Topic 2. Polynomials.

Level 1.
Problem 1. Find the zeros of P ( x ) : (a) over Q; (b) over R; (c) over C,
if (i) P ( x ) = x 4 − 4 x 2 + 3 = 0 ; (ii) P ( x ) = x 4 − 2 x 2 − 3 .

Answer: (i) (a) ±1, (b, c) ± 1, ± 3 .

(ii) (a) no zeros, (b) ± 3 , (c) ± i, ± 3 .

Explanation: ( i ) In order to solve the polynomial equation P ( x ) = x 4 − 4 x 2 + 3 = 0 , denote x 2

as t and use the quadratic formula. Then x 2 = 1 or x 2 = 3 . Hence P ( x ) = ( x 2 − 1)( x 2 − 3) .
( a ) Irreducible factors of P ( x ) over Q are P ( x ) = ( x − 1)( x + 1)( x 2 − 3) . Each linear factor gives
rise to a zero of P ( x ) . Hence the zeros of P ( x ) over Q are ± 1 .
( b, c ) Irreducible factors of P ( x ) over R and over C are P ( x ) = ( x − 1)( x + 1)( x − 3 )( x + 3 ) .
Hence the zeros of P ( x ) over R and over C are ± 1, ± 3 .
( ii ) Use the quadratic formula, then P ( x ) = x 4 − 2 x 2 − 3 = 0 ⇒ x 2 = −1 or x 2 = 3 .
Hence P ( x ) = ( x 2 + 1)( x 2 − 3) .
( a ) P ( x ) has no linear factors over Q and P ( x ) = 0 has no solutions in the field of rational
numbers. Hence P ( x ) has no zeros over Q.
( b ) Irreducible factors of P ( x ) over R are P ( x ) = ( x 2 + 1)( x − 3 )( x + 3 ) . Each linear factor
gives rise to a zero of P ( x ) . Hence the zeros of P ( x ) over R are ± 3 .
( c ) Irreducible factors of P ( x ) over C are P ( x ) = ( x − i )( x + i )( x − 3 )( x + 3 ) . Hence the zeros
of P ( x ) over C are ± i, ± 3 .

Problem 2. Show that 7 is a root of the equation 2 x 3 − 13x − 7 = 0 . Hence solve the
equation fully over R.

Explanation: Let P ( x ) = 2 x 3 − 13x − 7 , then P( 7 ) = 14 7 − 13 7 − 7 = 0 . Hence x − 7 is a

factor of P ( x ) . By polynomial division, P ( x ) = ( x − 7 )( 2 x 2 + 2 7 x + 1) .
Factorising 2 x 2 + 2 7 x + 1 ,
 5 − 7
P( x) = ( x − 7 ) x −  ( 2 x + 5 + 7 ) . Hence the roots of the equation P ( x ) = 0 are
 2 
− 7± 5
7, .
2

Problem 3. Divide P ( x ) = x 3 − x 2 + x − 1 by x − 1 .

Answer: P ( x ) = ( x − 1)( x 2 + 1) .

Explanation: x 2 + 1 ← quotient
x −1 x3 − x2 + x −1
 x3 − x2
x −1
x −1
0 ← remainder
⇒ ( x 3 − x 2 + x − 1) = ( x − 1)( x 2 + 1) . Also P ( x ) = x 3 − x 2 + x − 1 ⇒ P(1) = 1 − 1 + 1 − 1 = 0 .

Problem 4. Divide P ( x ) = ( x 3 − 3x 2 + 4 x − 2) .

Answer: P ( x ) = ( x + i ){x 2 − ( 3 + i ) x + ( 3i + 3)} + (1 − 3i ) .

Explanation: ( b ) x 2 − ( 3 + i ) x + ( 3i + 3) ← quotient
x + i x 3 − 3x 2 + 4 x − 2

x 3 + ix 2
( −3 − i ) x 2 + 4x − 2
( −3 − i ) x 2 + ( −3i + 1) x
(3i + 3) x − 2
(3i + 3) x + 3i − 3
1 − 3i ← remainder
⇒ ( x − 3x + 4 x − 2) = ( x + i ){x 2 − ( 3 + i ) x + ( 3i + 3)} + (1 − 3i ) .
3 2

Also P ( x ) = x 3 − 3x 2 + 4 x − 2 ⇒ P ( −i ) = i + 3 − 4i − 2 = 1 − 3i .

Problem 5. Express P ( x ) = x 3 + x 2 − 3x − 3 as a product of irreducible factors over

(i) Q; (ii) R; (iii) C.

Answer: (i) P ( x ) = ( x + 1)( x 2 − 3) ,

(ii, iii) P ( x ) = ( x + 1)( x − 3 )( x + 3 ) .

Explanation: (i) P ( x ) = x 3 + x 2 − 3x − 3 = x 2 ( x + 1) − 3( x + 1) = ( x + 1)( x 2 − 3) are irreducible

factors over Q.
(ii, iii) Irreducible factors of P ( x ) over R and also C are P ( x ) = ( x + 1)( x − 3 )( x + 3 ) .

Problem 6. Express P ( x ) = x 4 + 3x 3 − 6 x 2 − 6 x + 8 as a product of irreducible factors over (i) Q;

(ii) R; (iii) C.

Answer: (i) P ( x ) = ( x − 1)( x + 4)( x 2 − 2)

(ii-iii) P ( x ) = ( x − 1)( x + 4)( x − 2 )( x + 2 ) .

Explanation: (i) The only possible rational zeros of P ( x ) are ± 1, ± 2, ± 4, ± 8 (integer divisors of
the constant term 8). But of these, only 1 and − 4 satisfy P ( x ) = 0 . Hence ( x − 1) and ( x − 4) are
factors of P ( x ) . By polynomial division P ( x ) by ( x − 1)( x + 4) = x 2 + 3x − 4 we obtain
P ( x ) = ( x − 1)( x + 4)( x 2 − 2) and these are irreducible factors over Q.
(ii, iii) P ( x ) = ( x − 1)( x + 4)( x 2 − 2) = ( x − 1)( x + 4)( x − 2 )( x + 2 ) .

Problem 7. If P ( x ) = 2 x 3 − x 2 − 6 x + 3 has a rational zero, find all the zeros and factorise P ( x )
fully over the real numbers.

Answer: P ( x ) = ( 2 x − 1)( x − 3 )( x + 3 ) .

Explanation: P ( x ) = 2 x 3 − x 2 − 6 x + 3 . All rational zeros of P ( x ) have the form p / q , where

p and q are integer divisors of 3 and 2 respectively. Hence the only possible rational zeros of
P ( x ) are ± 1, ± 3, ± 3 / 2 . But of these, only 1 / 2 satisfies P ( x ) = 0 . Hence (2 x − 1) is a factor of
P( x) .
By polynomial division, P ( x ) = (2 x − 1)( x 2 − 3) = ( 2 x − 1)( x − 3 )( x + 3 ) , and these are
irreducible factors of P ( x ) over the real numbers. Each linear factor gives rise to a zero of P ( x ) .
Hence the zeros of P ( x ) are 1 / 2, ± 3 .

Problem 8. Given that P ( x ) = 4 x 4 + 8 x 3 + 5x 2 + x − 3 has two rational zeros, find these zeros
and factorise P ( x ) over the real numbers.

Answer: P ( x ) = (2 x − 1)( 2 x + 3)( x 2 + x + 1) .

1 3 3 1
Explanation: The only rational zeros of P ( x ) are ± 1, ± 3, ± , ± , ± . But of these, only and
2 2 4 2
3
− satisfy P ( x ) = 0 . Hence (2 x − 1) and ( 2 x + 3) are factors of P ( x ) . By polynomial division,
2
4 x 4 + 8 x 3 + 5x 2 + x − 3 = ( 2 x − 1)( 2 x + 3)( x 2 + x + 1) , and these are irreducible factors over R, as
( x 2 + x + 1) = 0 has no roots over R and so cannot be factored further over R.

Problem 9. Find P ( x ) , given that P ( x ) is monic, of degree 3, with 5 as a single zero and -2 as a
zero of multiplicity 2.

Answer: P ( x ) = x 3 − x 2 − 16 x − 20 .

Explanation: According to the condition of the problem and factor theorem factors of P ( x ) are
( x − 5) and ( x + 2) 2 .
Hence P ( x ) = ( x − 5)( x + 2) 2 = ( x − 5)( x 2 + 4 x + 4) = x 3 − x 2 − 16 x − 20 .

Problem 10. P ( x ) is an even monic polynomial of degree 4 with integer coefficients. If 2 is a

zero, and the constant term is 6, factorise P ( x ) fully over R.

Answer: P ( x ) = ( x − 2 )( x + 2 )( x − 3 )( x + 3 ) .

Explanation: P ( x ) = x 4 + ax 2 + 6 , as P ( x ) is an even monic polynomial of degree 4 . Then

P ( 2 ) = 0 ⇒ 4 + 2a + 6 = 0 . Hence a = −5 and P ( x ) = x 4 − 5x 2 + 6 .
P ( x ) is even. Hence P ( 2 ) = 0 ⇒ P( − 2 ) = 0 and then ( x − 2 )( x + 2 ) = x 2 − 2 is a factor of
P ( x ) . By inspection, P ( x ) = ( x 2 − 2)( x 2 − 3) . So the irreducible factors of P ( x ) are
P ( x ) = ( x − 2 )( x + 2 )( x − 3 )( x + 3 ) .

Problem 11. If P ( x ) = x 3 − 3x 2 + 4 , show that P ( x ) has a multiple zero and find this zero and
its multiplicity.

Answer: 2 is a multiple zero of P ( x ) .

Explanation: P ( x ) = x 3 − 3x 2 + 4 ⇒ P ′( x ) = 3x 2 − 6 x ⇒ P ′ ( 0) = 0 and P ′( 2) = 0 , but

P (0) ≠ 0, P( 2) = 0 . Hence 2 is a multiple zero of P ( x ) . As P ′′( 2) ≠ 0 , its multiplicity is two.

Problem 12. If P ( x ) = 4 x 3 + 12 x 2 − 15x + 4 has a double zero, find all the zeros and factorise
P ( x ) over the real numbers.

Answer: P ( x ) = ( x + 4)( 2 x − 1) 2 .

Explanation: P ( x ) = 4 x 3 + 12 x 2 − 15x + 4 ,
P ′( x ) = 12 x 2 + 24 x − 15 ,
P ′′( x ) = 24 x + 24 .
⇒ P ′(1 / 2) = 0, P ′ ( −5 / 2) = 0 . But P (1 / 2) = 0, P( −5 / 2) ≠ 0 ⇒ 1 / 2 is a multiple zero. As
P ′′(1 / 2) ≠ 0, 1 / 2 is double zero, and (2 x − 1) 2 is a factor of P ( x ) . By polynomial division
P ( x ) = ( x + 4)( 2 x − 1) 2 . These are irreducible factors over R.
− 4 is a single zero, 1 / 2 is a double zero.

Problem 13. If P ( x ) = x 3 − 3x 2 − 9 x + c has a double zero, find c and factorise P ( x ) over the
real numbers.

Answer: c = 27 P( x ) = ( x − 3) 2 ( x + 3) .

Explanation: P ( x ) = x 3 − 3x 2 − 9 x + c ,
P ′ ( x ) = 3x 2 − 6 x − 9 ,
P ′′( x ) = 6 x − 6 .
⇒ P ′ ( −1) = 0, P ′′( −1) ≠ 0, P ′ ( 3) = 0, P ′′ (3) ≠ 0 . Hence both − 1 and 3 can be a double zero of
P ( x ) . Let − 1 be a double zero of P ( x ). ⇒ P ( x ) = ( x + 1) 2 ( x + k ) for some constant k , as P ( x ) is
a monic polynomial of degree 3 . P (0) = c ⇒ k = c. P( −1) = 0 ⇒ c = −5 ⇒ P ( x ) = ( x + 1) 2 ( x − 5) .
Let 3 be a double zero of P ( x ). ⇒ P ( x ) = ( x − 3) 2 ( x + l ) for some constant l ,
c
P (3) = 0 ⇒ c = 27. P (0) = c ⇒ l = = 3. ⇒ P( x ) = ( x − 3) 2 ( x + 3) .
9

Problem 14. If ax 3 + cx + d = 0 has a double root, show that 4c 3 + 27ad 2 = 0 .

Explanation: P ( x ) = ax 3 + cx + d ,
P ′ ( x ) = 3ax 2 + c ,
P ′′( x ) = 6ax .
 − c −c
⇒ P′ e  = 0 , where e:= ±1 . Hence e can be a double root.
 3a  3a
3 −1/ 2
 − 3   − c  1/ 2   − c
1/ 2
 − c −2
P e  = 0 ⇒ a e     + ce   + d = 0 ⇒ d   = ec ⇒
        
3a  3a  3a 3a 3

⇒ 4c 3 + 27ad 2 = 0 .

Problem 15. Find the remainder when P ( x ) = x 3 + 2 x 2 + 1 is divided by (a) x + i ; (b) x 2 + 1 .

Answer: (a) R = −1 + i ; (b) R = − x −1 .

Explanation: ( a ) x + i is a linear divisor. Hence we can use a remainder theorem, and the
remainder is P ( −i ) = ( −i ) 3 + 2( −i ) 2 + 1 = i − 2 + 1 = −1 + i .
( b ) P ( x ) = x 3 + 2 x 2 + 1 and D( x ) = x 2 + 1 are polynomials over Q. By the division
transformation, P ( x ) ≡ D( x ) S ( x ) + R( x ) where R( x ) is a polynomial over Q, such that degree R<
< degree D =2. Thus P ( x ) ≡ ( x 2 + 1) S ( x ) + ax + b, a , b rational. ⇒ P (i ) = 0 + ai + b , and hence
ai + b = i 3 + 2i 2 + 1 = −1 − i ⇒ a = −1, b = −1 . Hence the remainder ax + b is − x − 1 .

Problem 16. When P ( x ) = x 4 + ax 2 + 2 x is divided by x 2 + 1 , the remainder is 2 x + 3 . Find the

value of a.

Answer: a = −2 .

Explanation: By the division transformation, x 4 + ax 2 + 2 x = ( x 2 + 1) S ( x ) + 2 x + 3 . Substituting

x = i , 1 − a + 2i = 2i + 3 ⇒ a = −2 .

Problem 17. When P ( x ) = x 4 + ax 2 + bx + 2 is divided by x 2 + 1 , the remainder is − x + 1 . Find

the values of a and b.

Answer: a = 2, b = −1 .

Explanation: By the division transformation, x 4 + ax 2 + bx + 2 = ( x 2 + 1) S ( x ) − x + 1 . Substituting

x = i , 1 − a + bi + 2 = −i + 1, that is − a + bi = −i − 2 ⇒ a = 2, b = −1 .

Problem 18. Find the zeros of P ( x ) = x 4 + x 3 − x 2 + x − 2 over C, given that i is a zero. Factor
P ( x ) fully over R.

Answer: P ( x ) = ( x 2 + 1)( x − 1)( x + 2) .

Explanation: P ( x ) has real coefficients. Hence P (i ) = 0 ⇒ P ( −i ) = 0 and then
( x − i )( x + i ) = x 2 + 1 is a factor of P ( x ) . By the division transformation
P ( x ) = x 4 + x 3 − x 2 + x − 2 = ( x 2 + 1)( x 2 + x − 2). ⇒ P( x ) = ( x 2 + 1)( x − 1)( x + 2) . This is the
factorization of P ( x ) into irreducible factors over R, and P ( x ) has zeros i , − i , − 2 and 1 over C.

Problem 19. If P ( x ) = x 4 − 2 x 3 − x 2 + 6 x − 6 has a zero 1 − i , find the zeros of P ( x ) over C, and

factorise P ( x ) fully over R.

Answer: P ( x ) = ( x 2 − 2 x + 2)( x − 3 )( x + 3 ) .

Explanation: P ( x ) has real coefficients. Hence P (1 − i ) = 0 ⇒ P(1 + i ) = 0 and then

[ x − (1 − i )][ x − (1 + i )] = x 2 − 2 x + 2 is a factor of P ( x ) . By polynomial division,
P ( x ) = ( x 2 − 2 x + 2)( x 2 − 3) . Hence P ( x ) = ( x 2 − 2 x + 2)( x − 3 )( x + 3 ) , this is the factorization
of P ( x ) into irreducible factors over R, and P ( x ) has zeros 1 ± i, ± 3 .

Problem 20. 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.

Answer: P ( x ) = ( x 2 + 4)( x − 2 )( x + 2 ) .

Explanation: P ( x ) is an even monic polynomial of degree 4 . Hence P ( x ) = x 4 + ax 2 + b. P ( x )

has real coefficients. Hence P (2i ) = 0 ⇒ P ( −2i ) = 0 and then ( x − 2i )( x + 2i ) = x 2 + 4 is a factor of
P ( x ) ⇒ P ( x ) = ( x 2 + 4)( x 2 + c) . The product of zeros of P ( x ) is − 8 . Hence 4c = −8 ⇒ c = −2 , and
P ( x ) = ( x 2 + 4)( x 2 − 2) = ( x 2 + 4)( x − 2 )( x + 2 ) . These are irreducible factors of P ( x ) over R,
and P ( x ) has zeros − 2i , + 2i , − 2 and 2 over C.

Problem 21. Find the monic polynomial of degree 3 with zeros 1, 2 and 3.

Answer: P ( x ) = x 3 − 6 x 2 + 11x − 6 .

Explanation: P ( x ) = x 3 + ax 2 + bx + c , since P ( x ) is the monic of degree three. If

α = 1, β = 2, γ = 3 denote the zeros of P ( x ) , then
a = − ∑ α = −(1 + 2 + 3) = −6,
b = ∑ α β = 2 + 3 + 6 = 11,
c = − ∑ α β γ = −6.
Hence P ( x ) = x 3 − 6 x 2 + 11x − 6.

Problem 22. Two of the roots of 3x 3 + ax 2 + 23x − 6 = 0 are reciprocals. Find the value of a and
the three roots.

1
Answer: a = −16 ; the roots are 3, , 2 .
3
 1 −6
Explanation: Let the roots of P ( x ) be α, , β. Then product of roots is β. ⇒ β = − = 2 . Sum of
α 3
2 23 1
products taken two at a time is 1 + + 2α = ⇒ α = 3, .
α 3 3
1 −a 1
Sum of the roots is 3 + + 2 = ⇒ a = −16 . Hence the roots are 3, , 2 and the coefficient a is
3 3 3
− 16 .

Problem 23. Two of the roots of x 3 − 3x 2 − 4 x + a = 0 are opposites. Find the value of a and the
three roots.

Answer: a = 12 ; the roots are − 2, 2, 3 .

Explanation: Let the roots of P ( x ) = x 3 − 3x 2 − 4 x + a be α, − α , β.

Then sum of the roots is β. ⇒ β = 3 . Sum of products taken two at a time is
− α 2 + 3α − 3α = −α 2 ⇒ − α 2 = −4 ⇒ α = −2, 2 .
Product of the roots is − 2 ⋅ 2 ⋅ 3 = − a ⇒ a = 12 .
Hence the roots are − 2, 2, 3 and the coefficient a = 12 .

Problem 24. The equation px 3 + qx 2 + rx + s = 0 has roots (a − c), a, (a + c), which are in
−q
arithmetic progression. Show that the a = and hence show that 2q 3 − 9 pqr + 27 p 2 s = 0 .
3p

−q −q
Explanation: The sum of roots is = ∑ α = ( a − c) + a + (a + c) = 3a ⇒ a = .
p 3p
 − q  − p ⋅ q 3 q ⋅ q 2 rq
Hence 0 = P( a ) = P = + − − s ⇒ 0 = P (a ) ⋅ 27 p 2
 3p  27 p 3
9p 2 3p
= 2q 3 − 9 pqr + 27 p 2 s .

Problem 25. Solve the equation 18x 3 + 27 x 2 + x − 4 = 0 . Given that the roots are in arithmetic
progression.

4 1 1
Answer: − ; − ; .
3 2 3

Explanation: Let the roots be a − c, a , a + c . Then

27 1
− = ∑ α = (a − c) + a + (a + c) = 3a ⇒ a = − .
18 2
4 11  5 5
= ∑ α ⋅ β ⋅ γ = ( a − c) a (a + c) = −  − c 2  ⇒ c = (or c = − that gives the same roots).
18 2 4  6 6
4 1 1
Hence the roots are a − c = − , a = − and a + c = .
3 2 3
 a
Problem 26. The equation px 3 + qx 2 + rx + s = 0 has the roots ac, a and , which are in
c
geometric progression. Show that a = 3 (− s / p) and hence show that pr 3 − q 3 s = 0 .

= ∑ α ⋅ β ⋅ γ = ac ⋅ a ⋅ = a 3 ⇒ a = 3 ( − s / p) .
s a
Explanation: The product of the roots is −
p c
−q a  1
The sum of the roots is = ∑ α = a ⋅ c + a + = a  c + 1 +  , and the product of the roots taken
p c  c
r a2  1
two at a time is = ∑ α ⋅ β = a 2c + a 2 + = a2 c + 1 +  .
p c  c
−q
a = ⇒ − q ⋅ 3 ( − s / p) = r ⇒ pr 3 − q 3 s = 0 .
r
Hence
p p

Problem 27. The equation x 3 + 3 x 2 − 2 x − 2 = 0 has roots α, β and γ . Find the equations with the
roots (a) α − 2, β − 2 and γ − 2 ; (b) α 2 , β 2 and γ 2 .

Answer: (a) x 3 + 9 x 2 + 22 x + 14 = 0 ; (b) x 3 + 9 x 2 + 22 x + 14 = 0 .

Explanation: (a ) 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 x 3 + 9 x 2 + 22 x + 14 = 0 .

( ) + 3( x )
3
1/ 2 2
( b ) The values α 2 , β 2 and γ 2 satisfy x 1/2 − 2 x 1/ 2 − 2 = 0 . Rearrangement gives

x 1/ 2 ( x − 2) = 2 − 3x . Squaring we obtain x ( x − 2) 2 = ( 2 − 3x ) 2 and hence the required equation is

x 3 + 9 x 2 + 22 x + 14 = 0 .

Problem 28. The equation x 3 + x 2 − 2 x − 3 = 0 has roots α, β and γ . Find the equations with
α β γ
roots (a) , and ; (b) α + 2, β + 2 and γ + 2 .
2 2 2

Answer: (a) 8 x 3 + 4 x 2 − 4 x − 3 = 0 ; (b) x 3 − 5x 2 + 6 x − 3 = 0 .

α β γ
Explanation: ( a ) , and satisfy (2 x ) 3 + (2 x ) 2 − 2(2 x ) − 3 = 0 . Hence the required equation
2 2 2
3 2
is 8 x + 4 x − 4 x − 3 = 0 .
( b ) α + 2, β + 2 and γ + 2 satisfy ( x − 2) 3 + ( x − 2) 2 − 2( x − 2) − 3 = 0 . Hence the required
equation is x 3 − 5x 2 + 6 x − 3 = 0 .

Problem 29. The equation x 3 + px + q = 0 has roots α, β and γ . Find the monic cubic equation
with roots α 2 , β 2 and γ 2 .

Answer: x 3 + 2 px 2 + p 2 x − q 2 = 0 .
Explanation: If α, β, and γ satisfy x 3 + px + q = 0 , then α 2 , β 2 , and γ 2 satisfy

(x )
1/ 2 3
+ px 1/ 2 + q = 0 .

Rearrangements gives x 1/ 2 ( x + p) = −q .
Squaring gives x ( x + p) 2 = q 2 .
Simplifying, we get x 3 + 2 px 2 + p 2 x − q 2 = 0 .

Problem 30. The equation x 3 + x 2 + 2 = 0 has roots α, β and γ . Evaluate (a) α + β + γ ; (b)
α 2 + β 2 + γ 2 ; (c) α 3 + β 3 + γ 3 ; (d) α 4 + β 4 + γ 4 .

Answer: (a) α + β + γ = −1 ; (b) α 2 + β 2 + γ 2 = 1 ; (c) α 3 + β 3 + γ 3 = −7 ; (d) α 4 + β 4 + γ 4 = 9 .

Explanation: ( a ) α + β + γ = ∑ α = −1 .

( ) + (x )
3 1/ 2 2
( b ) α 2 , β 2 , and γ 2 are roots of the equation x 1/ 2 +2 = 0.

Rearrangement gives x 1/2 x = −2 − x . Squaring and simplifying, x 3 − x 2 − 4 x − 4 = 0 .

Hence α 2 + β 2 + γ 2 = 1 .
( c ) α3 + α2 + 2 = 0 , (since α, β, γ are roots of the given equation)
β 3 + β 2 + 2 = 0 , γ 3 + γ 2 + 2 = 0 . Hence (α 3 + β 3 + γ 2 ) + (α 2 + β 2 + γ 2 ) + 6 = 0 .
From ( b ) α 2 + β 2 + γ 2 = 1 , therefore α 3 + β 3 + γ 3 = −7 .
( d ) From ( b ) α 2 , β 2 , and γ 2 satisfy x 3 − x 2 − 4x − 4 = 0 .

( ) ( ) ( ) ( ) ( )
2 2 2 3 2
Hence α 4 = α 2 , β4 = β2 and γ 4 = γ 2 satisfy x 1/2 − x 1/ 2 − 4 x 1/ 2 − 4 = 0 .

Rearrangement gives x 1/ 2 ( x − 4) = x + 4 . Squaring and simplifying, x 3 − 9 x 2 + 8 x − 16 = 0 .

α 4 + β 4 + γ 4 is the sum of roots of this equation. Hence α 4 + β 4 + γ 4 = 9 .

Problem 31. Show that -i is a zero of P ( x ) = x 3 + ix 2 − 4 x − 4i . Hence factorise P ( x ) over C.

Answer: P ( x ) = ( x + i )( x − 2)( x + 2) .

Explanation: P ( −i ) = i − i + 4i − 4i = 0 ⇒ ( x + i ) is a factor of P ( x ) . By inspection, or by

polynomial division, x 3 + ix 2 − 4 x − 4i = ( x + i )( x 2 − 4) . Hence P ( x ) = ( x + i )( x − 2)( x + 2) , and
these are irreducible factors over C.

Problem 32. If P ( x ) = 3x 4 + 10 x 3 + 6 x 2 + 10 x + 3 , solve P ( x ) = 0 over C, and factorise P ( x )

over R.

1
Answer: P ( x ) = ( x 2 + 1)( x + 3)( 3x + 1) ; the zeros of P ( x ) are − 3, − , ± i .
3

 10 3   1  1 
Explanation: P ( x ) = x 2  3x 2 + 10 x + 6 + +  = x 2 3 x 2 +  + 10 x +  + 6 .
 x x2    x 2  x 
 2   2
1  
2
Using  x +  = x 2 +
1 1 1 
+ 2, P ( x ) = x 2 3 x +  + 10 x +   . Since 0 is not a zero of
 x 2
  x  x 
x 
2
 1  1
P ( x ) , the solutions of P ( x ) = 0 are the solutions of 3 x +  + 10 x +  = 0 . By factorising this
 x   x
quadratic
 1   1 
P ( x ) = x 2  x +  3 x +  + 10 = ( x 2 + 1)( 3x 2 + 10 x + 3) . (1)
 x   x 
Hence
P( x) = 0 ⇒x 2 + 1 = 0 or 3x 2 + 10 x + 3 = 0
− 5± 4
x = ±i x=
3
1
x = −3 or x = − .
3
1
Therefore, the zeros of P ( x ) are − 3, − , ± i .
3
From (1) P ( x ) = ( x + 1) 3 ( x + 3)( x + 1 / 3) = ( x 2 + 1)( x + 3)(3x + 1) over R.
2

Problem 33. Show that the zeros of P ( x ) = x 4 + x 2 + 1 are included in the zeros of x 6 − 1 .
Hence factorise P ( x ) over R.

Answer: P ( x ) = ( x 2 − x + 1)( x 2 + x + 1). .

Explanation:
{
x 6 − 1 = ( x − 1)( x 5 + x 4 + x 3 + x 2 + x + 1) = ( x − 1) x ( x 4 + x 2 + 1) + ( x 4 + x 2 + 1)}
( ) ( )( )
3
= ( x − 1)( x + 1)( x 4 + x 2 + 1) or x 6 − 1 = x 2 − 1 = x 2 − 1 x 4 + x 2 + 1 .

Hence x 6 − 1 = 0 ⇒ x = ±1 or x 4 + x 2 + 1 = 0 . Further more, the sixth roots of unity are equally

π
spaced by around a circle of radius 1 and center (0,0) in the Argand diagram. The zeros of
3
P ( x ) = x 4 + x 2 + 1 are the non-real sixth roots of unity which are:

π π
z 2 and z6 = z 2 , where z2 = cos + i sin , y
3 3
2π 2π
z 3 and z5 = z3 , where z3 = cos + i sin . z3 1
z2
3 3
π/3
Hence z4 z1 x
-1 1
P ( x ) = x + x + 1 = ( x − z2 )( x − z 2 )( x − z 3 )( x − z3 )
4 2

π 2π
x + 1)(zx52 − 2 cos z 6 x + 1).
2 2
= ( x 2 − 2 Re z2 x + z 2 )( x 2 − 2 Re z 3 x + z 3 ) = ( x 2 − 2 cos
3 -1 3

Using
π 1 2π π 1
cos = , cos = − cos = − , x 4 + x 2 + 1 = ( x 2 − x + 1)( x 2 + x + 1).
3 2 3 3 2
These factors are irreducible over R.

Problem 34. Solve the equation z 5 − 16z = 0 over C.

Answer: The roots are 0, ± 2, ± 2i .

.
Explanation:

z 5 − 16z = 0 ⇒ z( z 4 − 16) = 0 . Hence z = 0 or z is a

y
complex root of 16 . Clearly, one such root is
z1
2, as 2 4 = 16 . The other three roots are equally spaced
π
by around a circle of radius 2 and center (0,0) in π/ 2
2 z2 0 x
the Argand diagram. The fourth roots of z0=2
16 are ± 2 and ± 2i . Hence z 5 − 16z has the roots
0, ± 2, ± 2i .
z3
Problem 35. Show that cos 4θ = 8 cos θ − 8 cos θ + 1 .
4 2

π 5π
Hence solve the equation 8 x 4 − 8 x 2 + 1 = 0 and deduce the exact values of cos and cos .
8 8

1 1
Answer: The roots are ± 2+ 2 ,± 2− 2 ;
2 2
π 1 5π 1
cos = 2 + 2 , cos =− 2− 2 .
8 2 8 2

Explanation: Let z = cos θ + i sin θ . Then by de Moivre’s theorem, z 4 = cos 4θ + i sin 4θ . But by
4
 4
the Binomial theorem, z 4 = (cos θ + i sin θ) 4 = ∑  k  i k sin k θ cos4 − k θ . Equating real parts,
k =0
cos 4θ = cos θ − 6 cos θ sin θ + sin θ = 8 cos θ − 8 cos 2 θ + 1 .
4 2 2 4 4

( a ) Let cosθ = x . Then cos 4θ = 0 ⇔ 8 x 4 − 8 x 2 + 1 = 0 . Hence if θ is a solution of cos 4θ , cosθ

is a root of 8 x 4 − 8 x 2 + 1 = 0 .
π π π
But cos 4θ = 0 ⇒ 4θ = ± + 2 πn, n integral θ = ± + n, n = 0,±1,±2, K
2 8 2
These values of θ give exactly four distinct values of cos θ , namely
π 5 3 9 π 13 3
cos , cos π = − cos π, cos π = − cos , cos π = cos π .
8 8 8 8 8 8 8
At the same time considering 8 x − 8 x + 1 = 0 as a quadratic in x 2 ,
4 2

4± 8 2± 2
x2 = = ,
8 4
1 1
x=± 2 + 2 or x = ± 2 − 2.
2 2
π 3π π 1 5π 9π
But x = cosθ . Since cos > cos > 0 , we deduce cos = 2 + 2 . Since 0 > cos > cos ,
8 8 8 2 8 8
5π 1
we deduce cos = − 2− 2 .
8 2
 1 1 1
( b ) Let cosθ = x . Then cos 4θ = ⇔ 8 x 4 − 8 x 2 + 1 = . Hence if θ is a solution of cos 4θ = ,
2 2 2
cosθ is a root of 16 x 4 − 16 x 2 + 1 = 0 .
1 π π π
But cos 4θ = ⇒ 4θ = ± + 2 πn, n integral θ = ± + n, n = 0,±1,±2, K
2 3 12 2
These values of θ give exactly four distinct values of cosθ , namely
π 5 7 5 13 π
cos ,cos π, cos π = − cos π ,cos π = − cos .
12 12 12 12 12 12
At the same time considering 16 x − 16 x + 1 = 0 as a quadratic in x 2 ,
4 2

8 ± 48 2 ± 3
x2 = = ,
16 4
1 1
x=± 2 + 3 or x = ± 2 − 3.
2 2
π 5π π 1 5π 1
But x = cosθ . Since cos > cos > 0 , we deduce that cos = 2 + 3 , cos = 2− 3.
12 12 12 2 12 2

4 tan θ − 4 tan 3 θ
Problem 36. Use de Moivre’s theorem to show that tan 4θ = .
1 − 6 tan 2 θ + tan 4 θ
(i) Find the general solution of tan 4θ = 1 .
(ii) Hence find the roots of the equation x 4 + 4 x 3 − 6 x 2 − 4 x + 1 = 0 in trigonometric form.

π(4n + 1) π 5 3 7
Answer: (i) θ = , n integral; (ii) tan , tan π , − tan π, − tan π .
16 16 16 16 16

Explanation: Let z = cos θ + i sin θ . Then by de Moivre’s theorem, z 4 = cos 4θ + i sin 4θ . But by
4
 4
the Binomial theorem, z 4 = ∑  k  i k sin k θ cos 4 − k θ . Equating real and imaginary parts,
k =0
 4
 0
 4
 2
( )  4
cos 4θ =   cos 4 θ +   − sin 2 θ cos 2 θ +   sin 4 θ ,
 4
 4
1
(
 4
)
sin 4θ =   sin θ cos 3 θ +   − sin 3 θ cos θ . Hence
 3
sin 4θ 4 sin θ cos 3 θ − 4 sin 3 θ cos θ 4 tan θ − 4 tan 3 θ
tan 4θ = = ⇒ tan 4θ = . (1)
cos 4θ cos 4 θ − 6 sin 2 θ cos 2 θ + sin 4 θ 1 − 6 tan 2 θ + tan 4 θ
π
(i) tan 4θ = 1 ⇒ 4θ = tan −1 1 + πn, n integer, ⇒ 4θ = + πn, n = 0, ± 1, ± 2,... ⇒
4
π(4n + 1)
θ= , n integral. (2)
16
4x − 4x3
(ii) Let tan θ = x . Then from (1) tan 4θ = 1 ⇔ = 1 ⇔ x 4 + 4 x 3 − 6x 2 − 4 x + 1 = 0 .
1 − 6x + x
2 4

π(4n + 1)
Hence from (2) x = tan , n = 0, ± 1, ± 2,... But of these there are only four distinct non-
16
π 5 3 7
zero values: α = tan , β = tan π , γ = − tan π , δ = − tan π .
16 16 16 16
α, β, γ and δ are the roots of x 4 + 4 x 3 − 6x 2 − 4 x + 1 = 0 .
 2 x + 10
Problem 37. Express as a sum of partial fractions .
( x − 1)( x + 3)

2 x + 10 3 1
Answer: = − .
( x − 1)( x + 3) x − 1 x + 3

2 x + 10 c c
Explanation: Let = 1 + 2 .
( x − 1)( x + 3) x − 1 x + 3
Then 2 x + 10 = c1 ( x + 3) + c2 ( x − 1) . Putting x = 1 gives c1 = 3 , while x = −3 gives c2 = −1 .
2 x + 10 3 1
Hence = − .
( x − 1)( x + 3) x − 1 x + 3

4x + 5
Problem 38. Express as a sum of partial fractions .
2 x + 5x + 3
2

4x + 5 2 1
Answer: = + .
2 x + 5x + 3
2 2x + 3 x + 1

Explanation: Using the quadratic formula, we get

 3
2 x 2 + 5x + 3 = 2 x +  ( x + 1) = (2 x + 3)( x + 1) .
 2
4x + 5 c c
Let = 1 + 2 . Then 4 x + 5 = c1 ( x + 1) + c2 (2 x + 3) .
( 2 x + 3)( x + 1) 2 x + 3 x + 1
Putting x = −1 gives c2 = 1 , while x = −3 / 2 gives c1 = 2 .
4x + 5 2 1
Hence = + .
2 x + 5x + 3 2 x + 3 x + 1
2

2x + 4
Problem 39. Express as a sum of partial fractions .
( x − 2)( x 2 + 4)

2x + 4 1 x
Answer: = − .
2
( x − 2)( x + 4) ( x − 2) ( x + 4)
2

2x + 4 c1 ax + b
Explanation: Let = + .
2
( x − 2)( x + 4) x − 2 x2 + 4
Then 2 x + 4 = c1 ( x 2 + 4) + (ax + b)( x − 2) . Putting x = 2 gives c1 = 1 .
Equate coefficients of x 2 : 0 = c1 + a ⇒ a = −1 . Put x = 0 : then 4 = 4c1 − 2b ⇒ b = 0 .
2x + 4 1 x
Hence = − .
2
( x − 2)( x + 4) ( x − 2) ( x + 4 )
2

5− x
Problem 40. Express as a sum of partial fractions .
( 2 x + 3)( x 2 + 1)
 5− x 2 1− x
Answer: = + .
2
( 2 x + 3)( x + 1) 2x + 3 x2 + 1

5− x c1 ax + b
Explanation: Let = + .
( 2 x + 3)( x 2 + 1) 2x + 3 x 2 +1
Then 5 − x = c1 ( x 2 + 1) + ( ax + b)(2 x + 3) . Putting x = −3 / 2 gives c1 = 2 .
Equate coefficients of x 2 : 0 = c1 + 2a ⇒ a = −1 . Put x = 0 : then 5 = c1 + 3b ⇒ b = 1 .
5− x 2 1− x
Hence = + .
( 2 x + 3)( x 2 + 1) 2x + 3 x2 + 1