Yimin Math Centre
4 Unit Math Homework for Year 12 (Worked Answers)
Grade:
Date:
Score:
nt
re
Student Name:
Table of contents
1
1
1.1.1
The Division transformation . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.1.2
The Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . . . .
5
1.1.3
Polynomials with Real Coefficients . . . . . . . . . . . . . . . . . . . . . . .
5
at
h
Factorisation of polynomials and fundamental theorem of algebra . . . . . . . . . . .
M
1.1
Ce
Topic 4 — Polynomials (Part 2)
1.2
The Relationship Between the Roots and Coefficients of a Polynomial . . . . . . . . .
6
1.3
Solutions of Polynomial Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
1.4
Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
Yi
m
in
1
This edition was printed on June 17, 2013.
Camera ready copy was prepared with the LATEX2e typesetting system.
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4 Unit Math Homework for Year 12
Year 12 Topic 1 Worked Answers
1
Page 1 of 18
Topic 4 — Polynomials (Part 2)
1.1
Factorisation of polynomials and fundamental theorem of algebra
The division transformation: If P (x) and D(x) are polynomials over a field F, the
process of polynomial division of P (x) by D(x) yields a quotient Q(x) and a remainder
R(x), which are both polynomials of F, such that deg R < degD. then:
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P (x) ≡ D(x)Q(x) + R(x),
Example 1.1.1 Consider P (x) = x4 − x + 2 and D(x) = x2 + 1.
x4 − x + 2 ≡ (x2 + 1) × (x2 − 1) + (−x + 3)
↑
↑
↑
↑
P (x) ≡
D(x) × Q(x) +
R(x)
x2
−1
x4
−x+2
4
2
−x −x
h
x2 + 1
Ce
Solution:
at
− x2 − x + 2
x2
+1
in
M
−x+3
Solution:
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Exercise 1.1.1 Divide 3x4 − 4x3 + 4x − 8 by x2 − 2.
x2 − 2
3x2 − 4x + 6
3x4 − 4x3
+ 4x − 8
− 3x4
+ 6x2
− 4x3 + 6x2 + 4x
4x3
− 8x
6x2 − 4x − 8
− 6x2
+ 12
− 4x + 4
Hence 3x4 − 4x3 + 4x − 8 = (x2 − 2)(3x2 − 4x + 6) + (−4x + 4)
Or
3x4 − 4x3 + 4x − 8
−4x + 4
= (3x2 − 4x + 6) + 2
.
2
x −2
x −2
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Year 12 Topic 1 Worked Answers
1.1.1
Page 2 of 18
The Division transformation
The identity P (x) ≡ D(x).Q(x) + R(x) is true for all complex numbers x, when Q(x) and
R(x) are polynomials such that deg R < deg D. The nature of the coefficients of Q(x) and
R(x) is determined by the nature of he coefficients of P (x) and D(x).
Example 1.1.2 Find the remainder when P (x) = x3 − 2x + 1 is divide by :
1. x − i,
x − i is a linear divisor.
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Solution:
Hence we can use the remiander theorem, and the remainder is :
Ce
P (i) = −i − 2i + 1 = 1 + 3i.
2. x2 + 1.
Both P (x) = x3 − 2x + 1 and D(x) = x2 + 1 are polunomials over Q.
h
Solution:
at
By the division transformation, P (x) ≡ (x2 + 1)Q(x) + R(x),
M
where Q(x) and R(x) are polynomials over Q, such that deg R < deg D = 2.
Thus P (x) ≡ (x2 + 1)Q(x) + ax + b, where both a and b are rational,
in
∴ P (i) = 0 + ai + b, ⇒ ∴ 1 − 3i = ai + b, ⇒ a = −3 and b = 1.
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Hence the remainder ax + b is − 3x + 1.
Exercise 1.1.2 If P (x) is divided by (x − 1)(x + 3), remainder is 2x − 11. What is the remainder
when P (x) is divided by (x − 1)?
Solution:
Let the quateint be Q1 (x) when P (x) is divided by (x − 1)(x + 3),
P (x)
(2x − 11)
= Q(x) +
(x − 1)(x + 3)
(x − 1)(x + 3)
P (x)
2x − 11
= Q(x)(x + 3) +
⇒
2
x−1
x−1
x−1
2x − 11
− 2x + 2
We have
−9
∴ remiander is − 9 when P (x) is divided by (x − 1).
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Year 12 Topic 1 Worked Answers
Page 3 of 18
Exercise 1.1.3 When x4 − kx + 1 is divided by x2 + 1, the remainder is 3x + 2. Find value of k.
Solution:
By the division transformation, x4 − kx + 1 ≡ (x2 + 1) × Q(x) + (3x + 2)
Substituting x = i, ⇒ 2 − ki = 0 + 3i + 2, ⇒ k = −3.
Exercise 1.1.4
1. Find the quotient and remainder when x4 − 2x3 + x2 − 5x + 7 is divided by x2 + x − 1.
Solution:
x2 + x − 1
x4 − 2x3 + x2 − 5x + 7
− x4 − x3 + x2
− 3x3 + 2x2 − 5x
3x3 + 3x2 − 3x
Ce
5x2 − 8x + 7
− 5x2 − 5x + 5
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x2 − 3x + 5
h
− 13x + 12
at
Therefore the quotient is x2 − 3x + 5, and the remaider is − 13x + 12.
When the remaider is zero, ⇒ P (x) = D(x) × Q(x),
in
Solution:
M
2. Find a and b if x4 − 2x3 + x2 + ax + b is exactly divided by x2 + x − 1.
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∴ x4 − 2x3 + x2 + ax + b = (x2 + x − 1)(x2 − 3x + 5)
= x4 − 3x3 + 52 + x3 − 3x2 + 5x − x2 + 3x − 5
= x4 − 2x3 + x2 + 8x − 5.
∴ a = 8 and b = −5.
3. Hence factor x4 − 2x3 + x2 + 8x − 5.
Solution:
the factors of the x4 − 2x3 + x2 + 8x − 5 is (x2 + x − 1)(x2 − 3x + 5)
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Year 12 Topic 1 Worked Answers
Page 4 of 18
Exercise 1.1.5
1. When P (x) = x4 + ax2 + 2x is divided by x2 + 1, the remainder is 2x + 3. find the value of a.
Solution:
By the division transformation, x4 + ax2 + 2x ≡ (x2 + 1)Q(x) + (2x + 3).
Substituting x = i, ⇒ 1 − a + 2i = 2i + 3, ⇒ a = −2.
2. When P (x) = x4 + ax2 + bx + 2 is divided by x2 + 1, the remainder is −x + 1. Find the values
of a and b.
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Solution: By the division transformation, x4 + ax2 + bx + 2 ≡ (x2 + 1)Q(x) + (−x + 1).
Substituting x = i, ⇒ 1 − a + bi + 2 = −i + 1,
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M
at
h
Ce
that is − a + bi = −i − 2, ⇒ a = 2, b = −1.
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Year 12 Topic 1 Worked Answers
Page 5 of 18
1.1.2
The Fundamental Theorem of Algebra
1.1.3
Polynomials with Real Coefficients
Exercise 1.1.6
1. Find the zeros of P (x) = x4 + x3 − x2 + x − 2 over C, given that i is a zero. Hence factor P(x)
fully over R.
Solution:
P (x) has real coefficients. Hence P (i) = 0 ⇒ P (−i) = 0,
and then (x − 1)(x + 1) = x2 + 1 is a factors of P (x).
By the division tansformation P (x) = x4 + x3 − x2 + x − 2
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= (x2 + 1)(x2 + x − 2)
= (x2 + 1)(x − 1)(x + 2)
This is the factorisation of P (x) into irreducible factor over R,
Ce
and P (x) has zeros i, −i, −2 and 1 over C.
P (x) has real coefficients. Hence P (1 − i) = 0 ⇒ P (1 + i) = 0,
M
Solution:
at
h
2. If P (x) = x4 − 2x3 − x2 + 6x − 6 has a zero 1 − i, find the zeros of P (x) over C, and factorise
P (x) fully over R.
and then [x − (1 − i)][x − (1 + i)] = x2 − 2x + 2 is a factor of P (x).
in
By polynomial division, P (x) = (x2 − 2x + 2)(x2 − 3).
√
√
Hence P (x) = (x2 − 2x + 2)(x − 3)(x + 3),
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this is the factorisation of P (x) into irreducible factors over R
√
∴ P (x) has zeros 1 ± i, ± 3.
3. P (x) is an even monic polynomial of degree 4 with integer coefficients. One zero is 2i and the
product of the zeros is −8. Factorise P (x) fully over R.
Solution: P (x) is an even monic polynomial of degree 4.Hence P (x) = x4 + ax2 + b.
P (x) has real coefficients. Hence P (2i) = 0, ⇒ P (−2i) = 0
and then (x − 2i)(x + 2i) = x2 + 4 is a factors of P (x) ⇒ P (x) = (x2 + 4)(x2 + c).
The product of zeros of P (x) is −8. Hence 4c = −8 ⇒ c = −2,
√
√
P (x) = (x2 + 4)(x2 − 2) = (x2 + 4)(x − 2)(x + 2).
These are irreducible factors of P (x) over R, and P (x) has zeros ± 2i and ±
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√
2 over C.
Year 12 Topic 1 Worked Answers
1.2
Page 6 of 18
The Relationship Between the Roots and Coefficients of a Polynomial
Let ax4 + bx3 + cx2 + dx + e = 0 have roots α, β, γ, δ over C
Then ax4 + bx3 + cx2 + dx + e ≡ a(x − α)(x − β)(x − γ)(x − δ).
P
α = ab ,
P
αβ = ac ,
P
αβγ = − ad , αβγδ = ae .
For P (x) = an xn + an−1 xn−1 + . . . + a1 x + a0
The sum of the products of roots taken r at a time = (−1)r an−r
.
an
Example 1.2.1 Expand P (x) = (x − 1)(x + 2)(x − 3)(x + 1)
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Solution: P (x) has zeros 1, −1, −2, 3 and form x4 + bx3 + cx2 + dx + e,
Ce
Since P (x) is monic of degree 4. If α, β, γ, δ denote the zeros of P (x), then
X
α = 1 ⇒ b = −1
X
αβ = −7 ⇒ c = +(−7)
X
αβγ = −1 ⇒ d = −(−1)
αβγδ = 6 ⇒ e = +6
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h
Hence P (x) = x4 − 3x − 7x2 + x + 6.
M
Exercise 1.2.1
1. Find the monic polynomial of degree 3 with zeros 1, 2, and 3.
in
Solution: P (x) = x3 + ax2 + bx + c, since P (x) is the monic of degree three.
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If α = 1, β = 2, andγ = 3 denote the zeros of P (x), then
X
a=−
α = −(1 + 2 + 3) = −6,
X
b=
αβ = 2 + 3 + 6 = 11,
X
c=−
αβγ = −6. Hence P (x) = x3 − 6x + 11x − 6.
2. Two of the roots of 3x3 + ax2 + 23x − 6 = 0 are reciprocals. Find the value of a and the three
roots.
Solution: Let the roots of P (x) be α, 1 , β. Then product of roots is β ⇒ β = − −6 = 2.
α
3
2
23
1
Sum of products taken two at a time is 1 + + 2α =
⇒ 3, .
α
3
3
1
−a
Sum of the roots is 3 + + 2 =
⇒ a = −16.
3
3
1
Hence the roots are 3, and 2, and the coefficient a is − 16.
3
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Year 12 Topic 1 Worked Answers
Page 7 of 18
Exercise 1.2.2
1. Two of the roots of x3 − 3x2 − 4x + a = 0 are opposites. Find the value of a and the three roots.
Solution:Let the roots of P (x) = x3 − 3x2 − 4x + a be α, −α, β.
Then sum of the roots of β, ⇒ β = 3.
Sum of products taken two at a time is − α2 + 3α − 3α = −α2 = −4, ⇒ α = ±2.
Product of the roots is − 2 × 2 × 3 = −a ⇒ a = 12.
Hence the roots are 2, −2, 3 and the coefficient a = 12.
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2. Find the monic polynomial of degrees 4 with zeros −3, −1, 1 and 3.
Ce
Solution: P (x) = x4 + ax3 + bx2 + cx + d, since P (x) is monic of degree four.
If α = −3, β = −1, γ = 1andδ = 3 denote the zeros of P (x), then:
a = −Σα = −(−3 − 1 + 1 + 3) = 0,
h
b = Σαβ = 3 − 3 − 9 − 1 − 3 + 3 = −10,
at
c = −Σαβγ = −(3 + 9 − 9 − 3) = 0,
M
d = Σαβγδ = 9.
in
Hence P (x) = x4 − 10x2 + 9.
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3. Two of the zeros of the polynomial P (x) = x4 + bx3 + cx2 + dx + e, where b, c, d, and e are
real, are 2 + i and 1 − 3i. Find the other two zeros and hence find the values of b and e.
Solution: P (x) = x4 + bx3 + cx2 + dx + e, is the polynomial with real coeficients.
Hence z1 = (2 + i) is zero of P (x) ⇒ z2 = z1 = 2 − i is zero of P (x) too,
z3 = (1 − 3i) is zero of P (x) ⇒ z4 = z3 = 1 + 3i is zero of P (x) too.
The product of the roots of the roots of P (x) is
z1 × z2 × z3 × z4 = (z1 z1 )(z3 z3 )
= |z1 |2 |z3 |2 = (4 + 1)(1 + 9) = 50, ⇒ e = 50.
The sum of the roots of P (x) is
z1 + z2 + z3 + z4 = (z1 + z1 ) + (z3 + z3 )
= 2Rez1 + 2Rez3 = 2 × 2 + 2 × 1 = 6, ⇒ b = −6.
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Year 12 Topic 1 Worked Answers
Page 8 of 18
Exercise 1.2.3
1. The equation px3 + qx2 + rx + s = 0 has roots (a − c), a, (a + c), which are in arithmetic
and hence show that 2q 3 − 9pgr + 27p2 s = 0.
progression. Show that the a = −q
3p
−q
−q
= Σα = (a − c) + a + (a + c) = 3a, ⇒ a =
.
The sum of roots is
p
3p
−q
−p × q 3 q × q 2 rq
Hence 0 = P (a) = P
=
− s ⇒ 0 = P (a) × 27p2
+
−
3p
27p3
9p2
3p
Solution:
= 2q 3 − 9pqr + 27p2 s.
Let the roots be a − c, a, a + c. then
27
1
−
= Σα = (a − c) + a + (a + c) = 3a ⇒ a = − .
18
2
5
1 1
4
= Σα.β.γ = (a − c) × a × (a + c) = −
− c2 ⇒ c = ± .
18
2 4
6
4
1
1
1
Hence the roots are a − c = − = −1 , a = − , and a + c = .
3
3
2
3
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Ce
Solution:
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2. Solve the equation 18x3 + 27x2 + x − 4 = 0. given that the roots are in arithmetic progression.
in
3
3. The equation pxq
+qx2 +rx+s = 0 has the roots ac, a and ac , which are in geometric progression.
show that a = 3 (− ps ) and hence show that pr3 − q 3 s = 0.
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r
Solution: The product of the roots is −s = Σα.β.γ = ac × a × a = a3 , ⇒ a = 3 (− s ).
p
c
p
q
a
1
The sum of the roots is − = Σα = ac + a + = a(c + 1 + ),
p
c
c
r
a2
The product of the roots taken two at a time is = Σα.β = a2 c + a2 +
p
c
1
2
=a c+1+
.
c
r
q
r
s
Hence − a =
⇒ −q × 3 (− ) = r ⇒ pr3 − q 3 s = 0.
p
p
p
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Year 12 Topic 1 Worked Answers
Page 9 of 18
Exercise 1.2.4
1. The equation x3 + 3x2 − 2x − 2 = 0 has roots α, β, and γ. Find the equation with the roots (I)
α − 2, β − 2 and γ − 2; (II) α2 , β 2 and γ 2 .
Solution:
(I) The values α − 2, β − 2 and γ − 2 are satisfy
(x + 2)3 + 3(x + 2)3 − 2(x + 2) − 2 = 0
and hence the required equation is x3 + 9x2 + 22x + 14 = 0.
1
1
1
(II) The values α2 , β 2 and γ 2 satisfy (x 2 )3 + 3(x 2 )2 − 2x 2 − 2 = 0.
1
Reaangement gives x 2 (x − 2) = 2 − 3x.
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Squaring both sides we obtain x(x − 2)2 = (2 − 3x)2
Hence the required equation is x3 + 9x2 + 22x + 14 = 0.
(I)
α β
γ
, and satisfy (2x)3 + (2x)2 − 2(2x) − 3 = 0.
2 2
2
Hence the required equation is 8x3 + 4x2 − 4x − 3 = 0.
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Solution:
h
Ce
2. The equation x3 + x2 − 2x − 3 = 0 has roots α, β and γ. Find the equation s with roots (I)
α β
, 2 and γ2 ; (II) α + 2, β + 2 and γ + 2.
2
(II) α + 2, β + 2 and γ + 2 satisfy (x − 2)3 + x − 2)2 − 2(x − 2) − 3 = 0.
in
Hence the required is x3 − 5x2 + 6x − 3 = 0.
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3. The equation x3 + px + q = 0 has roots α, β and γ. Find the monic cubic equation with roots
α2 , β 2 and γ 2 .
Solution:
If α, β and γsatisf yx3 + px + q = 0,
1
1
then α2 , β 2 and γ 2 satisfy (x 2 )3 + px 2 + q = 0.
1
Rearrangement gives x 2 (x + p) = −q.
Squaring both sides we obtain x(x + p)2 = q 2 .
Simplifying, we get x3 + 2px2 + p2 x − q 2 = 0.
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Year 12 Topic 1 Worked Answers
Page 10 of 18
Exercise 1.2.5
1. The equation x3 − 6x2 + ax + 10 = 0 has roots that are in arithmetic progression. Find the value
of a and solve the equation.
Solution:
Let the roots be b − c, b, b + c.
Then 6 = Σα = (b − c) + b + (b + c) = 3b ⇒ b = 2.
− 10 = σα.β.γ = (b − a) × b × (b + c) = 2(4 − c)2 ⇒ c = 3.
Hence the roots are b − c = −1, b = 2, b + c = 5 and a = Σα.β = −2 − 5 + 10 = 3.
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2. Solve the equation 2x63−13x2 −26x+16 = 0, given that the roots arte in geometric progression.
in
M
at
h
Let the roots be ac, a,
Ce
a
.
c
16
a
Then −
= Σα.β.γ = ac × a × = a3 ⇒ a = −2,
2
c
13
a
1
1
17
= Σα = a.c + a + = −2(c + 1 + ) ⇒ c + = −
2
c
c
c
4
1
This gives 4c2 + 17c + 4 = 0 ⇒ c = − (or c = −4 that gives the same roots).
4
1
a
Hence the roots are a.c = , a = −2, = 8.
2
c
Solution:
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3. The equation x3 + 3x2 − 2x − 2 = 0 has roots α, β and γ. Find the equations with roots (I)
2α, 2β and 2γ; (II) α1 , β1 and γ1 .
Solution: (I) The values 2α, 2β and 2γ satisfy
x 3
x 2
−2
x
− 2 = 0,
2
2
2
and hence the required equation is x3 + 6x2 − 8x − 16 = 0.
3
2
1 1 1
1
1
1
(II) The values , , satisfy
+3
−2
− 2 = 0,
α β γ
x
x
x
+3
and hance the required equation is 2x3 + 2x2 − 3x − 1 = 0.
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Year 12 Topic 1 Worked Answers
1.3
Page 11 of 18
Solutions of Polynomial Equations
Example 1.3.1
1. Show that i is a zero of P (x) = x3 − 5x + 6i . Hence solve x3 − 5x + 6i = 0.
Solution:
P (i) = −i − 5i + 6i = 0 ⇒ (x − i) is a factor of P (x).
By inspection, x3 − 5x + 6i = (x − i)(x2 + ix − 6)
∴ P (x) = 0 ⇒ x = i or x2 + ix − 6 = 0,
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√
−i ± 23
.
Using the quadratic formula, x = i orx =
2
Ce
2. Solve x4 − 4x3 + 5x2 − 4x + 1 = 0 over C. Hence factorise P (x) = x4 − 4x3 + 5x2 − 4x + 1
over Q and over R.
1
1
4
1
2
2
2
P (x) = x x − 4x + 5 − + 2 = x
+5 .
x + 2 −4 x+
x x
x
x
"
#
2 2
1
1
1
1
Using x +
= x2 + 2 + 2, ⇒ P (x) = x2
−4 x+
+3 .
x+
x
x
x
x
2
at
h
Solution:
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Since 0 is not a zero of P (x),
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in
2
1
1
the solutions of P (x) = 0 are the solutions of x +
−4 x+
+ 3 = 0.
x
x
1
1
by factorising this quadratic P (x) = x2 (x + − 3)(x + − 1)
x
x
2
2
= (x − 3x + 1)(x − x + 1)
∴ P (x) = 0 ⇒ x2 − 3x + 1 = 0 or x2 − x + 1 = 0,
√
√
1±i 3
3± 5
x=
or x =
.
2
2
3. Show that -i is a zero of P (x) = x3 + ix2 − 4x − 4i. Hence factorise P (x) over C.
Solution:
P (−i) = i − i + 4i − 4i = 0 ⇒ (x + i) is a factor of P (x).
By inspection, or by polynomial division, x3 + ix2 − 4x − 4i = (x + i)(x2 − 4).
Hence P (x) = (x + i)(x − 2)(x + 2), and these are irreducible facotr over C.
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Year 12 Topic 1 Worked Answers
Page 12 of 18
Exercise 1.3.1
1. If P (x) = 3x4 + 10x3 + 6x2 + 10x + 3, solve P (x) = 0 over C and factorise P (x) over R.
Solution:
10
3
P (x) = x2 3x2 + 10x + 6 +
+ 2
x
x
1
1
2
2
= x 3 x + 2 + 10 x +
+6 .
x
x
" #
1 2
1
1 2
1
2
2
Using x +
= x + 2 + 2 , ⇒ P (x) = x 3 x +
+ 10 x =
.
x
x
x
x
Since 0 is not a zero of P (x),
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1
1 2
+ 10 x +
= 0.
The solutions of P (x) = 0 are the solutions of 3 x +
x
x
1
1
2
By factoreising this quadratic P (x) = x x +
3 x+
+ 10
x
x
= (x2 + 1)(3x2 + 10x + 3).
M
at
h
Ce
Hence P (x) = 0 ⇒ x2 + 1 = 0 ⇒ x = ±i,
−5 ± 4
1
or 3x2 + 10x + 3 = 0, ⇒ x =
⇒ x = −3 or x = − .
3
3
1
Therefore the zeros of P (x) are − 3, − , ±i.
3
1
2
2
∴ P (x) = (x + 1)3(x + 3)(x + = (x + 1)(x + 3)(3x + 1) over R.
3
in
2. If P (x) = 2x4 + 7x3 + 2x2 − 7x + 2 solve P (x) = 0 over C, and factorise P (x) over R.
1
1
2
Solution: P (x) = x
=x 2 x − 2 +7 x−
+2
x
x
" #
1 2
1
1 2
1
2
2
= x − 2 + 2 , ⇒ P (x) = x 2 x −
+7 x−
+6 .
Using x −
x
x
x
x
2
7
2x + 7x + 2 − + 2
x x
2
2
Yi
m
2
Since 0 is not a zero of P (x).
1 2
1
The solutions of P (x) = 0 are the solutions of 2 x −
+7 x−
+ 6 = 0.
x
x
1
3
1
2
By factorising this quadratic: P (x) = x .2 x −
+2
x−
+
x
x
2
= (x2 + 2x − 1)(2x2 + 3x − 2).
√
Hence P (x) = 0 ⇒ x2 + 2x − 1 = 0 ⇒ x = −1 ± 2,
−3 ± 5
1
or 2x2 + 3x − 2 = 0, ⇒ x =
⇒ x = −2 orx = .
4
2
√
1
Therefoer, the roots of P (x) = 0 are − 2, , −1, ± 2.
2
√
√
∴ P (x) = (x + 1 − 2)(x + 1 + 2)(x + 2)(2x − 1) over R.
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Year 12 Topic 1 Worked Answers
Page 13 of 18
Definition: Let z = cos θ + i sin θ. then by Demoivre’s theorem,
z n = cos nθ + i sin nθ, n = 1, 2, 3, . . .
But by the binomial theorem,
n
X
zn =
n
k
k=0
!
ik sink θ cosn−k θ
Equating real and imaginary part,
cos nθ = cos θ −
n
2
!
n−2
sin θ cos
n
3
θ+
!
n
4
!
sin4 θ cosn−4 θ − . . .
sin3 θ cosn−3 θ +
n
5
!
sin5 θ cosn−5 θ − . . .
Ce
n
sin nθ = ( ) sin θ cosn−1 θ −
1
2
nt
re
n
Using the above method for n = 3, we obtain by De Moivre’s theorem
in
cos 3θ = cos3 θ − 3 sin2 θ cos θ = 4 cos3 θ − 3 cos θ.
1
1
Let x = cos θ. Then cos 3θ =
⇔ 4x3 − 3x =
⇒ 8x3 − 6x − 1 = 0.
2
2
1
Hence if θ is a solution of cos 3θ = , cos θ is a root of 8x3 − 6x − 1 = 0.
2
1
π
cos 3θ =
⇒ 3θ = 2nπ ± , n integral,
2
3
π
θ = (6n ± 1) , n = 0, ±1, ±2, . . .
9
these values of θ give exactly three distinct values of cos θ,
π
5π
4π
7π
2π
namely cos , cos
= − cos and cos
= − cos .
9
9
9
9
9
π
4π
2pi
3
Hence the roots of 8x − 6x − 1 = 0are cos , − cos
and − cos
.
9
9
9
But the coefficient of x2 is zero and hence the sum of the roots is zero, giving
π
2π
4π
cos = cos
+ cos .
9
9
9
Yi
m
Solution:
M
at
h
Example 1.3.2 Use De Moivre’s theorem to show that cos 3θ = 4 cos3 θ − 3 cos θ. Hence solve
8x3 − 6x − 1 = 0. Deduce that cos π9 = cos 2π
+ cos 4π
.
9
9
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Year 12 Topic 1 Worked Answers
Page 14 of 18
Exercise 1.3.2 Use De Moivre’s theorem to show that tan 4θ =
4 tan θ−4 tan3 θ
.
1−6 tan2 θ+tan4 θ
1. Find the general solution of tan 4θ = 1.
Solution:
Let z = cos θ + i sin θ.
Then by de Moivre’s theorem, z 4 = cos 4θ + i sin 4θ.
!
4
X
4
But by the binormial theorem, z 4 =
ik sink θ cos4−k θ.
k
k=0
nt
re
Equating real and imaginary parts,
!
!
!
4
4
4
cos 4θ =
cos4 θ +
(− sin2 θ) cos2 θ +
sin4 θ,
0
2
4
!
!
4
4
sin 4θ =
sin θ cos3 θ +
(− sin3 θ) cos θ.
1
3
h
Ce
sin 4θ
4 sin θ cos3 θ − 4 sin3 θ cos θ
Hence tan 4θ =
=
cos 4θ
cos4 θ − 6 sin2 θ cos2 θ + sin4 θ
4 tan θ − 4 tan3 θ
∴ tan 4θ =
.
1 − 6 tan2 θ + tan4 θ
Thus tan 4θ = 1 ⇒ 4θ = tan−1 1 + nπ, n integer,
at
π
π(4π + 1)
+ nπ, n = 0, ±1, ±2, . . . ⇒ θ =
, n integral.
4
16
M
∴ 4θ =
Let tan θ = x. ⇒ Then tan 4θ = 1 ⇒,
Yi
m
Solution:
in
2. Hence find the roots of the equation x4 + 4x3 − 6x2 − 4x + 1 = 0 in trigonometric form.
4x − 4x3
=1
1 − 6x2 + x4
∴ x4 + 4x3 − 6x2 − 4x + 1 = 0.
π(4n + 1)
, n = 0, ±1, ±2, . . .
16
But of these are only four distict non-zero.
π
5π
3π
7π
values: α = tan , β = tan , γ = − tan , δ = − tan .
16
16
16
16
4
3
2
α, β, γ, and δ are the roots of x + 4x − 6x − 4x + 1 = 0.
from above x = tan
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Year 12 Topic 1 Worked Answers
1.4
Page 15 of 18
Partial Fractions
Definition: The highest common factor of two polynomials A(x) and B(x) is the monic
polynomial of highest degree which is a factor of both A(x) and B(x). It can be found by
repeated use of the division transformation.
Example 1.4.1 Find the highest common factor of A(x) = x4 − 3x3 + 2x2 − 3x − 9 and
B(x) = x3 − 4x2 + 5x − 6.
(1) A(x) ÷ B(x) ⇒
x+1
x3 − 4x2 + 5x − 6
x4 − 3x3 + 2x2 − 3x − 9
− x4 + 4x3 − 5x2 + 6x
nt
re
x3 − 3x2 + 3x − 9
− x3 + 4x2 − 5x + 6
x2 − 2x − 3
Ce
⇒ x4 − 3x3 + 2x2 − 3x − 9 = (x3 − 4x2 + 5x − 6)(x + 1) + (x2 − 2x − 3)
↑
A(x)
=
↑
↑
B(x)
.Q1 (x)
R1 (x)
x3 − 4x2 + 5x − 6
at
x2 − 2x − 3
↑
x −2
h
(2) B(x) ÷ R1 (x) ⇒
− x3 + 2x2 + 3x
in
M
− 2x2 + 8x − 6
2x2 − 4x − 6
4x − 12
⇒ x3 − 4x2 + 5x − 6 = (x2 − 2x − 30)(x − 2) + 4(x − 3)
Yi
m
Solution:
↑
B(x)
=
↑
↑
R1 (x)
.Q2 (x)
1
4x
(3) R1 (x) ÷ R2 (x) ⇒
4x − 12
+
↑
R2 (x)
1
4
x2 − 2x − 3
− x2 + 3x
x −3
−x +3
0
1
⇒ x2 − 2x − 3 = 4(x − 3). (x + 1) + 0
4
↑
↑
↑
R1 (x) =
R2 (x)
.Q3 (x)
The last non-zero remainder 4(x − 3) gives the highest common factor as (x − 3).
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Year 12 Topic 1 Worked Answers
Page 16 of 18
P (x)
Definition: Consider Q(x)
, where deg P < deg Q, and
Q(x) ≡ (x − α1 )(x − α2 ) . . .(x − αn ), α1 , α2 , . . .αn distinct.
Then we can find a constant c1 and a polynomial L1 (x) with deg L1 < deg B.
P (x)
c1
1 (x)
≡ x−α
+ LB(x)
. continuing this process
Such that Q(x)
1
P (x)
c2
c1
cn
+ x−α
+ . . . + x−α
,
≡ x−α
Q(x)
n
1
2
In practice, c1 , c2 , . . . are obtained from the identity by substitution
or by equating coefficients in the usual way.
Solution:
Let
3x−2
(x−1)(x−2)
as a sum of partial fractions.
3x − 2
c−1
c2
≡
+
.
(x − 1)(x − 2)
x−1 x−2
Then 3x − 2 ≡ c1 (x − 2) + c − 2(x − 1).
nt
re
Example 1.4.2 Express
Ce
Putting x = 1 gives c1 = −1, while x = 2 gives c − 2 = 4.
−1
4
3x − 2
≡
+
Hence
(x − 1)(x − 2)
x−1 x−2
An alternative method of obtaining c1 , c2 . .cn in the identity
Yi
m
in
M
at
h
c1
P (x)
c2
cn
≡
+
+ ... +
where Q(x) ≡ (x − α1 )(x − α2 ) . . .(x − αn )
Q(x)
x − α1 x − α2
x − αn
x − α1
x − α1
x − α1
≡ c1 + c2
is to note that P (x).
+ . . .cn
Q(x) − Q(α1 )
x − α2
x − αn
P (α1 )
Since Q(α1 ) = 0.T akinglimitsof bothsidesasx → α1 , c1 = 0
.
Q (α1 )
P (αk )
, where k = 1, 2, . . .n.
Similarly, ck = 0
Q (αk )
3x − 2
c1 = d 2
when x = 1 ⇒ c1 = −1.
(x − 3x + 2)
dx
3x − 2
c2 = d 2
when x = 2 ⇒ c2 = 4.
(x − 3x + 2)
dx
Exercise 1.4.1 Express
Solution:
Let
x2 +1
(x+2)(x−1)(x2 +x+1)
as a sum of partial fractions over R.
x2 + 1
c1
c2
ax + b
≡
+
+ 2
2
(x + 2)(x − 1)(x + x + 1)
x+2 x−1 x +x+1
Then x2 + 1 ≡ c1 (x − 1)(x2 + x + 1) + c2 (x + 2)(x2 + x + 1) + (ax + b)(x + 2)(x + 1)
2
5
Put x = 1 : ⇒ 2 = 9c2 ⇒ c2 = ; Put x = −2 : ⇒ 5 = −9c1 ⇒ c1 = −
9
9
1
Equate coefficients of x3 : ⇒ 0 = c1 + c2 + a ⇒ a =
3
2
x +1
−5
2
1
∴
≡
+
+ 2
2
(x + 2)(x − 1)(x + x + 1)
x+2 x−1 x +x+1
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Year 12 Topic 1 Worked Answers
Page 17 of 18
Exercise 1.4.2
1. Express as a sum of partial fractions
Solution:
Let
2x+10
.
(x−1)(x+3)
2x + 10
c1
c2
=
+
.
(x − 1)(x + 3)
x−1 x+3
Putting x = 1 : ⇒ c1 = 3, Puttting x = −3 : ⇒ c2 = −1.
2x + 10
3
1
∴
=
−
.
(x − 1)(x + 3)
x−1 x+3
4x+5
.
2x2 +5x+3
nt
re
2. Express as a sum of partial fractions
Solution:
x+3
x4 −2x2 +1
Solution:
as a sum of partial fractions.
in
3. Write
M
at
h
Ce
Using the quadratic formula we obtain
3
2
(x + 1) = (2x + 3)(x + 1)
2x + 5x + 3 = 2 x +
2
4x + 5
c1
c2
Let 2
=
+
2x + 5x + 3
2x + 3 x + 1
3
Puttingx = −1 gives c2 = 1, while x = − gives c1 = 2.
2
4x + 5
2
1
=
Hence 2
+
2x + 5x + 3
2x + 3 x + 1
Yi
m
x4 − 2x2 + 1 = (x2 − 1)2 = (x − 1)2 (x + 1)2
x+3
c1
c2
c3
c4
∴ 4
≡
+
+
+
. c1 , c2 , c3 , c4 constants.
2
2
x − 2x + 1
x − 1 (x − 1)
x + 1 (x + 1)2
x + 3 ≡ c1 (x − 1)(x + 1)2 + c2 (x + 1)2 + c3 (x + 1)(x − 1)2 + c4 (x − 1)2 .
1
Put x = 1 : then c2 = 1; Put x = −1 : then c4 =
2
3
Equate coefficients of x :
0 = c1 + c3
3
3
⇒ c3 =
and c1 = −
4
4
Put x = 0 :
3 = −c1 + c3 + c2 + c4
∴
x4
x+3
−3
1
3
1
≡
+
+
+
.
2
2
− 2x + 1
4(x − 1) (x − 1)
4(x + 1) 2(x + 1)2
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Year 12 Topic 1 Worked Answers
Page 18 of 18
Exercise 1.4.3
1. Express as a sum of patrials
Solution:
Let
2x+4
.
(x−2)(x2 +4)
2x + 4
c1
ax + b
=
+ 2
.
2
(x − 2)(x + 4)
x−2 x +4
Then 2x + 4 = c1 (x2 + 4) + (ax + b)(x − 2).
Putting x = 2 gives c1 = 1.
Equate coefficients of x2 : ⇒ 0 = c1 + a = −1 ⇒ a = −1.
Solution:
Let
5−x
.
(2x+3)(x2 +1)
Ce
2. Express as a sum of partial fractions
nt
re
Putting x = 0 : then 4 = 4c1 − 2b ⇒ b = 0.
1
x
2x + 4
=
+ 2
.
∴
2
(x − 2)(x + 4)
x−2 x +4
5−x
c1
ax + b
=
+ 2
2
(2x + 3)(x + 1)
2x + 3 x + 1
M
at
h
Then 5 − x = c − 1(x2 + 1) + (ax + b)(2x + 3).
3
Putting x = − gives c1 = 2.
2
Equate coefficients of x2 : 0 = c1 + 2a ⇒ a = −1.
Yi
m
in
Putting x = 0 : then 5 = c1 + 3b ⇒ b = 1.
5−x
2
1−x
Hence
=
+
.
(2x + 3)(x2 + 1)
2x + 3 x2 + 1
3. Express as a sum of a partial fractions
Solution:
Let
3x2 −3x+2
.
(2x−1)(x2 +1)
3x2 − 3x + 2
c1
ax + b
=
+ 2
2
(2x − 1)(x + 1)
2x − 1 x + 1
Then 3x2 − 3x + 2 = c − 1(x2 + 1) + (ax + b)(2x − 1)
1
Putting x =
⇒ sc1 = 1.
2
Equate coefficients of x2 : 3 = c1 + 2a ⇒ a = 1.
Putting x = 0 : ⇒ 2 = c1 − b ⇒ b = −1.
3x2 − 3x + 2
1
x−1
∴
=
+ 2
.
2
(2x − 1)(x + 1)
2x − 1 x + 1
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