j z1¤ z2¤
g (z1 ¡ z2 )¤
³ ´¤
z1
k
n (z1¤ )2
o (z23 )¤
m (z12 )¤
z2
l
z1¤
z2¤
2 Repeat 1 with z1 and z2 of your choice.
3 Examine your results from 1 and 2, and hence suggest some rules for complex conjugates.
From the Investigation you should have discovered the following rules for complex conjugates:
² (z ¤ )¤ = z
² (z1 + z2 )¤ = z1¤ + z2¤
and (z1 ¡ z2 )¤ = z1¤ ¡ z2¤
µ ¶¤
z1
z¤
and
= 1¤ , z2 6= 0
z2
z2
² (z1 z2 )¤ = z1¤ £ z2¤
² (z n )¤ = (z ¤ )n
² z
+ z¤
for positive integers n
and zz ¤
are real.
Self Tutor
Example 9
Show that for all complex numbers z1 and z2 :
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a (z1 + z2 )¤ = z1¤ + z2¤
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182
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
CONJUGATE GENERALISATIONS
Notice that (z1 + z2 + z3 )¤ = (z1 + z2 )¤ + z3¤
= z1¤ + z2¤ + z3¤
ftreating z1 + z2 as one complex numberg
.... (1)
Likewise (z1 + z2 + z3 + z4 )¤ = (z1 + z2 + z3 )¤ + z4¤
= z1¤ + z2¤ + z3¤ + z4¤
ffrom (1)g
There is no reason why this process cannot continue for the conjugate of 5, 6, 7, .... complex numbers.
We can therefore generalise the result:
(z1 + z2 + z3 + :::: + zn )¤ = z1¤ + z2¤ + z3¤ + :::: + zn¤
The process of proving the general case using the simpler cases when n = 1, 2, 3, 4, .... requires
mathematical induction. Formal proof by the Principle of Mathematical Induction is discussed in
Chapter 9.
EXERCISE 6B.4
1 Show that (z1 ¡ z2 )¤ = z1¤ ¡ z2¤
for all complex numbers z1 and z2 .
2 Simplify the expression (w¤ ¡ z)¤ ¡ (w ¡ 2z ¤ ) using the properties of conjugates.
3 It is known that a complex number z satisfies the equation z ¤ = ¡z.
Show that z is either purely imaginary or zero.
4 Suppose z1 = a + bi and z2 = c + di are complex numbers.
³ ´¤ z ¤
z
z1
a Find 1 in the form X + iY .
b Show that
= 1¤
z2
z2
³
5
a An easier way of proving
´
z¤
z1 ¤
= 1¤
z2
z2
z2
³
is to start with
for all z1 and z2 6= 0.
´
z1 ¤
£ z2¤ .
z2
Show how this can be done, remembering we have already proved that “the conjugate of a
product is the product of the conjugates” in Example 9.
b Let z = a + bi be a complex number. Prove the following:
i If z = z ¤ , then z is real.
ii If z ¤ = ¡z, then z is purely imaginary or zero.
6 Prove that for all complex numbers z and w:
a zw¤ + z ¤ w
b zw¤ ¡ z ¤ w
is always real
is purely imaginary or zero.
a If z = a + bi, find z 2 in the form X + iY .
7
b Hence, show that (z 2 )¤ = (z ¤ )2
for all complex numbers z.
3
c Repeat a and b but for z instead of z 2 .
8 Suppose w =
z¡1
z¤ + 1
where z = a + bi. Find the conditions under which:
a w is real
b w is purely imaginary.
a Assuming (z1 z2 )¤ = z1¤ z2¤ , explain why (z1 z2 z3 )¤ = z1¤ z2¤ z3¤ .
9
b Hence show that (z1 z2 z3 z4 )¤ = z1¤ z2¤ z3¤ z4¤ .
c Generalise your results from a and b.
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d Given your generalisation in c, what is the result of letting all zi values be equal to z?
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
C
183
REAL POLYNOMIALS
Up to this point we have studied linear and quadratic functions at some depth, with perhaps occasional
reference to cubic functions. These are part of a larger family of functions called the polynomials.
A polynomial function is a function of the form
P (x) = an xn + an¡1 xn¡1 + :::: + a2 x2 + a1 x + a0 , a1 , ...., an constant, an 6= 0.
We say that:
x is the variable
a0 is the constant term
an is the leading coefficient and is non-zero
ar is the coefficient of xr for r = 0, 1, 2, ...., n
n is the degree of the polynomial, being the highest power of the variable.
In summation notation, we write P (x) =
n
P
ar xr ,
r=0
which reads: “the sum from r = 0 to n, of ar xr ”.
A real polynomial P (x) is a polynomial for which ar 2 R , r = 0, 1, 2, ...., n.
The low degree members of the polynomial family have special names, some of which you are already
familiar with. For these polynomials, we commonly write their coefficients as a, b, c, ....
Polynomial function
Degree
Name
ax + b, a 6= 0
1
linear
ax + bx + c, a 6= 0
2
quadratic
ax3 + bx2 + cx + d, a 6= 0
3
cubic
4
quartic
2
4
3
2
ax + bx + cx + dx + e, a 6= 0
ADDITION AND SUBTRACTION
To add or subtract two polynomials, we collect ‘like’ terms.
Self Tutor
Example 10
If P (x) = x3 ¡ 2x2 + 3x ¡ 5 and Q(x) = 2x3 + x2 ¡ 11, find:
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b P (x) ¡ Q(x)
a P (x) + Q(x)
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184
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
SCALAR MULTIPLICATION
To multiply a polynomial by a scalar (constant) we multiply each term by the scalar.
Self Tutor
Example 11
If P (x) = x4 ¡ 2x3 + 4x + 7 find:
a 3P (x)
b ¡2P (x)
POLYNOMIAL MULTIPLICATION
To multiply two polynomials, we multiply each term of the first polynomial by each
term of the second polynomial, and then collect like terms.
Self Tutor
Example 12
If P (x) = x3 ¡ 2x + 4 and Q(x) = 2x2 + 3x ¡ 5, find P (x)Q(x).
SYNTHETIC MULTIPLICATION (OPTIONAL)
Polynomial multiplication can be performed using the coefficients only.
multiplication.
We call this synthetic
coefficients of x3 + 2x ¡ 5
coefficients of 2x + 3
x2
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(x3 + 2x ¡ 5)(2x + 3)
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For example, for (x3 + 2x ¡ 5)(2x + 3) we detach coefficients and multiply. It is different from the
ordinary multiplication of large numbers because we sometimes have negative coefficients, and because
we do not carry tens into the next column.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
185
EXERCISE 6C.1
1 If P (x) = x2 + 2x + 3 and Q(x) = 4x2 + 5x + 6, find in simplest form:
a 3P (x)
c P (x) ¡ 2Q(x)
b P (x) + Q(x)
d P (x)Q(x)
2 If f(x) = x2 ¡ x + 2 and g(x) = x3 ¡ 3x + 5, find in simplest form:
a f(x) + g(x)
b g(x) ¡ f (x)
c 2f(x) + 3g(x)
d g(x) + xf(x)
e f (x) g(x)
f [f (x)]2
a (x2 ¡ 2x + 3)(2x + 1)
b (x ¡ 1)2 (x2 + 3x ¡ 2)
c (x + 2)3
d (2x2 ¡ x + 3)2
e (2x ¡ 1)4
f (3x ¡ 2)2 (2x + 1)(x ¡ 4)
3 Expand and simplify:
4 Find the following products:
a (2x2 ¡ 3x + 5)(3x ¡ 1)
b (4x2 ¡ x + 2)(2x + 5)
c (2x2 + 3x + 2)(5 ¡ x)
d (x ¡ 2)2 (2x + 1)
e (x2 ¡ 3x + 2)(2x2 + 4x ¡ 1)
f (3x2 ¡ x + 2)(5x2 + 2x ¡ 3)
g (x2 ¡ x + 3)2
h (2x2 + x ¡ 4)2
i (2x + 5)3
j (x3 + x2 ¡ 2)2
DIVISION OF POLYNOMIALS
The division of polynomials is only useful if we divide a polynomial of degree n by another of degree n
or less.
DIVISION BY LINEARS
Consider (2x2 + 3x + 4)(x + 2) + 7.
If we expand this expression we obtain (2x2 + 3x + 4)(x + 2) + 7 = 2x3 + 7x2 + 10x + 15.
Dividing both sides by (x + 2), we obtain
2x3 + 7x2 + 10x + 15
(2x2 + 3x + 4)(x + 2) + 7
=
x+2
x+2
(2x2 + 3x + 4)(x + 2)
7
+
x+2
x+2
7
= 2x2 + 3x + 4 +
where
x+2
=
x + 2 is the divisor,
2
2x + 3x + 4 is the quotient,
and 7 is the remainder.
If P (x) is divided by ax + b until a constant remainder R is obtained, then
P (x)
R
= Q(x) +
ax + b
ax + b
where
ax + b is the divisor, D(x),
Q(x) is the quotient,
and R is the remainder.
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Notice that P (x) = Q(x) £ (ax + b) + R.
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186
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
DIVISION ALGORITHM
We can divide a polynomial by another polynomial using an algorithm similar to that used for division
of whole numbers:
What do we multiply x by to get 2x3 ?
The answer is 2x2 ,
and 2x2 (x + 2) = 2x3 + 4x2 .
| {z }
Step 1:
2x2 + 3x + 4
x+2
Step 2:
Subtract 2x3 + 4x2 from 2x3 + 7x2 .
The answer is 3x2 .
Step 3:
Bring down the 10x to obtain 3x2 + 10x.
2x3 + 7x2 + 10x + 15
¡ (2x3 + 4x2 )
3x2 + 10x
¡ (3x2 + 6x)
4x + 15
¡ (4x + 8)
Return to Step 1 with the question:
“What must we multiply x by to get 3x2 ?”
7
The answer is 3x, and 3x(x + 2) = 3x2 + 6x ....
We continue the process until we are left with a constant.
2
The division algorithm can also be performed by leaving
out the variable, as shown alongside.
Either way,
2x3
1 2
+ 7x2
+ 10x + 15
7
= 2x2 + 3x + 4 +
x+2
x+2
2
¡ (2
3
4
7 10 15
4)
3 10
¡ (3 6)
4 15
¡ (4 8)
7
Self Tutor
Example 13
Find the quotient and remainder for
x3
¡ x2
¡ 3x ¡ 5
.
x¡3
Hence write x3 ¡ x2 ¡ 3x ¡ 5 in the form Q(x) £ (x ¡ 3) + R.
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Check your answer by
expanding the RHS.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
187
Self Tutor
Example 14
x4 + 2x2 ¡ 1
.
x+3
Perform the division
Hence write x4 + 2x2 ¡ 1 in the form Q(x) £ (x + 3) + R.
Notice the insertion
of 0x 3 and 0x.
EXERCISE 6C.2
1 Find the quotient and remainder for the following, and hence write the division in the form
P (x) = Q(x) D(x) + R, where D(x) is the divisor.
a
x2 + 2x ¡ 3
x+2
b
x2 ¡ 5x + 1
x¡1
c
2x3 + 6x2 ¡ 4x + 3
x¡2
2 Perform the following divisions, and hence write the division in the form
P (x) = Q(x) D(x) + R.
a
x2 ¡ 3x + 6
x¡4
b
x2 + 4x ¡ 11
x+3
c
2x2 ¡ 7x + 2
x¡2
d
2x3 + 3x2 ¡ 3x ¡ 2
2x + 1
e
3x3 + 11x2 + 8x + 7
3x ¡ 1
f
2x4 ¡ x3 ¡ x2 + 7x + 4
2x + 3
3 Perform the divisions:
a
x2 + 5
x¡2
b
2x2 + 3x
x+1
c
3x2 + 2x ¡ 5
x+2
d
x3 + 2x2 ¡ 5x + 2
x¡1
e
2x3 ¡ x
x+4
f
x3 + x2 ¡ 5
x¡2
SYNTHETIC DIVISION (OPTIONAL)
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Click on the icon for an exercise involving a synthetic division process for the division
of a polynomial by a linear.
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SECTION
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188
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
DIVISION BY QUADRATICS
As with division by linears we can use the division algorithm to divide polynomials by quadratics. The
division process stops when the remainder has degree less than that of the divisor, so
If P (x) is divided by ax2 + bx + c then
P (x)
ex + f
= Q(x) + 2
ax2 + bx + c
ax + bx + c
ax2 + bx + c is the divisor,
where
Q(x) is the quotient,
ex + f
is the remainder.
ax2 + bx + c
and
The remainder will be linear if e 6= 0, and constant if e = 0.
Self Tutor
Example 15
x4 + 4x3 ¡ x + 1
.
x2 ¡ x + 1
Find the quotient and remainder for
Hence write x4 + 4x3 ¡ x + 1 in the form Q(x) £ (x2 ¡ x + 1) + R(x).
EXERCISE 6C.3
1 Find the quotient and remainder for:
a
x3 + 2x2 + x ¡ 3
x2 + x + 1
b
3x2 ¡ x
x2 ¡ 1
3x3 + x ¡ 1
x2 + 1
c
d
x¡4
x2 + 2x ¡ 1
2 Carry out the following divisions and also write each in the form P (x) = Q(x) D(x) + R(x):
a
x2 ¡ x + 1
x2 + x + 1
b
x3
x2 + 2
c
x4 + 3x2 + x ¡ 1
x2 ¡ x + 1
d
2x3 ¡ x + 6
(x ¡ 1)2
e
x4
(x + 1)2
f
x4 ¡ 2x3 + x + 5
(x ¡ 1)(x + 2)
3 Suppose P (x) = (x ¡ 2)(x2 + 2x + 3) + 7. Find the quotient and remainder when P (x) is
divided by x ¡ 2.
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4 Suppose f(x) = (x ¡ 1)(x + 2)(x2 ¡ 3x + 5) + 15 ¡ 10x. Find the quotient and remainder when
f(x) is divided by x2 + x ¡ 2.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
D
189
ZEROS, ROOTS, AND FACTORS
A zero of a polynomial is a value of the variable which makes the polynomial equal to zero.
® is a zero of polynomial P (x) , P (®) = 0.
The roots of a polynomial equation are the solutions to the equation.
® is a root (or solution) of P (x) = 0 , P (®) = 0.
The roots of P (x) = 0 are the zeros of P (x) and the x-intercepts of the graph of y = P (x).
P (x) = x3 + 2x2 ¡ 3x ¡ 10
P (2) = 23 + 2(2)2 ¡ 3(2) ¡ 10
= 8 + 8 ¡ 6 ¡ 10
=0
Consider
)
An equation has roots.
A polynomial has zeros.
² 2 is a zero of x3 + 2x2 ¡ 3x ¡ 10
² 2 is a root of x3 + 2x2 ¡ 3x ¡ 10 = 0
² the graph of y = x3 + 2x2 ¡ 3x ¡ 10 has the x-intercept 2.
This tells us:
If P (x) = (x + 1)(2x ¡ 1)(x + 2), then (x + 1), (2x ¡ 1), and (x + 2) are its linear factors.
Likewise P (x) = (x + 3)2 (2x + 3) has been factorised into 3 linear factors, one of which is repeated.
(x ¡ ®) is a factor of the polynomial P (x) , there exists a polynomial Q(x)
such that P (x) = (x ¡ ®)Q(x).
Self Tutor
Example 16
a x2 ¡ 4x + 53
Find the zeros of:
b z 3 + 3z
EXERCISE 6D.1
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f z 4 = 3z 2 + 10
100
e z 3 + 5z = 0
d x3 = 5x
50
c ¡2z(z 2 ¡ 2z + 2) = 0
75
b (2x + 1)(x2 + 3) = 0
25
2 Find the roots of:
a 5x2 = 3x + 2
0
c z 2 ¡ 6z + 6
f z 4 + 4z 2 ¡ 5
5
b x2 + 6x + 10
e z 3 + 2z
100
1 Find the zeros of:
a 2x2 ¡ 5x ¡ 12
d x3 ¡ 4x
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190
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
Self Tutor
Example 17
a 2x3 + 5x2 ¡ 3x
Factorise:
3 Find the linear factors of:
a 2x2 ¡ 7x ¡ 15
d 6z 3 ¡ z 2 ¡ 2z
b z 2 + 4z + 9
b z 2 ¡ 6z + 16
e z 4 ¡ 6z 2 + 5
c x3 + 2x2 ¡ 4x
f z4 ¡ z2 ¡ 2
4 If P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °) then ®, ¯, and ° are its zeros.
Verify this statement by finding P (®), P (¯), and P (°).
Self Tutor
Example 18
Find all cubic polynomials with zeros
1
2
and ¡3 § 2i.
Example 19
Find all quartic polynomials with zeros 2,
¡ 13 ,
p
and ¡1 § 5.
Self Tutor
5 Find all cubic polynomials with zeros:
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6 Find all quartic polynomials with zeros of:
p
p
a §1, § 2
b 2, ¡1, §i 3
c 3, ¡1 § i
75
b ¡2, §i
25
a §2, 3
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
191
POLYNOMIAL EQUALITY
Two polynomials are equal if and only if they have the same degree (order) and corresponding
terms have equal coefficients.
If we know that two polynomials are equal then we can equate coefficients to find unknown coefficients.
For example, if 2x3 + 3x2 ¡ 4x + 6 = ax3 + bx2 + cx + d, where a, b, c, d 2 R , then
a = 2, b = 3, c = ¡4, and d = 6.
Self Tutor
Example 20
Find constants a, b, and c given that:
6x3 + 7x2 ¡ 19x + 7 = (2x ¡ 1)(ax2 + bx + c) for all x.
Self Tutor
Example 21
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Find constants a and b if z 4 + 9 = (z 2 + az + 3)(z 2 + bz + 3) for all z.
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192
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
Self Tutor
Example 22
(x + 3) is a factor of P (x) = x3 + ax2 ¡ 7x + 6. Find a 2 R and the other factors.
Self Tutor
Example 23
(2x + 3) and (x ¡ 1) are factors of 2x4 + ax3 ¡ 3x2 + bx + 3.
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Find constants a and b and all zeros of the polynomial.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
193
EXERCISE 6D.2
1 Find constants a, b, and c given that:
a 2x2 + 4x + 5 = ax2 + [2b ¡ 6]x + c for all x
b 2x3 ¡ x2 + 6 = (x ¡ 1)2 (2x + a) + bx + c for all x.
2 Find constants a and b if:
a z 4 + 4 = (z 2 + az + 2)(z 2 + bz + 2) for all z
b 2z 4 + 5z 3 + 4z 2 + 7z + 6 = (z 2 + az + 2)(2z 2 + bz + 3) for all z.
3 Show that z 4 + 64 can be factorised into two real quadratic factors of the form z 2 + az + 8 and
z 2 + bz + 8, but cannot be factorised into two real quadratic factors of the form z 2 + az + 16
and z 2 + bz + 4.
4 Find real numbers a and b such that x4 ¡ 4x2 + 8x ¡ 4 = (x2 + ax + 2)(x2 + bx ¡ 2).
Hence solve the equation x4 + 8x = 4x2 + 4.
a (2z ¡ 3) is a factor of 2z 3 ¡ z 2 + az ¡ 3. Find a 2 R and all zeros of the cubic.
5
b (3z + 2) is a factor of 3z 3 ¡ z 2 + (a + 1)z + a. Find a 2 R and all the zeros of the cubic.
a (2x + 1) and (x ¡ 2) are factors of P (x) = 2x4 + ax3 + bx2 ¡ 12x ¡ 8.
Find constants a and b, and all zeros of P (x).
6
b (x + 3) and (2x ¡ 1) are factors of 2x4 + ax3 + bx2 + ax + 3.
Find constants a and b, and hence determine all zeros of the quartic.
a x3 + 3x2 ¡ 9x + c, c 2 R , has two identical linear factors. Prove that c is either 5 or ¡27,
and factorise the cubic into linear factors in each case.
b 3x3 + 4x2 ¡ x + m, m 2 R , has two identical linear factors. Find the possible values of m,
and find the zeros of the polynomial in each case.
7
E
POLYNOMIAL THEOREMS
There are many theorems about polynomials, some of which we look at now. Some of the theorems are
true for all polynomials, while others are true only for real polynomials.
THE REMAINDER THEOREM
Consider the cubic polynomial P (x) = x3 + 5x2 ¡ 11x + 3.
If we divide P (x) by x ¡ 2, we find that
x3 + 5x2 ¡ 11x + 3
9
= x2 + 7x + 3 +
x¡2
x¡2
A real polynomial is a
polynomial with real
coefficients.
remainder
So, when P (x) is divided by x ¡ 2, the remainder is 9.
Notice that P (2) = 8 + 20 ¡ 22 + 3
= 9, which is the remainder.
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By considering other examples like the one above, we formulate
the Remainder theorem.
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194
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
The Remainder Theorem
When a polynomial P (x) is divided by x ¡ k until a constant remainder R
is obtained, then R = P (k).
P (x) = Q(x)(x ¡ k) + R
P (k) = Q(k) £ 0 + R
P (k) = R
By the division algorithm,
Letting x = k,
)
Proof:
When using the Remainder theorem, it is important to realise that the following statements are equivalent:
² P (x) = (x ¡ k)Q(x) + R
² P (k) = R
² P (x) divided by x ¡ k leaves a remainder of R.
Self Tutor
Example 24
Use the Remainder theorem to find the remainder when x4 ¡ 3x3 + x ¡ 4 is divided by x + 2.
Self Tutor
Example 25
2
2
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When P (x) is divided by x ¡ 3x + 7, the quotient is x + x ¡ 1 and the remainder R(x)
is unknown.
When P (x) is divided by x ¡ 2 the remainder is 29. When P (x) is divided by x + 1 the
remainder is ¡16.
Find R(x) in the form ax + b.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
195
EXERCISE 6E.1
1 For P (x) a real polynomial, write two equivalent statements for each of:
a If P (2) = 7, then ....
b If P (x) = Q(x)(x + 3) ¡ 8, then ....
c If P (x) divided by x ¡ 5 has a remainder of 11 then ....
2 Without performing division, find the remainder when:
a x3 + 2x2 ¡ 7x + 5 is divided by x ¡ 1
b x4 ¡ 2x2 + 3x ¡ 1 is divided by x + 2.
3 Find a 2 R given that:
a when x3 ¡ 2x + a is divided by x ¡ 2, the remainder is 7
b when 2x3 + x2 + ax ¡ 5 is divided by x + 1, the remainder is ¡8.
4 When x3 + 2x2 + ax + b is divided by x ¡ 1 the remainder is 4, and when divided by x + 2
the remainder is 16. Find constants a and b.
5 2xn + ax2 ¡ 6 leaves a remainder of ¡7 when divided by x ¡ 1, and 129 when divided by x + 3.
Find a and n given that n 2 Z + .
6 When P (z) is divided by z 2 ¡ 3z + 2 the remainder is 4z ¡ 7.
Find the remainder when P (z) is divided by:
a z¡1
b z ¡ 2.
7 When P (z) is divided by z + 1 the remainder is ¡8, and when divided by z ¡ 3 the remainder
is 4. Find the remainder when P (z) is divided by (z ¡ 3)(z + 1).
8 If P (x) is divided by (x ¡ a)(x ¡ b), where a 6= b, a, b 2 R , prove that the remainder is:
³
´
P (b) ¡ P (a)
£ (x ¡ a) + P (a).
b¡a
THE FACTOR THEOREM
For any polynomial P (x), k is a zero of P (x) , (x ¡ k) is a factor of P (x).
k is a zero of P (x) , P (k) = 0
,R=0
, P (x) = Q(x)(x ¡ k)
, (x ¡ k) is a factor of P (x)
Proof:
fdefinition of a zerog
fRemainder theoremg
fdivision algorithmg
fdefinition of a factorg
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The Factor theorem says that if 2 is a zero of P (x) then (x ¡ 2) is a factor of P (x), and vice versa.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
Self Tutor
Example 26
Find k given that (x ¡ 2) is a factor of x3 + kx2 ¡ 3x + 6.
Hence, fully factorise x3 + kx2 ¡ 3x + 6.
EXERCISE 6E.2
1 Find the constant k and hence factorise the polynomial if:
a 2x3 + x2 + kx ¡ 4 has the factor (x + 2)
b x4 ¡ 3x3 ¡ kx2 + 6x has the factor (x ¡ 3).
2 Find constants a and b given that 2x3 + ax2 + bx + 5 has factors (x ¡ 1) and (x + 5).
a Suppose 3 is a zero of P (z) = z 3 ¡ z 2 + (k ¡ 5)z + (k2 ¡ 7).
Find the possible values of k 2 R and all the corresponding zeros of P (z).
3
b Show that (z ¡ 2) is a factor of P (z) = z 3 + mz2 + (3m ¡ 2)z ¡ 10m ¡ 4 for all values
of m 2 R . For what value(s) of m is (z ¡ 2)2 a factor of P (z)?
a Consider P (x) = x3 ¡ a3
4
where a is real.
i Find P (a). What is the significance of this result?
ii Factorise x3 ¡ a3 as the product of a real linear and a quadratic factor.
b Now consider P (x) = x3 + a3 , where a is real.
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i Find P (¡a). What is the significance of this result?
ii Factorise x3 + a3 as the product of a real linear and a quadratic factor.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
a Prove that “x + 1 is a factor of xn + 1, n 2 Z
5
197
, n is odd”.
b Find the real number a such that (x ¡ 1 ¡ a) is a factor of P (x) = x3 ¡ 3ax ¡ 9.
THE FUNDAMENTAL THEOREM OF ALGEBRA
The theorems we have just seen for real polynomials can be generalised
in the Fundamental Theorem of Algebra:
a Every polynomial of degree n > 1 has at least one zero which
can be written in the form a + bi where a, b 2 R .
b If P (x) is a polynomial of degree n, then P (x) has exactly n
zeros, some of which may be either irrational numbers or complex
numbers.
Gauss proved this
theorem in 1799 as
his PhD dissertation.
Using the Fundamental Theorem of Algebra, the following properties of real polynomials can be
established:
² Every real polynomial of degree n can be factorised into n complex linear factors, some of which
may be repeated.
² Every real polynomial can be expressed as a product of real linear and real irreducible quadratic
factors (where ¢ < 0).
² Every real polynomial of degree n has exactly n zeros, some of which may be repeated.
² If p + qi (q 6= 0) is a zero of a real polynomial then its complex conjugate p ¡ qi is also a
zero.
² Every real polynomial of odd degree has at least one real zero.
Self Tutor
Example 27
Suppose ¡3 + i is a zero of P (x) = ax3 + 9x2 + ax ¡ 30 where a is real.
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Find a and hence find all zeros of the cubic.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
Self Tutor
Example 28
One zero of ax3 + (a + 1)x2 + 10x + 15, a 2 R , is purely imaginary.
Find a and the zeros of the polynomial.
EXERCISE 6E.3
1 Find all real polynomials of degree 3 with zeros ¡ 12 and 1 ¡ 3i.
2 p(x) is a real cubic polynomial for which p(1) = p(2 + i) = 0 and p(0) = ¡20.
Find p(x) in expanded form.
3 2 ¡ 3i is a zero of P (z) = z 3 + pz + q
where p and q are real.
a Using conjugate pairs, find p and q and the other two zeros.
b Check your answer by solving for p and q using P (2 ¡ 3i) = 0.
4 3 + i is a root of z 4 ¡ 2z 3 + az 2 + bz + 10 = 0, where a and b are real. Find a and b and the
other roots of the equation.
5 One zero of P (z) = z 3 + az 2 + 3z + 9, a 2 R , is purely imaginary. Find a and hence factorise
P (z) into the product of linear factors.
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6 At least one zero of P (x) = 3x3 + kx2 + 15x + 10, k 2 R , is purely imaginary. Find k and
hence factorise P (x) into the product of linear factors.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
199
SUM AND PRODUCT OF ROOTS THEOREM
We have seen that for the quadratic equation ax2 + bx + c = 0, a 6= 0, the sum of the roots is ¡
and the product of the roots is
b
a
c
.
a
For the polynomial equation an xn + an¡1 xn¡1 + :::: + a2 x2 + a1 x + a0 = 0, an 6= 0
n
P
which can also be written as
ar xr = 0,
r=0
the sum of the roots is
¡an¡1
, and the product of the roots is
an
(¡1)n a0
.
an
We can explain this result as follows:
Consider the polynomial equation an (x ¡ ®1 )(x ¡ ®2 )(x ¡ ®3 )::::(x ¡ ®n ) = 0
with roots ®1 , ®2 , ®3 , ...., ®n .
Expanding the LHS we have
an (x ¡ ®1 )(x ¡ ®2 )(x ¡ ®3 )(x ¡ ®4 )::::(x ¡ ®n )
|
{z
}
= an (x2 ¡ [®1 + ®2 ]x + (¡1)2 ®1 ®2 )(x ¡ ®3 )(x ¡ ®4 )::::(x ¡ ®n )
|
{z
}
= an (x3 ¡ [®1 + ®2 + ®3 ]x2 + :::: + (¡1)3 ®1 ®2 ®3 )(x ¡ ®4 )::::(x ¡ ®n )
fthe term of order x is no longer important as it will not contribute to
either the xn¡1 term or the constant termg
..
.
= an (xn ¡ [®1 + ®2 + ®3 + :::: + ®n ]xn¡1 + :::: + (¡1)n ®1 ®2 ®3 :::: ®n )
= an xn ¡ an [®1 + ®2 + ®3 + :::: + ®n ]xn¡1 + :::: + (¡1)n ®1 ®2 ®3 :::: ®n an
Equating coefficients,
an¡1 = ¡an [®1 + ®2 + ®3 + :::: + ®n ] and a0 = (¡1)n ®1 ®2 ®3 :::: ®n an
) ¡
an¡1
= ®1 + ®2 + ®3 + :::: + ®n
an
(¡1)n a0
= ®1 ®2 ®3 :::: ®n
an
and
Self Tutor
Example 29
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Find the sum and product of the roots of 2x3 ¡ 7x2 + 8x ¡ 1 = 0.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
Self Tutor
Example 30
A real polynomial has the form P (x) = 3x4 ¡ 12x3 + cx2 + dx + e. The graph of y = P (x)
has y-intercept 180. It cuts the x-axis at 2 and 6, and does not meet the x-axis anywhere else.
Suppose the other two zeros are m § ni, n > 0. Use the sum and product formulae to find m
and n.
If the other two zeros are m § ni, the sum of the zeros is
EXERCISE 6E.4
1 Find the sum and product of the roots of:
a 2x2 ¡ 3x + 4 = 0
c x4 ¡ x3 + 2x2 + 3x ¡ 4 = 0
e x7 ¡ x5 + 2x ¡ 9 = 0
b 3x3 ¡ 4x2 + 8x ¡ 5 = 0
d 2x5 ¡ 3x4 + x2 ¡ 8 = 0
f x6 ¡ 1 = 0
p
2 A real cubic polynomial P (x) has zeros 3 § i 2 and
2
3.
It has a leading coefficient of 6. Find:
b the coefficient of x2
a the sum and product of its zeros
c the constant term.
3 A real polynomial of degree 5 has leading coefficient ¡1 and zeros of ¡2, 3 § i, and
The y-intercept is 18. Find:
p
k § 1.
b the coefficient of x4 .
a k
4 A real polynomial of degree 5 has leading coefficient 2 and the coefficient of x4 is 3. When the
p
polynomial is graphed, the y-intercept is 5, and it cuts the x-axis at 12 and 1 § 2 only.
Suppose the other two zeros are m § ni, n > 0. Use the sum and product formulae to find m
and n.
5 A real quartic polynomial has leading coefficient 1 and zeros of the form a § i and 3 § a, where
a 2 R . Its constant term is 25. What are the possible values that a may take?
6 x3 ¡ px2 + qx ¡ r = 0 has non-zero roots p, q, and r, where p, q, r 2 R .
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b Hence find p, q, and r.
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a Show that q = ¡r
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
F
201
GRAPHING REAL POLYNOMIALS
Any polynomial with real coefficients can be graphed on the Cartesian plane.
Use of a graphics calculator or the graphing package provided will help in this section.
CUBIC POLYNOMIALS
INVESTIGATION 2
CUBIC GRAPHS
Every real cubic polynomial can be categorised into one of four types. In each case a 2 R , a 6= 0,
and the zeros are ®, ¯, °.
Type 1:
Three real, distinct zeros: P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °)
Type 2:
Two real zeros, one repeated: P (x) = a(x ¡ ®)2 (x ¡ ¯)
Type 3:
One real zero repeated three times: P (x) = a(x ¡ ®)3
Type 4:
One real and two complex conjugate zeros:
P (x) = (x ¡ ®)(ax2 + bx + c), ¢ = b2 ¡ 4ac < 0.
What to do:
1 Experiment with the graphs of Type 1 cubics. State the effect of changing both the size and
sign of a. What is the geometrical significance of ®, ¯, and °?
2 Experiment with the graphs of Type 2 cubics. What is the geometrical significance of the squared
factor?
3 Experiment with the graphs of Type 3 cubics. What is the geometrical significance
of ®?
GRAPHING
PACKAGE
4 Experiment with the graphs of Type 4 cubics. What is the geometrical significance
of ® and the quadratic factor which has complex zeros?
From Investigation 2 you should have discovered that:
² If a > 0, the graph has shape
or
. If a < 0 it is
or
.
² All cubics are continuous smooth curves.
² Every cubic polynomial must cut the x-axis at least once, and so has at least one real zero.
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² For a cubic of the form P (x) = a(x ¡ ®)(x ¡ ¯)(x ¡ °),
®, ¯, ° 2 R , the graph has three distinct x-intercepts
corresponding to the three distinct zeros ®, ¯, and °. The
graph crosses over or cuts the x-axis at these points, as
shown.
² For a cubic of the form P (x) = a(x ¡ ®)2 (x ¡ ¯), ®,
¯ 2 R , the graph touches the x-axis at the repeated zero ®
and cuts it at the other x-intercept ¯, as shown.
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®
°
¯
¯
®
x
x
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
² For a cubic of the form P (x) = a(x ¡ ®)3 , x 2 R , the
graph has only one x-intercept, ®. The graph is horizontal
at this point, and the x-axis is a tangent to the curve even
though the curve crosses over it.
®
² For a cubic of the form P (x) = (x ¡ ®)(ax2 + bx + c)
where ¢ < 0, there is only one x-intercept, ®. The graph
cuts the x-axis at this point. The other two zeros are complex
and so do not appear on the graph.
x
®
x
Self Tutor
Example 31
Find the equation of the cubic with graph:
a
b
y
2
y
4
6
x
-1
-3
We
-8
x
2
a The x-intercepts are ¡1, 2, 4.
) y = a(x + 1)(x ¡ 2)(x ¡ 4)
But when x = 0, y = ¡8
) a(1)(¡2)(¡4) = ¡8
) a = ¡1
So, y = ¡(x + 1)(x ¡ 2)(x ¡ 4)
When determining a polynomial function from a given graph, if we are not given all the zeros, or if some
of the zeros are complex, we write a factor in general form.
² If an x-intercept is not given, use
P (x) = (x ¡ k)2 (ax + b) .
| {z }
For example:
most general form of a linear
x
2
Using P (x) = a(x ¡ k) (x + b) is
more complicated.
?
k
² If there is clearly only one x-intercept
and that is given, use
P (x) = (x ¡ k) (ax2 + bx + c) .
|
{z
}
x
k
most general form of a quadratic
¢<0
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Either graph is possible.
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203
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
Self Tutor
Example 32
Find the equation of the cubic which cuts the x-axis at 2 and ¡3, cuts the y-axis at ¡48, and
which passes through the point (1, ¡40).
EXERCISE 6F.1
1 For a cubic polynomial P (x), state the geometrical significance of:
a a single real linear factor such as (x ¡ ®), ® 2 R
b a squared real linear factor such as (x ¡ ®)2 , ® 2 R
c a cubed real linear factor such as (x ¡ ®)3 , ® 2 R .
2 Find the equation of the cubic with graph:
a
b
y
c
y
12
- Qw
y
-4
6
Qw
-3
3
x
x
x
-1
2
3
-12
d
e
y
f
y
-3
9
-5
5
y
x
-2
x
-2 - Qw
-12
-5
3
-4
x
3 Find the equation of the cubic whose graph:
a cuts the x-axis at 3, 1, and ¡2, and passes through (2, ¡4)
b cuts the x-axis at ¡2, 0, and 12 , and passes through (¡3, ¡21)
c touches the x-axis at 1, cuts the x-axis at ¡2, and passes through (4, 54)
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d touches the x-axis at ¡ 23 , cuts the x-axis at 4, and passes through (¡1, ¡5).
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204
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
4 Match the given graphs to the corresponding cubic function:
a y = 2(x ¡ 1)(x + 2)(x + 4)
b y = ¡(x + 1)(x ¡ 2)(x ¡ 4)
c y = (x ¡ 1)(x ¡ 2)(x + 4)
d y = ¡2(x ¡ 1)(x + 2)(x + 4)
e y = ¡(x ¡ 1)(x + 2)(x + 4)
f y = 2(x ¡ 1)(x ¡ 2)(x + 4)
A
B
y
C
y
y
x
2
8
-4
16
x
1
2
D
-1
x
-4
1
E
y
-8
2
F
y
y
16
8
-2
x
-2
-4
4
x
-2
1
-4
1
x
-4
1
-16
5 Find the equation of a real cubic polynomial which:
a cuts the x-axis at
1
2
and ¡3, cuts the y-axis at 30, and passes through (1, ¡20)
b cuts the x-axis at 1, touches the x-axis at ¡2, and cuts the y-axis at (0, 8)
c cuts the x-axis at 2, cuts the y-axis at ¡4, and passes through (1, ¡1) and (¡1, ¡21).
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QUARTIC POLYNOMIALS
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205
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
From Investigation 3 you should have discovered that:
² For a quartic polynomial in which a is the coefficient of x4 :
I
If a > 0 the graph opens upwards.
I
If a < 0 the graph opens downwards.
² If a quartic with a > 0 is fully factorised into real linear factors, then:
I
I
for a single factor (x ¡ ®), the
for a square factor (x ¡ ®)2 , the
graph cuts the x-axis at ®
graph touches the x-axis at ®
®
y ®
y
x
x
for a cubed factor (x ¡ ®)3 , the
graph cuts the x-axis at ® and is
‘flat’ at ®
I
y
I
®
y
x
or
for a quadruple factor (x ¡ ®)4 ,
the graph touches the x-axis and is
‘flat’ at that point.
®
y
®
x
x
I
I
If a quartic with a > 0 has one
real quadratic factor with ¢ < 0
we could have:
If a quartic with a > 0 has two real
quadratic factors both with ¢ < 0
we have:
y
y
x
x
The graph does not meet the x-axis
at all.
Self Tutor
Example 33
Find the equation of the quartic
with graph:
y
-1
3
x
-3
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
Self Tutor
Example 34
Find the quartic which touches the x-axis at 2, cuts the x-axis at ¡3, and also passes
through (1, ¡12) and (3, 6).
EXERCISE 6F.2
1 Find the equation of the quartic with graph:
a
b
y
c
y
-3
2
x
-1
y
-2
-1
x
-6
x
1
2
-1
We
-16
d
e
y
f
y
(-3, 54)
4
x
-1
-3
y
x
Ew
-1
3
x
-2
-9
3
-16
2 Match the given graphs to the corresponding quartic functions:
a y = (x ¡ 1)2 (x + 1)(x + 3)
b y = ¡2(x ¡ 1)2 (x + 1)(x + 3)
c y = (x ¡ 1)(x + 1)2 (x + 3)
d y = (x ¡ 1)(x + 1)2 (x ¡ 3)
e y = ¡ 13 (x ¡ 1)(x + 1)(x + 3)2
f y = ¡(x ¡ 1)(x + 1)(x ¡ 3)2
B
y
x
x
-1
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207
COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
D
E
y
F
y
1
-1
-1
1
y
x
3
3
1 x
-3
x
-1
3 Find the equation of the quartic whose graph:
a cuts the x-axis at ¡4 and 12 , touches it at 2, and passes through the point (1, 5)
b touches the x-axis at
c cuts the x-axis at
§ 12
2
3
and ¡3, and passes through the point (¡4, 49)
and §2, and passes through the point (1, ¡18)
d touches the x-axis at 1, cuts the y-axis at ¡1, and passes through (¡1, ¡4) and (2, 15).
ACTIVITY
Click on the icon to run a card game for graphs of cubic and quartic functions.
CARD GAME
GENERAL POLYNOMIALS
We have already seen that every real cubic polynomial must cut the x-axis at least once, and so has at
least one real zero.
If the exact value of the zero is difficult to find, we can use technology to help us. We can then factorise
the cubic as a linear factor times a quadratic, and if necessary use the quadratic formula to find the other
zeros.
This method is particularly useful if we have one rational zero and two irrational zeros that are radical
conjugates.
DISCUSSION
GENERAL POLYNOMIALS
Consider the general polynomial P (x) = an xn + an¡1 xn¡1 + :::: + a1 x + a0 , an 6= 0.
² Discuss the behaviour of the graph as x ! ¡1 and x ! 1 depending on
I
I
the sign of an
whether n is odd or even.
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² Under what circumstances is P (x) an:
I
I
odd function
even function?
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
Self Tutor
Example 35
Find exactly the zeros of P (x) = 3x3 ¡ 14x2 + 5x + 2.
Self Tutor
Example 36
3
2
Find exactly the roots of 6x + 13x + 20x + 3 = 0.
For a quartic polynomial P (x) we first need to establish if there are any x-intercepts at all. If there
are not then the polynomial must have four complex zeros. If there are x-intercepts then we can try to
identify linear or quadratic factors.
EXERCISE 6F.3
1 Find exactly all zeros of:
a x3 ¡ 3x2 ¡ 3x + 1
c 2x3 ¡ 3x2 ¡ 4x ¡ 35
e 4x4 ¡ 4x3 ¡ 25x2 + x + 6
b x3 ¡ 3x2 + 4x ¡ 2
d 2x3 ¡ x2 + 20x ¡ 10
f x4 ¡ 6x3 + 22x2 ¡ 48x + 40
2 Find exactly the roots of:
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5
95
100
50
b 2x3 + 3x2 ¡ 3x ¡ 2 = 0
d 2x3 + 18 = 5x2 + 9x
f 2x4 ¡ 13x3 + 27x2 = 13x + 15
75
25
0
5
95
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a x3 + 2x2 + 3x + 6 = 0
c x3 ¡ 6x2 + 12x ¡ 8 = 0
e x4 ¡ x3 ¡ 9x2 + 11x + 6 = 0
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
3 Factorise into linear factors with exact values:
a x3 ¡ 3x2 + 4x ¡ 2
c 2x3 ¡ 9x2 + 6x ¡ 1
e 4x3 ¡ 8x2 + x + 3
g 2x4 ¡ 3x3 + 5x2 + 6x ¡ 4
209
x3 + 3x2 + 4x + 12
x3 ¡ 4x2 + 9x ¡ 10
3x4 + 4x3 + 5x2 + 12x ¡ 12
2x3 + 5x2 + 8x + 20
b
d
f
h
4 The following cubics will not factorise neatly. Find their zeros using technology.
a x3 + 2x2 ¡ 6x ¡ 6
b x3 + x2 ¡ 7x ¡ 8
5 A scientist is trying to design a crash test barrier with the characteristics shown graphically below.
The independent variable t is the time
f(t)
¡
after impact, measured in milliseconds, 120
100
such that 0 6 t 6 700.
The dependent variable is the distance 80
60
the barrier is depressed during the 40
impact, measured in millimetres.
20
t (ms)
200
400
600
800
a The equation for this graph has the form f (t) = kt(t ¡ a)2 , 0 6 t 6 700.
Use the graph to find a. What does this value represent?
b If the ideal crash barrier is depressed by 85 mm after 100 milliseconds, find the value of k.
Hence find the equation of the graph given.
6 Last year, the volume of water in a particular reservoir could be described by the model
V (t) = ¡t3 + 30t2 ¡ 131t + 250 ML, where t is the time in months.
The dam authority rules that if the volume falls below 100 ML, irrigation is prohibited. During
which months, if any, was irrigation prohibited in the last twelve months? Include in your answer
a neat sketch of any graphs you may have used.
7 A ladder of length 10 metres is leaning against a wall so that it is just
touching a cube of edge length one metre as shown.
What height up the wall does the ladder reach?
xm
10 m
1m
REVIEW SET 6A
NON-CALCULATOR
1 Find real numbers a and b such that:
a a + ib = 4
b (1 ¡ 2i)(a + bi) = ¡5 ¡ 10i
c (a + 2i)(1 + bi) = 17 ¡ 19i
2 If z = 3 + i and w = ¡2 ¡ i, find in simplest form:
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a 2z ¡ 3w
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3 Find exactly the real and imaginary parts of z if z =
p
3
p + 3.
i+ 3
4 Find a complex number z such that 2z ¡ 1 = iz ¡ i. Write your answer in the form
z = a + bi where a, b 2 R .
5 Prove that zw¤ ¡ z ¤ w
6 Given w =
z+1
z¤ + 1
is purely imaginary or zero for all complex numbers z and w.
where z = a + bi, a, b 2 R , write w in the form x + yi, x, y 2 R .
Hence determine the conditions under which w is purely imaginary.
a (3x3 + 2x ¡ 5)(4x ¡ 3)
7 Expand and simplify:
8 Carry out the following divisions:
a
b (2x2 ¡ x + 3)2
x3
x+2
b
x3
(x + 2)(x + 3)
9 Find the sum and product of the zeros of:
a 3x4 ¡ 4x3 + 3x2 + 8
b 2x6 + 2x4 ¡ x3 + 7x ¡ 10
10 State and prove the Remainder theorem.
11 ¡2 + bi is a solution to z 2 + az + (3 + a) = 0. Find constants a and b given that they are
real.
12 P (x) has remainder 2 when divided by x ¡ 3, and remainder ¡13 when divided by x + 2.
Find the remainder when P (x) is divided by x2 ¡ x ¡ 6.
p
p
13 Find all quartic polynomials with real, rational coefficients, having 2 ¡ i 3 and 2 + 1 as
two of the zeros.
14 If f(x) = x3 ¡ 3x2 ¡ 9x + b has (x ¡ k)2 as a factor, show that there are two possible
values of k. For each of these two values of k, find the corresponding value for b, and hence
solve f (x) = 0.
p
p
15 Find all real quartic polynomials with rational coefficients, having 3 ¡ i 2 and 1 ¡ 2 as
two of the zeros.
16 When P (x) = xn + 3x2 + kx + 6 is divided by (x + 1) the remainder is 12. When P (x)
is divided by (x ¡ 1) the remainder is 8. Find k and n given that 34 < n < 38.
17 If ® and ¯ are two of the roots of x3 ¡ x + 1 = 0, show that ®¯ is a root of x3 + x2 ¡ 1 = 0.
Hint: Let x3 ¡ x + 1 = (x ¡ ®)(x ¡ ¯)(x ¡ °).
REVIEW SET 6B
³
1 If z =
CALCULATOR
´3
5
¡ 3 ¡ 2i , x, y 2 Z , express z in the form z = x + yi.
2¡i
2 Without using a calculator, find z if z 2 = 5 ¡ 12i. Check your answer using a calculator.
3 Prove that if z is a complex number then both z + z ¤ and zz ¤ are real.
4 If z = 4 + i and w = 3 ¡ 2i find 2w¤ ¡ iz.
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2 ¡ 3i
= 3 + 2i.
2a + bi
5 Find rationals a and b such that
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211
6 a + ai is a root of x2 ¡ 6x + b = 0 where a, b 2 R .
Explain why b has two possible values. Find a in each case.
7 Find the remainder when x47 ¡ 3x26 + 5x3 + 11 is divided by x + 1.
8 Factorise 2z 3 + z 2 + 10z + 5 as a product of linear factors with exact terms.
9 A quartic polynomial P (x) has graph y = P (x) which touches the x-axis at (¡2, 0), cuts
the x-axis at (1, 0), cuts the y-axis at (0, 12), and passes through (2, 80).
Find an expression for P (x) in factored form, and hence sketch the graph of y = P (x).
10 Find all zeros of 2z 4 ¡ 5z 3 + 13z 2 ¡ 4z ¡ 6.
11 Factorise z 4 + 2z 3 ¡ 2z 2 + 8 into linear factors.
12 Find the general form of all real polynomials of least degree which have zeros 2 + i and
¡1 + 3i.
13 3 ¡ 2i is a zero of z 4 + kz 3 + 32z + 3k ¡ 1, where k is real.
Find k and all zeros of the quartic.
14 Find all the zeros of the polynomial z 4 + 2z 3 + 6z 2 + 8z + 8 given that one of the zeros is
purely imaginary.
15 When a polynomial P (x) is divided by x2 ¡ 3x + 2 the remainder is 2x + 3.
Find the remainder when P (x) is divided by x ¡ 2.
16 Find possible values of k if the line with equation y = 2x + k
equation x2 + y2 + 8x ¡ 4y + 2 = 0.
does not meet the circle with
p
17 P (x) = 2x4 ¡ 8x3 + ax2 + bx ¡ 110, a, b 2 R , has zeros m § 2i and 1 § n 3 where
m, n 2 R .
a Find m and n.
b Sketch the graph of y = P (x).
REVIEW SET 6C
1 Find real numbers x and y such that (3x + 2yi)(1 ¡ i) = (3y + 1)i ¡ x.
2 Solve the equation: z 2 + iz + 10 = 6z.
3 Prove that zw¤ + z ¤ w
is real for all complex numbers z and w.
4 Find real x and y such that:
c (x + iy)2 = x ¡ iy
b (3 ¡ 2i)(x + i) = 17 + yi
a x + iy = 0
5 z and w are non-real complex numbers with the property that both z + w and zw are real.
Prove that z ¤ = w.
6 Find the sum and product of the roots of:
a 2x3 + 3x2 ¡ 4x + 6 = 0
7 Find z if
p
z=
b 4x4 = x2 + 2x ¡ 6
2
+ 2 + 5i.
3 ¡ 2i
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8 Find the remainder when 2x17 + 5x10 ¡ 7x3 + 6 is divided by x ¡ 2.
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COMPLEX NUMBERS AND POLYNOMIALS (Chapter 6)
9 5 ¡ i is a zero of 2z 3 + az 2 + 62z + (a ¡ 5), where a is real. Find a and the other two
zeros.
10 Find, in general form, all real polynomials of least degree which have zeros:
p
b 1 ¡ i and ¡3 ¡ i.
a i 2 and 12
11 P (x) = 2x3 + 7x2 + kx ¡ k
is the product of 3 linear factors, 2 of which are identical.
a Show that k can take 3 distinct values.
b Write P (x) as the product of linear factors for the case where k is greatest.
12 Find all roots of 2z 4 ¡ 3z 3 + 2z 2 = 6z + 4.
13 Suppose a and k are real. For what values of k does z 3 + az 2 + kz + ka = 0 have:
a one real root
b 3 real roots?
14 (3x + 2) and (x ¡ 2) are factors of 6x3 + ax2 ¡ 4ax + b. Find a and b.
15 Find the exact values of k for which the line y = x¡k
(x ¡ 2)2 + (y + 3)2 = 4.
is a tangent to the circle with equation
16 Find the quotient and remainder when x4 + 3x3 ¡ 7x2 + 11x ¡ 1 is divided by x2 + 2.
Hence, find constants a and b for which x4 + 3x3 ¡ 7x2 + (2 + a)x + b is exactly divisible
by x2 + 2.
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17 P (x) is a real polynomial of degree 5 with leading coefficient 1 and zeros m § 2i, 1 § mi,
and 2 (m 2 R ). The graph of y = P (x) cuts the y-axis at ¡56.
p
a Show that m = § 3.
b Find the coefficient of x4 .
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