CHAPTER 2 SOLUTIONS TO REINFORCEMENT EXERCISES IN
ALGEBRA
2.3.1 Multiplication of linear expressions
2.3.1A.
Identify which of the following are algebraic expressions in the variables
involved
i)
14
ii)
x–3
iii)
2t + 1 = 0
iv) 2x + y
3
2
v)
x – 2x + x – 1
vi) ex + e–x
ix)
x2 – 3x + 2 = 0
2
x+1
xi)
s3 – 2s2 + s = 0
xii)
4x + 1
x – 2x + 4
2 cos x = 1
x–y
x+y = 0
xiv)
u2 + 2uv – 3v
xvi)
x ln x + x2 = 0
vii)
viii)
x)
1
xiii)
xv)
2 x – x2
sin x – 3 cos x
2
Solution
i)
14, or any other number can be regarded as an algebraic expression
in any variable.
ii)
x  3 uses only numbers, symbols (x) and the arithmetic operation
() and is therefore algebraic. It is a linear expression or function.
iii)
2t + 1 = 0 is not an expression, but an algebraic equation that
equates the linear algebraic expression 2t + 1 to zero. It tells us that
1
t=2 .
iv)
2x + y is a linear algebraic expression in both x and y.
v)
x3  2x2 + x  1 contains only powers (repeated multiplication) and +
and  and so is algebraic  it is in fact a cubic polynomial.
vi)
The exponential function ex (Chapter 4) cannot be expressed in
terms of arithmetic operations on x except by infinite processes
(such as an infinite power series). Therefore neither ex or its inverse
ex = 1/ex are algebraic expressions, so neither is ex + ex. ex is in fact
called a transcendental function which is any function that cannot
be expressed algebraically in terms of the variable x.
vii)
x2  3x + 2 = 0 is an algebraic equation equating the quadratic
expression x2  3x + 2 to zero. It tells us that x must be either 1 or 2.
viii)
ix)
x  x2 = 2x
2
1
2
 x2 is an algebraic expression  roots are allowed.
2
x + 1 requires only division and addition and so is algebraic  it is
called a rational function.
x)
Similarly to ex, sin x and cos x are not algebraic expressions  they
are in fact called trigonometric functions or circular functions
(Chapter 6).
xi)
1
3
s  2s2 + s = 0 is an algebraic equation in s obtained by equating
1
3
the algebraic expression s  2s2 + s to zero. This is not so easy to
solve for s!
xii)
xiii)
x2
4x + 1
is rational (algebraic) function.
 2x + 4
2 cos x = 1 is an equation, equating the transcendental expression 2
1
cos x (not algebraic) to 1. It tells us that cos x = 2 .
xiv)
u2 + 2uv  3v is algebraic in both u and v.
xv)
xy
x + y = 0 is an algebraic equation in x and y. It tells us that x = y
(note that we cannot have x + y = 0 here, so x = y = 0 is not allowed).
xvi)
x ln x + x2 = 0 is an equation (which can only be solved for x by
numerical methods  the ‘obvious’ solution x = 0 is not allowed
because ln x is not defined for x = 0. You can check that it does have
a solution by sketching the graph of y = ln x and y =  x and looking
for the intersection of the graphs. x ln x + x2 is not algebraic).
2.3.1B
Identify the algebraic equations in 2.3.1A.
Solution
As noted in the solution to Question 2.3.1A the algebraic equations are iii),
vii), xi), xv).
2.3.1C
For each of the pair of expressions, insert brackets in the one on the left to
make it identically equal to the one on the right:i) a+bc+d
a+bc+bd
viii) a+bc+bd
a+bc+b2d
ii) a+bc+d
ac+bc+d
ix) a+b c+bd ad+bcd+bd
iii) a+bc+d
ac+bc+ad+bd
x) a–bc–d
ac–bc–ad+bd
iv) a–bc+d
a–bc–bd
xi) a–b c–d
a–bc+bd
2
v) a–bc–d
ac–bc–d
xii) x –3 x+4
x3–3x+4
vi) a–bc+d
ac–bc+d
2
2
vii) x –3x+4
x –3x–12
Solution
This exercise essentially tests the distributive rule and use of brackets.
Basically it requires finding ways to insert brackets on the left hand side to
produce the right hand side.
i)
a + b (c + d) = a + bc + bd
ii)
(a + b)c + d = ac + bc + d
iii)
(a + b)(c + d)
iv)
a  b(c + d) = a  bc  bd
v)
(a  b)c  d = ac  bc  d
vi)
(a  b)c + d = ac  bc + d
vii)
x2  3(x + 4) = x2  3x  12
viii)
a + b(c + bd)
= ac + bc + ad + bd
= a + bc + b2d
ix)
(a + bc + b)d = ad + bcd + bd
x)
(a  b)(c  d) = ac  bc  ad + bd
xi)
a  b(c  d) = a  bc + bd
xii)
(x2  3)x + 4 = x3  3x + 4
2.3.1D
Remove the brackets in the following expressions
i)
iii)
v)
vii)
ix)
xi)
xiii)
xv)
2(x + 2)
3t(t – 1)
a2(a – 3)
– 2u(u2 + 3)
(a2 – 1)(a + 2) – 3(a – 3)
– (x – x2)(x – 2)
(1 – t)(1 – s)(1 – u)
(x – y)2 + (x + y)2
ii)
iv)
vi)
viii)
x)
xii)
xiv)
3(x – 1) – (x – 4)
(s – t)(s + 2t)
(x2 + 2x – 1)(x – 1)
9(x2 – 3) – 2(x + 4)
x(x – 1)(x + 2) – 3x2
– [(x2 – 1)(x – 2) – (x – 3)(x + 2)]
(a – 2b)2 – (a + 2b)2
Solution
This
‘removing the brackets’ is essentially repeated application of the
distributive rule.
a(b + c) = ab + ac
Just be careful with signs and squares.
i)
ii)
2(x + 2)
= 2x + 2  2 = 2x + 4
3(x  1)  (x  4) = 3x  3  x + 4
(note the sign change)
= 2x + 1
collecting ‘like terms’ together.
iii)
3t(t  1) = 3t  t  3t = 3t2  3t
(using power notation)
iv)
(s  t)(s + 2t) = s2  ts + s(2t)  t(2t)
= s2  st + 2st  st2
=
s2 + st  2t2
v)
a2(a  3)
= a3  3a2
vi)
(x2 + 2x  1) (x  1) = x3 + 2x2  x  x2  2x + 1
= x3 + x2  3x + 1
vii)
 2u(u2 + 3) =  2u3  6u
viii)
9(x2  3)  2(x + 4) = 9x2  27  2x  8
= 9x2  2x  35
ix)
(a2  1)(a + 2)  3(a  3) = a3 + 2a2  a  2  3a + 9
= a3 + 2a2  4a + 7
x)
x(x  1) (x + 2)  3x2 = x(x2 + x  2)  3x2
= x3 + x2  2x  3x2
= x3  2x2  2x
xi)
 (x  x2) (x  2) = (x2  x) (x  2)
(switching signs makes it slightly easier)
= x3  2x2  x2 + 2x
= x3  3x2 + 2x
xii)
 [(x2  1) x  2)  (x  3) (x + 2)] = (x  3) (x + 2)  (x2  1) (x  2)
(again, note the tidying up of the signs before proceeding)
= x2  x  6  (x3  2x2  x + 2)
(note keeping the sign in mind by retaining the brackets in the
expanded expression)
  
  
= x2  x  6  x3 + 2x2 + x  2
=  x3 + 3x2  8
(note  if it helps, tick off like terms as you gather them, as
indicated)
xiii)
(1  t) (1  s) (1  u) = (1  t) (1  s  u + su)
= 1  s  u + su  t + ts +tu  tsu
= 1  s  t  u + st + su + tu  sut
xiv)
(a  2b)2  (a + 2b)2 = a2  2(a) (2b) + (2b)2
 (a2 + 2(a) (2b) + (2b)2)
(using (x + y)2 = x2 + 2xy + y2 and (x  y)2 = x2  2xy + y2)
=  4ab  4ab
=  8ab
If you really know your algebraic identities well there is an easier
approach using the difference of two squares: (a  2b)2  (a + 2b)2 = (a  2b  (a + 2b)) (a  2b + a + 2b)
= ( 4b) (2a) =  8ab
xv)
Similarly to xiv)
(x  y)2 + (x + y)2 = x2  2xy + y2 + x2 + 2xy + y2
= 2x2 + 2y2
This time the difference of two squares won’t help.
2.3.1E
Factorise each of your answers to Question 2.3.1D as far as possible.
Solution
This is the difficult part! In some cases we simply reverse the removal of
brackets, in others factorisation is more difficult, or indeed impossible.
i)
2x + 4 = 2x + 2  2 = 2(x + 4)
ii)
2x + 1 doesn’t factorise into anything simpler  leave as it is.
iii)
3t2  3t = 3(t2  t) = 3t(t  1)
iv)
This is factorising a quadratic  it helps to think of the t as a number
and regard the expression as a quadratic in s.
s2 + st  2t2 = (s  t) (s + 2t)
(for more on this see section 2.2.3)
v)
a3  3a2 = a2  a  3a2 = a2(a  3)
vi)
x3 + x2  3x + 1 yields no obvious factors, although you might notice
that x = 1 is a root of the equation
x3 + x2  3x + 1 = 0
so (x  1) must be a factor of the polynomial on the LHS, from
Section 2.2.3. For the moment be content with the fact that we know
x3 + x2  3x + 1  (x2 + 2x  1) (x  1)
because that is precisely how we obtained it. Note that x2 + 2x  1
factorises no further  the roots of this quadratic are not integers.
vii)
 2u3  6u =  2(u3 + 3u)
(watch the signs!)
=  2u(u2 + 3)
viii)
Another one that doesn’t factorise further  the resulting quadratic,
9x2  2x  35, does not have integer roots.
ix)
Again a3 + 2a2  4a + 7 does not factorise  its roots, if integers,
must be factors of 7 ( 1,  7) neither of which is in fact a root.
x)
x3  2x2  2x = x(x2  2x  2) which factorises no further since x2  2x
 2 does not have integer roots.
xi)
x3  3x2 + 2x = x(x2  3x + 2)
= x(x  1) (x  2)
xii)
 x3 + 3x2  8 =  (x3  3x2 + 8)
is again as simple as it gets.
Note: it is not easy to spot when such a polynomial is factorisable or
not  we address this later in the book.
xiii)
It is not easy to spot the factors of the final expression
1  s  t  u + st + su + tu  sut
by rearranging terms, but noticing that the whole expression
vanishes for s = t = u = 1 suggests the answer
(1  s)(1  t)(1  u)
which of course we already know to be correct.
xiv)
 8ab is already nicely factorised!
xv)
2x2 + 2y2 = 2(x2 + y2) is as simple as it gets.
2.3.2 Polynomials
2.3.2A.
Which of the following are polynomials? For those that are give the degree
and list the coefficients.
i)
iv)
t2 – t + 4
ii) 0
u+1
iii) u – 1
7t3 – 2t + 1
1
v) 4x4 – 2x3 + 3x – x
vi) 27x4 – 3x2 + 1
vii)
x)
x3 + 2x
x
x2y +
viii) x +
ix) 3x2 + t3
x
y
Solution
i)
t2  t + 4 is a polynomial of degree 2 (a quadratic) in t, with
coefficients 1,  1, 4. (Note that the minus sign is included).
ii)
0 is a polynomial of degree 0 (a constant) with coefficient 0.
iii)
u+1
is not a polynomial, it is a rational function in u.
u1
iv)
7t3  2t + 1 is a third degree polynomial with coefficients 7, 0,  2, 1
(note the zero coefficient).
v)
1
1
4x4  2x3 + 3x  x is not a polynomial because of the x part.
vi)
27x4  3x2 + 1 is a polynomial of degree 4 with coefficients
27, 0,  3, 0, 1.
vii)
x3 + 2x
x
is actually a rational function, not a polynomial, even
though we might be tempted to cancel the x to get
x3 + 2x
x(x2 + 2)
=
= x2 + 2
x
x
we can only do this if x  0. So
x3 + 2x
exists only provided x  0
x
while the polynomial x2 + 2 exists for all x.
viii)
The
x means that x +
x is not a polynomial, as
x = x1/2 it has a
non-integer power.
ix)
3x2 + t3 is a polynomial in both x and t, of degree 3. Coefficients all
zero except that of x2, 3, and t3, 1.
x)
x2y +
y is a polynomial in x, but not in y.
2.3.2B.
Expand the following brackets, collecting like terms
i) (x – 1)(x + 2)
ii)
(x – 1)(x + 2)(x + 4)
(x – 1)(x + 1)2
iii)
2
v)
vii)
ix)
xi)
xiii)
xv)
2
(u – 1) (u + 1)
(t + 1)(t – 2)(t + 2)
(s – 2)4
(x + 2)2(x – 3)2
(3s – 1)(s + 2)(4s + 3)
(3x + 1)(3x – 1)(x + 3)
iv)
(x – 2)(x – 3)(x + 1)(x + 2)
vi)
viii)
x)
xii)
xiv)
(x – 1)3 (x + 2)
(u – 2)(u + 3)(u – 3)
(x – 1)(x + 2)(x – 3)(x + 4)
(2t + 1)(3t – 4)
(3x + 2)2(2x – 1)(x + 2)
Check each expansion with suitable numerical values.
Solution
This exercise provides plenty of practice in multiplying brackets out. In each
case choosing a value of the variable that makes one of the brackets vanish
should also make the resulting expression vanish, which provides a check on
the result.
i)
(x  1) (x + 2) = x2 + 2x  x  2
= x2 + x  2
RHS = 0 = LHS if x = 1
ii)
(x  1) (x + 2) (x + 4) = (x2 + x  2) (x + 4)
= x3 + x2  2x + 4x2 + 4x  8
= x3 + 5x2 + 2x  8
RHS = 0 = LHS if x = 1
iii)
(x  1) (x + 1)2 = (x  1) (x + 1) (x + 1)
= (x2  1) (x + 1)
= x3 + x2  x  1
the easy way, or
(x  1) (x + 1)2 = (x  1) (x2 + 2x + 1)
= x3 + 2x2 + x  x2  2x  1
= x3 + x2  x  1
the long way, which also provides a check.
Both sides = 0 when x = 1
iv)
A slog  made easier by difference of two squares:(x  2) (x  3) (x + 1) ( x + 2) = (x  2) (x + 2) (x  3) (x + 1)
= (x2  4) (x2  2x  3)
= x4  2x3  3x2  4x2 + 8x + 12
= x4  2x3  7x2 + 8x + 12
LHS = RHS = 0 when x =  1
v)
(u  1)2 (u + 1)2 = ((u  1) (u + 1)) = (u2  1)2
= (u2)2  2u2 + 1
= u4  2u2 + 1
LHS = RHS = 0 when x = 1
vi)
(x  1)3 (x + 2) = (x3  3x2 + 3x  1) (x + 2)
(by a simple binomial theorem if you are familiar with that)
= x4  3x3 + 3x2  x + 2x3  6x2 + 6x  2
= x4  x3  3x2 + 5x  2
vii) (t + 1) (t  2) (t + 2) = (t + 1) (t2  4) = t3  4t + t2  4
= t3 + t2  4t  4
viii) (u  2) (u + 3) (u  3) = (u  2) (u2  9) = u2  9u  2u2 + 18
= u3  2u2  9u + 18
ix) (s  2)4 can be expanded by the binomial theorem. Here, simply use
(a  b)2 = a2  2ab + b2
(s  2)4 = ((s  2))2 = (s2  4s + 4)2
= (s2)2 + (4s)2 + 42 + 2s2(4s) + 2 s2(4) + 2( 4s) 4
(using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc)
= s4 + 16s2 + 16  8s3 + 8s2  32s
= s4  8s3 + 24s2  32s + 16
x) (x  1) (x + 2) (x  3) (x + 4) = (x2 + x  2) (x2 + x  12)
= x4 + x3  12x2 + x3 + x2  12x  2x2  2x + 24
= x4 + 2x3  13x2  14x + 24
xi) (x + 2)2 (x  3)2 = (x2 + 4x + 4) (x2  6x + 9)
= x4  6x3 + 9x2 + 4x3  24x2 + 36x + 4x2  24x + 36
= x4  2x3  11x2 + 12x + 36
xii) (2t + 1) (3t  4) = (2t) (3t)  4(2t) + 3t  4
= 6t2  8t + 3t  4
= 6t2  5t  4
xiii) (3s  1) (s + 2) (4s  3) = (3s  1) (4s2 + 11s + 6)
= 12s3 + 23s2 + 18s  4s2  11s  6
= 12s3 + 29s2 + 7s  6
xiv) (3x + 2)2 (2x  1) (x + 2) = (9x2 + 12x + 4) (2x2 + 3x  2)
= 18x4 + 27x3  18x2 + 24x3 + 36x2  24x + 8x2 + 12x  8
= 18x4 + 51x3 + 26x2  12x  8
xv) (3x + 1) (3x  1) ( x + 3) = ((3x)2  1) (x + 3)
= (9x2  1) (x + 3)
= 9x3 + 27x2  x  3
2.3.3 Factorisation of polynomials by inspection
2.3.3A.
Factorise the following, retaining only real coefficients
i) x2 + x
ii) 3x3 – 2x2
iii) – 7x2 + 42x4
iv) t4 – 3t3 + t2
v) u2 – 9
vi) t2 – 121
vii)
s24 – 16s22
viii) 4x12 – 64x8
Solution
i) x2 + x = x(x + 1)
ii) 3x3  2x2 = x2(3x  2)
iii)  7x2+ 42x4 = 7x2 ( 1) + 7x2 (6x2)
= 7x2 (6x2  1) = 7x2 ( 6 x  1) ( 6 x + 1)
iv) t4  3t3 + t2 = t2(t2  3t + 1)
(the quadratic does not factorise using real coefficients).
v) u2  9 = u2  32 = (u  3) (u + 3)
vi) t2  121 = t2  (11)2 = (t  11) (t + 11)
vii) s24  16522 = s22(s2  16) = s22(s2  42)
= s22(s  4) (s + 4)
viii) 4x12  64x8 = 4x8(x4  16) = 4x8((x2)2 42)
= 4x8(x2  4)(x2 + 4)
= 4x8(x  2) (x + 2) (x2 + 4)
2.3.3B.
Factorize the following polynomial expressions (hint: look back at 2.3.2B)
t2 + 5t + 6
ii) t3 + t2 – 4t – 4
9x3 + 27x2 – x – 3
i)
iv)
iii)
y4 – 2y3 – 7y2 + 8y + 12
Solution
This question does not really require you to actually factorise the expressions
given, but simply to be aware that factorising entails looking for the factors,
which in some cases are conveniently provided by the results of Exercise
2.3.2B.
i) t2 + 5t + 6 = (t + 2) (t + 3)
‘by inspection’
ii) t3 + t2  4t  4 = t2(t + 1)  4(t + 1) = (t2  4) (t + 1) = (t  2) (t + 2)(t + 1)
(See RE 2.3.2B iv))
iii) From RE 2.3.2B iv) with x replaced by y:y4  2y3  7y2 + 8y + 12 = (y  2) (y  3) (y + 1) (y + 2)
iv) RE 2.3.2B xv) shows that
9x3 + 27x2  x  3  (3x + 1) (3x  1) (x + 3)
or, not so bad:9x3 + 27x2  x  3 = 9x2(x + 3)  (x + 3)
= (9x2  1) (x + 3)
= (3x  1) (3x + 1) (x + 3)
We will see more powerful methods for factorising polynomials in
Section 2.2.6.
2.3.4 Simultaneous equations
2.3.4A.
Solve the following systems of linear equations, verifying your solution by
back substitution in each case.
i)
iv)
x–y = 1
x + 2y = 0
ii)
A+B = 0
3A – B = 1
iii)
s + 3t = 1
s – 2t = 1
3x + 2y = 2
– 2x + 3y = 1
v)
u + 4v = 1
u–v = 2
vi)
7x1 – 2x2 = 1
3x1 – 2x2 = 0
Solution
In this question we will not be systematic – we will just obtain the result in the
quickest way.
i) x  y = 1
x + 2y = 0
 (1)
 (2)
Subtracting (1) from (2) removes x to give
3y =  1
so
1
y=3
Then from (2)
2
x =  2y = 3
Substituting these values in the equations gives
2  1
x  y = 3   3 = 1
 
2
 1
x + 2y = 3 + 2   3 = 0


as required.
ii) A + B = 0
(1)
3A  B = 1
(2)
1
(1) + (2) gives 4A = 1 so A = 4
1
(1) gives B =  A =  4
1 1
Then A + B = 4  4 = 0
1
1
and 3A  B = 3  4  ( 4 ) = 1
iii) s + 3t = 1
(1)
s  2t = 1
(2)
(1)  (2) gives 5t = 0, so t = 0
(1) then gives s = 1
Substituting back into the equations, we have
s + 3t = 1 + 3(0) = 1
s  2t = 1  2(0) = 1
iv) 3x + 2y = 2
 2x + 3y = 1
(1)
(2)
Eliminate y by making its coefficient identical in both equations.
Thus:3  (1)
9x + 6y = 6
2  (2)
 4x + 6y = 2
Subtract these equations to get
13x = 4
so
4
x = 13
From the original equation (1) we now obtain
12 14
2y = 2  3x = 2  13 = 13
7
y = 13
so
Checking :-
12
14
26
3x + 2y = 13 + 13 = 13 = 2
8
21
13
 2x + 3y =  13 + 13 = 13 = 1
v) u + 4v = 1
uv=2
(1)
(2)
1
(1)  (2) gives 5v =  1 so v =  5
Then
1
9
u=2+v=25 = 5
Checking:
9
 1
u + 4v = 5 + 4  5 = 1
 
9
 1
u  v = 5   5 = 2
 
vi) 7x1  2x2 = 1
(1)
3x1  2x2 = 0
(2)
Note that this notation, using variables xi denoted by a subscript i is a
common way to represent variables in systems of equations. To solve
the system subtract (2) from (1)
1
4x1 = 1, so x1 = 4
Then (2) gives
3
3
x2 = 2 x1 = 8
1
3
7x1  2x2 = 7 4  2 8 = 1
 
 
1
3
3x1  2x2 = 3 4  2 8 = 0
 
 
2.3.4B.
Comment on the following systems of equations
i)
x+y = 1
ii)
2x – y = 3
3x + 3y = 3
iv)
2A + B = 1
4A + 2B = – 1
iii)
4x – 2y = 1
v)
u+v= –1
3u + 3v = – 3
x+y = 0
x–y = 0
vi)
x+y = 0
x2 – y2 = 1
Solution
Here you are asked to say something about the properties of the systems, not
all of which actually have unique solutions.
i)
x+y=1
(1)
3x + 3y = 3
(2)
We can cancel the 3 from (2) to reveal a repetition of (1), so we really
have just one equation in two variables:x+y=1
This has an infinite number of solutions, since for any value of y, say
y = s, we can satisfy the equation by choosing x = 1  y = 1  s. We
say the general solution is x = 1  s, y = s where s is arbitrary.
ii)
2x  y = 3
(1)
4x  2y = 1
(2)
Dividing (2) by 2 gives the system
2x  y = 3
1
2x  y = 2
1
Such a system has no solutions, since it would require 3 = 2 . We say
the system is inconsistent.
iii)
x+y = 0
(1)
xy = 0
(2)
(1) + (2) gives 2x = 0 or x = 0, then (2) gives y = 0 also. This system
therefore has only the trivial solution x = y = 0.
iv)
2A + B = 1
4A + 2B =  1
This system is inconsistent, and can have no solution for A and B.
v)
u+v=1
(1)
3u + 3v =  3
(2)
Dividing (2) by 3 shows that it is equivalent to (1) so we again have
the situation of i), an infinity of solutions, with the general solution
u =  (1 + s) ,
vi)
v=s
x+y=0
(1)
x2  y2 = 1
(2)
s arbitrary.
That this system is inconsistent and has no solutions is most easily
seen by rewriting (2) as
(x  y) (x + y) = 1
From (1) this gives 0 = 1, which is not possible.
2.3.5 Equalities and identities
2.3.5A.
Determine the real values of A, B, C, D in the following identities
i)
ii)
iii)
(s – 1)(s + 2)  As2 + Bs + C
(x – 1)3  Ax3 + 2Bx2 – 3Cx + D
(x – A)(x + B) x2 – 4
iv)
(x + 2)(x – 3)(2x – 1) A(x – 1)2 + Bx + Cx3 – Dx2
v)
vi)
A(x – 1)+ B(x + 2) x – 3
(x + A)2 + B2 x2 – 2x + 5
Solution
In an identity the result must be true for all values of the variable, s, x,
whatever. This enables us to determine the unknown coefficients A, B, C,
D by a number of means. The method is important in partial fractions.
i)
(s  1) (s + 2)  As2 + Bs + C
LHS  s2 + s  2  As2 + Bs + C
Since this is an identity the coefficients of powers of s must be the
same on both sides, giving
A = 1, B = 1, C =  2
ii)
(x  1)3  x3  3x2 + 3x  1  Ax3 + 2Bx2  3Cx + D
So
A=1
3
2B =  3 or
B=2
 3C = 3
or C =  1
D=1
iii)
(x  A) (x + B)  x2  4
There is no need to equate coefficients here if you remember your
difference of two squares:(x  A) (x + B)  (x  4) (x + 4)
so we can take
A=4
and B = 4
or A =  4 and B =  4 (Note that the answer given in the book is
wrong)
So there are two possible solutions in this case.
iv)
This one requires a bit more trickery if you want to avoid expanding
both sides out. Here we can use the fact that the identity must hold
for all values of x. So choose a few suitable ones to help us find A, B,
C, D.
Choosing x = 0 will give us A, since B, C, D drop out:(2) (3) (1) = 6 = A(1)2 = A
Also note that Cx3 is the only cubic terms on the right and so must
equate to the cubic term on the left, which even without full
expansion is clearly 2x3 so C = 2.
We now only have B and D to find. For these we can substitute any
two values of x to get two equations. x = 1 is an obvious choice since
it knocks out A to give:(3) ( 2) (1) =  6 = B + C  D
or since C = 2:BD=8
(1)
x = 3 will knock out the LHS and give
0 = A(2)2 + 3B + 27C  9D
or, with A = 6 and C = 2:3B  9D =  78
(2)
(2)  3(1) gives
 6D =  54
or
D=9
Then (1) gives B = 1.
This exercise demonstrates how good facility with algebra can save a
lot of work.
v)
A(x  1) + B(x + 2)  x  3
is the sort of problem one gets in partial fractions. Choosing suitable
values works well here.
x = 1 knocks out A to give
3B =  2
so
2
B=3
x =  2 knocks out B to give
5
 3A =  5 so A = 3
Or, you can equate coefficients:A(x  1) + B(x + 2)  (A + B) x  A + 2B
 x3
gives
A+B=1
 A + 2B =  3
which you can solve to check the results obtained above.
vi)
(x + A)2 + B2  x2  2x + 5
This is actually an example of completing the square for the quadratic
on the RHS (We cover this in Section 2.2.11). The ‘slickest’ way to do
it is to note that
x2  2x + 5  (x  1)2 + 4
so
A =  1 and B =  2
Alternatively expand both sides and equate coefficients:(x + A)2 + B2  x2 + 2Ax + A2 + B2
 x2  2x + 5
so
2A =  2
and
or
A=1
A2 + B2 = 5
so
B2 = 4
or
B=2
2.3.5B.
Given that
A(x – a) + B(x – b) ax + b
determine expressions for A, B in terms of a, b by two different methods
Solution
This question is really solving the partial fractions case for general
coefficients.
First Method
A(x  a) + B(x  b)  ax + b
Put
x = b:-
A(b  a) = ab + b
so
A=
(a+1)b
ba
Put x = a so
B(a  b) = a2 + b
and
B=
a2 + b
ab
Second method
Rearrange and equate coefficients:A(x  A) + B(x  b)  (A + B) x  (aA + bB)
 ax + b
So
A+B=a
aA + bB =  b
(2)  b(1) gives
(a  b)A =  b  ab
(1)
(2)
or
A=
b(a + 1)
ba
as before, and
B= aA = a
=
b(a + 1)
b1
a(b  a)  b(a + 1)
ba
=
=
 a2  b
ba
a2 + b
ab
as before.
2.3.6 Roots and factors of a polynomial
Use the factor theorem to factorize the following polynomials
i) u3 – u2 – 4u + 4
ii) x3 + 4x2 + x – 6
iii)
v)
t4 – t3 – 7t2 + t + 6
x3 + 3x2 – 10x – 24
iv)
x3 – x2 – x + 1
Solution
Here we factorise polynomials by using trial and error and the factor
theorem to discover roots and factors. Trials for the roots come from
the factors of the constant term.
i)u3  u2  4u + 4
This case is actually easy enough by inspection:u3  u2  4u + 4  u2(u 1)  4(u  1)
 (u2  4) (u  1)
= (u  2) (u + 2) (u 1)
But also the roots are fairly obvious:-
u=1
13  12  4 + 4 = 0
u=2
23  22  4  2 + 4 = 0
u = 2
 23  22 + 4  2 + 4 = 0
ii)
Let x3 + 4x2 + x  6  p(x)
Trying low values first:p(1) = 1 + 4 + 1  6 = 0
so x = 1 is a root and by the factor theorem this means (x  1) is a
factor.
Without evaluation it is clear that p(2)  0. On the other hand
p(2) =  8 + 16  2  6 = 0
so x =  2 is a root and x  2(2) = x + 2 is a factor.
As the constant term is  6 we suspect x =  3 will also do, and in fact
p( 3) =  27 + 38  3  6 = 0
So x =  3 is a root and x + 3 a factor.
As the expression is a cubic we know that three factors should do it.
Also, since the coefficient of x3 is 1 we suspect a simple product of
factors will do, and we can now confirm that
x3 + 4x2 + x  6  (x  1)(x + 2)(x + 3)
iii) p(t)  t4  t3  7t2 + t + 6
Again, for t we try low values that are factors of 6.
p(1) = 1  1  7 + 1 + 6 = 0
so (t  1) is a factor
p( 1) = 1 + 1  7  1 + 6 = 0
so (t + 1) is a factor
p(2) = 24  23  7  22 + 2 + 6  0
so (t  2) is not a factor
p( 2) = 24 + 23  7  22  2 + 6 = 0
so (t + 2) is a factor
p(3) = 81  27  63 + 3 + 6 = 0
so (t  3) is a factor.
Thus
t4  t3  7t2 + t + 6  (t  1)(t + 1)(t + 2)(t  3)
iv)
p(x)  x3  x2  x + 1  x2(x  1)  (x  1)
 (x2  1)(x  1)
 (x  1)2 (x + 1)
Again, p(1) = 0 = p( 1) is easy to spot  but does not tell us that (x 
1) is a double factor, so the remainder theorem needs a bit of help
here.
v)
x3 + 3x2  10x  24  p(x)
‘By inspection’
p(1)  0,
p(1)  0, p(2)  0
p( 2) = 0 so (x + 2) is a factor
p(3) = 0 so (x  3) is a factor.
Since the constant term is  24 we suspect that the final factor is (x +
4) and indeed:p( 4) = 0
confirms this. Thus
x3 + 3x2  10x  24  (x + 2)(x  3)(x + 4)
2.3.7 Rational functions
2.3.7A.
Identify the rational functions and state
numerator.
i)
x+1
x–1
ii)
x+ 2
x2 – 1
iv)
x3 + 2x – 1
x4 – 2x + 3
v)
2x4 – 1
x4 + 1
a) the denominator,
iii)
b) the
x+3
x – 2x + 1
3
Solution
i)
ii)
Rational function a) x  1
b) x + 1.
Rational function (despite the deliberately misleading
2! ) a) x2  1
b) x + 2 .
iii)
Not a rational function because the numerator is not a polynomial.
iv)
Rational function
v)
Rational function
a) x4  2x + 3
b) x3 + 2x  1
a) x4 + 1
b) 2x4  1.
2.3.7B.
1
1
For what values of x does x – 1 – x – 1 exist?
Solution
It may appear that
1
1

= 0
x1
x1
but of course this assumes the existence of both terms in the
difference.
1
x – 1 exists for all values of x except for x = 1
So
1
1

exists (and = 0) for all values of x except x = 1.
x1
x1
We should strictly write
1
1

= 0
x1
x1
x1
2.3.8 Algebra of rational functions
2.3.8A.
Put the following over a common denominator and check your results with x
= 0 and one other appropriate value.
i)
2
3
–
x+2
x–1
ii)
1
2
–
x+3
x–4
iii)
2
3
x+3 – x–2
iv)
2
3
x–2 + x–3
1
1
1
+
+
x–1
x–2 x–3
vi)
3
3
4
–
–
x+2
x+3
x–1
v)
vii)
x
1
+ x–1
x –1
ix)
3
2
+ 2
x +1
x +2
viii)
2
2
x)
1
2
x– x+1 + x+2
2
3
2x – 1 + x – 1 – x + 2
Solution
i)
2
3
2(x  1)  3(x + 2)

=
x+2
x1
(x + 2)(x  1)
=
2x  2  3x  6
(x + 2)(x  1)
=
x8
(x + 2)(x  1)
Check: x = 0 on LHS gives 1 + 3 = 4 and on the RHS
8
= 4
2
Notes:

The ‘=’ here should really be ‘’ , but it is common practice not to
bother – after all ‘=’ is two-thirds the work of ‘’ !

We have cross-multiplied to obtain the numerator. Again, this is
usual practice but for the cases where we have more than two
fractions to combine it is probably best to get into the habit of
constructing the common denominator, as in:2
3
2(x  1)
3(x + 2)
x + 2  x  1 = (x + 2)(x  1)  (x + 2)(x  1)
=
2(x  1)  3(x + 2)
(x + 2)(x  1)
etc
ii)
1
2
x  4  2(x + 3)

=
x+3 x4
(x + 3)(x  4)
=
 x  10
(x + 3)(x  4)
1 1
5
LHS (0) = 3 + 2 = 6
iii)
10
5
= 6
12
2
3
2(x  2)  3(x + 3)
x+3  x2 =
(x + 3)(x  2)
=
 x  13
(x + 3)(x  2)
2 3
13
LHS (0) = 3 + 2 = 6
iv)
RHS (0) =
RHS (0) =
 13
13
= 6
6
2
3
2(x  3) + 3(x  2)
+
=
x 2
x3
(x  2)(x  3)
=
5x  12
(x  2)(x  3)
 12
LHS (0) =  1  1 =  2 RHS (0) = + 6
v)
=2
1
1
1
+
+
x1
x2
x3
Here the common denominator is (x  1)(x  2)(x  3), so we convert
each fraction to have this denominator.
1
1
1
+
+
x1
x2
x3
=
(x  2)(x  3)
+
(x  1)(x  2)(x  3)
(x  1)(x  3)
(x  1)(x  2)(x  3)
=
(x  2)(x  3) + (x  1)(x  3) + (x  1)(x  2)
(x  1)(x  2)(x  3)
(x  1)(x  2)
(x  1)(x  2)(x  3)
+
(With practice you will soon be able to jump direct to the last line)
=
x2  5x + 6 + x2  4x + 3 + x2  3x + 2
(x  1)(x  2)(x  3)
=
3x2  12x + 11
(x  1)(x  2)(x  3)
1 1
11
LHS (0) =  1  2  3 =  6
vi)
RHS (0) =
3
3
4
3(x + 3)(x  1)  3(x + 2)(x  1)  4(x + 2)(x + 3)


=
x+2 x+3 x1
(x + 2)(x + 3)(x  1)
=
=
3(x2 + 2x  3)  3(x2 + x  2)  4(x2 + 5 + 6)
(x + 2)(x + 3)(x  1)
3x2 + 6x  9  3x2  3x + 6  4x2  20x  24
(x + 2)(x + 3)(x  1)
Note care needed with signs and brackets.
=
3
9
LHS (0) = 2  1 + 4 = 2
vii)
11
6
x
1
+
1 x1
x2
 4x2  17x  27
(x + 2)(x + 3)(x  1)
RHS (0) =
 27
9
= 2
6
In this case care is needed – the common denominator obtained by
cross multiplication is not the last word:-
x2
x
1
x(x  1) + x2  1
2x2  x  1
+
=
=
2
1
x1
(x  1)(x  1)
(x  1)(x2  1)
Noticing that 2x2  x  1  (2x + 1)(x  1), this becomes
2x2  x  1
(2x + 1)(x  1)
2x + 1
=
= 2
2
2
(x  1)(x  1)
(x  1)(x  1)
x 1
What has happened here is that in fact the LCD is not (x  1)(x2  1)
but (x  1)(x + 1):x2
x
1
x
1
+
=
+
1 x1
(x  1)(x + 1)
x1
=
x
x+1
+
(x  1)(x + 1)
(x  1)(x + 1)
=
2x + 1
(x  1)(x + 1)
Although we get there in the end by cross multiplication the use of
the correct LCD is clearly much quicker. We could have also (again
quicker) written:x2
x
1
x
1
+
=
+
1
x1
(x  1)(x + 1)
x1
=
1
 x

 x + 1 + 1
(x  1) 

=
1
(2x + 1)
(x  1) (x + 1)
etc.
The important message here is how we really need the basic results
such as difference of two squares and the factorisation of a quadratic
at our fingertips to be thoroughly competent in more advanced
algebra.
viii)
1
2
x (x + 1)(x + 2)  (x + 2) + 2(x + 1)
xx+1+x+2=
(x + 1)(x + 2)
=
x(x2 + 3x + 2)  x  2 + 2x + 2
(x + 1)(x + 2)
x3 + 3x2 + 3x
= (x + 1) (x + 2)
Notice how the numerator is of degree greater than the denominator
 i.e. the fraction is improper.
LHS (0) =  1 + 1 = 0
ix)
x2
RHS (0) = 0
3
2
3(x2 + 2) + 2(x2 + 1)
+
=
2
+1
x +2
(x2 + 1)(x2 + 1)
5x2 + 8
= (x2 + 1)(x2 + 2)
The x2 makes no difference to the methodology here.
8
RHS (0) = 2 = 4
LHS (0) = 3 + 1 = 4
x)
2x  1 +
2
3
(2x  1)(x  1)(x + 2) + 2(x + 2) 3(x  1)
 x+2 =
x1
(x  1)(x + 2)
=
=
(2x  1)(x2 + x  2) + 2x + 4  3x + 3
(x  1)(x + 2)
=
2x3 + 2x2  4x  x2  x + 2  x + 7
(x  1)(x + 2)
2x3 + x2  6x + 9
(x  1)(x + 2)
3
9
LHS (0) =  1  2  2 = 2
2.3.8B.
Put over a common denominator
2
3
i) x – 4 + x – 1
RHS (0) =
9
2
4
2
ii) x – 4 – x – 4
iii)
x–1
3
2 + x–1
(x – 2)
iv)
2x – 1
4
– x–2
2
x +1
3
v) 1 + x – 2
Solution
i)
2
3
2x  2 + 3x  12
+
=
x4
x1
(x  4)(x  1)
=
5x  14
(x  4)(x  1)
ii)
4
2
2

=
x4
x4
x4
iii)
x1
3
(x  1)2 + 3(x  2)2
+
=
(x  2)2
x1
(x 1)(x  2)2
iv)
v)
=
x2  2x + 1 + 3(x2  4x + 4)
(x  1)(x  2)2
=
4x2  14x + 13
(x  1)(x  2)2
2x  1
4
(2x  1)(x  2)  4(x2 + 1)

=
x2 + 1
x2
(x2 + 1)(x  2)
1+
=
2x2  5x + 2  4x2  4
(x2 + 2)(x  2)
=
 2x2  5x  2
(x2 + 1)(x  2)
3
x2+3
x+1
=
=
x2
x2
x2
2.3.9 Division and the remainder theorem
2.3.9A.
Perform the following divisions, whenever permissible:
i)
iii)
v)
x2
x2
ii)
x2 – 2x + 2
(x – 1)2
x4 – 2x3 + 4x – 1
x–3
iv)
x3 + 1
x+1
vi)
x4 – y4
x2 + y 2
2x2 – 3x + 4
x2 – 1
Solution
The object in each case is to convert the improper fraction to a polynomial and
a proper fraction. There is no need to use long division  just algebraic
manipulation of polynomials is all that is needed. The important thing is to
recognise when division is permissible.
i)
Apparently easy:x2
x2 = 1
ii)
but only if x  0
x2  2x + 2
x2  2x + 1 + 1
=
(x  1)2
(x  1)2
=
(x  1)2 + 1
(x  1)2
=1+
iii)
1
(x  1)2
provided x  1
x4  2x3 + 4x  1
x4  3x3 + x3 + 4x  1
=
x3
x3
=
x3(x  3) + x3 + 4x  1
x3
= x3 +
x3  3x2 + 3x2 + 3x  1
x3
= x3 + x2 +
3(x2  3x)+ 9x + 4x  1
x3
= x3 + x2 + 3x +
= x3 + x2 + 3x +
13x  1
x3
13(x  3)+ 38
x3
= x3 + x2 + 3x + 13 +
38
(x  3)
x3
Check: by the remainder theorem the remainder when x4  2x3 + 4x 
1 is divided by x  3 should be 34  2  33 + 4  3  1 = 38.
iv)
x3 + 1
x3 + x2  x2 + 1
=
x+1
x+1
=
x2(x + 1)  x2 + 1
x+1
x2  1
= x2  x + 1
(Notice how it is best to pull the sign out in such cases.)
= x2 
(x  1)(x + 1)
x+1
= x2  (x  1)
= x2  x + 1
(x   1)
which also follows of course from the cubic sum identity:a3 + b3  (a + b) (a2  ab + b2)
v)
2x2  3x + 4
2(x2  1) + 2  3x + 4
=
x2  1
x2  1
= 2
3(x  2)
x1
x2  1
2.3.9B.
Find the remainders when the following polynomials are divided by:a) x – 1
b) x + 2
c) x
i) 3x3 + 2x – 1
ii) x5 – 2x2 + 2x – 1
iii) x4 – x2
iv) 2x7 – 3x5 + 4x3 – 2x2 + 1
Solution
By the remainder theorem the remainder when the polynomial p(x) is divided
by x  a is p(a).
i)
p(x) = 3x3 + 2x  1
a)
Remainder when divided by x  1 is p(1) = 3 + 2  1 = 4
b)
Remainder = p( 2) (note the sign)
= 3( 2)3 + 2(2)  1
=  24  4  1 =  29
c)
ii)
Remainder = p(0) =  1
p(x) = x5  2x2 + 2x  1
a)
p(1) = 1  2 + 2  1 = 0
So (x  1) actually divides, i.e. is a factor of, x5  2x2 + 2x  1
b)
p( 2) = ( 2)5  2( 2)2 + 2( 2)  1
=  32  8  4  1 =  45
c)
iii)
p(0) =  1
p(x) = x4  x2
a)
p(1) = 0 so x  1 divides p(x)
b)
p( 2) = ( 2)4  ( 2)2 = 32  4 = 28
c)
p(0) = 0 so x divides p(x)
and indeed:x4  x2 = x2(x2  1) = x2 (x  1)(x + 1)
iv)
a)
b)
p(1) = 2  3 + 4  2 + 1 = 2
p( 2) = 2( 2)7  3( 2)5 + 4 ( 2)3  2( 2)2 + 1
=  199
c)
p(0) = 1
2.3.10 Partial fractions
2.3.10A.
For RE 2.3.8A, B check your results by reversing the operation and resolving
your answer into partial fractions. Usually, the answers should of course be
what you started with in those questions. There are however a couple of
cases where this is not so - explain these cases.
Solution
We will do one question in full detail and the rest by the cover-up rule and
various other short cuts.
i)
We first express the rational function in the general partial fraction
form
x8
A
B
A(x  1) + B(x + 2)
 x+2 +

(x +2)(x  1)
x1
(x + 2)( x  1)
So, equating numerators on both sides we have
 x  8  A(x  ) + B(x + 2)  (A + B)x  A + 2B
We can now either substitute values for x (since this is an identity, true
for all values of x), or we can equate coefficients of powers of x and solve
the resulting equations. Choosing convenient values of x, we find
x = 1 gives
 9 = B(3)
so B =  3
x =  2 gives
 6 = A( 3)
so A = 2
So
x8
2
3
 x+2 
(x +2)(x  1)
x1
which is the question 2.3.8A i).
Or, equating coefficients
A+B=1
 A + 2B =  8
Adding gives 3B =  9, etc, as above.
Either of the above approaches are the best to adopt when you are new
to this topic, but eventually you might aim to use short cuts such as the
cover-up rule, described in the text. In this case we would get directly:
x8
( ( 2) 8)/(  2  1)
( 1  8)/(1 + 2)

+
x
+
2
(x +2)(x  1)
x1
( 6)/(  3)
( 9)/(3)
2
3
+
 x+2 
as above
x+2
x1
x1
ii)
From now on we will use the cover-up rule where convenient.
 x  10
( ( 3) 10)/(  3  4)
( 4  10)/(4 + 3)

+
x
+
3
(x +3)(x  4)
x4

( 7)/(  7)
( 14)/(7)
1
2
+
 x+3 
x+3
x4
x4
as in 2.3.8A ii).
iii)
 x  13
( 10)/(  5)
( 15)/(5)
2
3

+
 x+3 
x+3
(x +3)(x  2)
x2
x2
as in 2.3.8A iii)
iv)
5x  12
( 2)/(  1)
( 3)/(1)
2
3

+

+
(x  2)(x  3)
x2
x3
x2
x3
as in 2.3.8A iv)
v) In this case we have three linear factors in the denominator. Following
the general procedure we would write
3x2  12x + 11
A
B
C

+
+
(x  1)(x  2)(x  3) x  1 x  2
x3
then take a common denominator on the right hand side and equate
numerators, obtaining an identity involving quadratic powers of x and
the coefficients A, B, C. By equating like powers of x or substituting
appropriate values of x (x = 1, 2, 3 are the obvious choices here) we
could then determine the values of A, B, C. This is a direct extension of
the case of two factors considered in i). However, with three parameters
to find we now have more work to do. In particular, the method of
equating coefficients of powers of x to find and solve linear equations in
A, B, C now becomes more messy. It is therefore better to use the
method of substituting in values of x in such cases. It is even better to
use the cover-up rule, which applies equally well in the case of any
number of linear factors. Thus, we get
3x2  12x + 11
2/2
(12  24 + 11)/(  1)
(27  36 + 11)/2

+
+
(x  1)(x  2)(x  3) x  1
x2
x3

1
1
1
+
+
x1 x2
x3
as in 2.3.8A v).
vi)
+
 4x2  17x  27
( 16 + 34  27)/(  3) ( 36 +51  27)/( 4)

+
x+2
x+3
(x + 2)(x + 3)(x  1)
( 4  17  27)/12
x1
3
3
4
x+2 x+3 
x1
as in 2.3.8A vi).
vii)
In this case we will not get the original result. Direct cover-up
rule gives
2x + 1
2x + 1
3
1


+ 2(x +1)
2
(x  1) (x  1)(x + 1)
2(x  1)
which you may think looks nothing like the original:
x
1
+
(x2  1) (x  1)
However, let’s write this as
x
1
x
1
1
1
+

+

+
+
(x2  1) (x  1)
(x  1)(x + 1)
(x  1)
2(x  1) 2(x +1)
1
(x  1)
using the cover-up rule
3
1
+ 2(x +1)
2(x  1)

as above.
viii)
In this case we must first convert to a polynomial and a proper
fraction before finding partial fractions – so we must divide out.
x3 + 3x2 + 3x x3 + 3x2 + 3x
(x + 1)(x + 2)  x2 + 3x + 2

x3 + 3x2 + 2x + x
x2 + 3x + 2
x
 x + x2 + 3x + 2
x
 x + (x + 1)(x + 2)
( 1)
2
 x + (x + 1) + (x + 2)
1
2
 x  (x + 1) + (x + 2)
as in RE 2.3.8A vii)
5x2 + 8
ix) ( x2 + 1)( x2 + 2)
is an interesting one. The general rules for
irreducible quadratic functions in the denominator would suggest quite
a complicated partial fractions form. However, because only x2 occurs
we can regard this as the primary variable and treat the fractions as
linear – even using the cover-up rule despite the fact that we can’t really
equate, say, x2 + 1 to zero for real x. The method does actually work in
fact because all the processes we use are equally valid in complex
algebra. So, applying the cover-up rule with x2 as the variable we get
5x2 + 8
8 + 5( 1)
5( 2) + 8
3
2
( x2 + 1)( x2 + 2)  x2 + 1 
x2 + 2  x2 + 1 + x2 + 2
as 2.3.8A ix).
x)
Once we have divided out we obtain a simple partial fractions to
do
2x3 + x2  6x + 9
x+7
 2x  1 +
(x  1)(x + 2)
(x  1)(x + 2)
 2x  1 +
2
3
 (x + 2)
(x  1)
as 2.3.8A x).
xi)
5x  14
6/3
9/3
2
3

+

+
(x  4)(x  1)
x4
x1
x4
x1
as 2.3.8B i).
xii) Nothing much you can do with
xiii)
In the case of
2
!
x4
4x2  14x + 13
we could use the decomposition
(x  1)(x  2)2
A
B
C
+
+
x1 x2
(x  2)2
and determine the coefficients. However, we will take the hint from the
original form given in RE 2.3.8B iii), and assume
4x2  14x + 13
A
Bx + C

+
2
(x  1)(x  2)
x1
(x  2)2

A(x  2)2 + (Bx + C)(x  1)
(x  1)(x  2)2
So
A(x  2)2 + (Bx + C)(x  1)  4x2  14x + 13
Putting x = 1 (equivalent to using the cover-up rule to find A)
A = 4  14 + 13 = 3
To get B and C it is probably best to equate coefficients:
x2
A + B = 4, so B = 1
Constants
4A  C = 13, so C = 4A  13 =  1
So
4x2  14x + 13
3
x1

+
2
(x  1)(x  2)
x1
(x  2)2
as 2.3.8B iii).
xiv) We assume the form
 2x2  5x  2
A
Bx + C

+ x2 + 1
(x2 + 1)(x  2) x  2
We can get A from the cover-up rule:
A=
 2(2)2  5(2)  2
=4
5
So
 2x2  5x  2
Bx + C
4
 x2 + 1 
2
(x + 1)(x  2)
x2

(Bx + C)(x  2)  4(x2 + 1)
(x2 + 1)(x  2)
So
(Bx + C)(x  2)  4(x2 + 1)   2x2  5x  2
Equating coefficients gives:
x2
B  4 =  2, so B = 2
Constant
 2C  4 =  2, so C =  1
So
 2x2  5x  2
2x  1
4
 x2 + 1 
2
(x + 1)(x  2)
x2
as in RE 2.3.8B iv).
xv) This is just a matter of dividing out
x+1 x2+3
3

1+
x2
x2
x2
as in 2.3.8B v).
2.3.10B.
Split into partial fractions
i)
x–1
(x + 2)(x – 3)
ii)
4
x –1
iii)
x+1
x(x – 3)
iv)
2x + 1
(x – 1)2(x + 2)
v)
3
(x + 1)(x + 1)
vi)
5x – 4
(x + 4)(x – 2)2
2
2
2
x+1
(x – 1)(x + 3)(x – 4)
vii)
Solution
x1
( 3/ 5)
2/5
3
2

+

+
x+2
5(x + 2)
(x + 2)(x  3)
x3
5(x  3)
i)
ii)
4
4
4
2
2



x+1
x2  1
(x  1)(x + 1)
(x  1)(x + 1)
x1
iii)
x+1
 1/3
4/3
4
1
= x +
=
 3x
x(x  3)
x  3 3(x  3)
iv)
2x + 1
A
B
C
 x+2 +
+
(x  1)2(x + 2)
x1
(x  1)2
Note that C as well as A can be found using the cover-up rule
A=
2( 2) + 1
=3
(1)2
3
C=3 =1
So
2x + 1
3
B
1
 x+2 +
+
(x  1)2(x + 2)
x1
(x  1)2
We can get B easily by putting x = 0, giving
1
3
=
1

B

2
2
So B =  1 and therefore
2x + 1
3
1
1
 x+2 +
+
2
(x  1) (x + 2)
x1
(x  1)2

v)
(x2
1
1
3

 x+2
2
(x  1)
x1
3
A
Bx + C
3/2
Bx + C

+

+
2
+ 1)(x + 1)
x+1
x +1
x+1
x2 + 1
by the cover up rule

3(x2 + 1)/2 + (Bx + C)(x + 1)
(x2 + 1)(x + 1)
Equating coefficients in the numerators now gives
x2
3
3
+
B
=
0
and
so
B
=

2
2
Constant
3
3
+
C
=
3
and
so
C
=
2
2
Hence
(x2
vi)
3
3
3(x  1)


+ 1)(x + 1)
2(x + 1)
2(x2 + 1)
Write
(x2
5x  4
Ax + B
C
D
 x2 + 4 +
+
2
+ 4)(x  2)
x2
(x  2)2
We can get D from the cover-up rule
5(2)  4 3
D = 22 + 4 = 4
And then we get the other coefficients from the identity obtained by
taking the common denominator:
4(5x  4)  4(Ax + B)(x  2)2 + 4C(x  2)(x2 + 4) + 3(x2 + 4)
Equating coefficients of:
x3
4A + 4C = 0 so A =  C
Constant
 16 = 16 B  32C + 12 or 4B  8C =  7
x
20 = 4(4A  4B) + 16C or 4A 4B + 4C = 5
So we now have three equations to solve for A, B, C:
A+C=0
(1)
4B  8C =  7
(2)
4A  4B + 4C = 5
(3)
Using (1) in (3) gives
5
 4B = 5 so B =  4
1
1
Substituting in (2) gives C = 4 then A =  4
So finally
(x2
vii)
5x  4
x+5
1
3
 4(x2 + 4) +
+
2
+ 4)(x  2)
4(x  2)
4(x  2)2
The cover-up rule works well here
x+1
2/4(3) ( 2)/(  4)(7) 5/(4)(7)

+
+
x+3
(x  1)(x + 3)(x  4)
x1
x4

1
1
5
 14(x + 3) +
6(x  1)
28(x  4)
2.3.11 Properties of quadratic expressions and equations
2.3.11A.
Factorise the quadratics
x2 + x – 2
ii)
x2 + 6x + 9
iii)
x2 – 81
iv)
2x2 + 5x – 3
v)
2x2 – 8x
i)
Solution
i) We look for factors of  2 that add to give  1. 1 and  2 work,
suggesting the factors x  1 and x  ( 2) = x + 2, and indeed we can
check that
(x  1)(x + 2) = x2 + x  2
ii)(Hopefully) we recognise the perfect square:
x2 + 6x + 9 = x2 + 2  3x + 32 = (x + 3)2
iii)You should have the difference of two squares at your fingertips now:
x2  81 = x2  92 = (x  9)(x + 9)
iv)Things get complicated when the coefficients of x2 are not unity. An
algorithm to help (although trial and error is just as good) is:
For ax2 + bx + c, write down factors of a (a1, a2) and c (c1, c2) in an
array:
a1, c1
a2, c2
With signs included this may yield a number of possible arrays. The aim
is to find one such that the ‘cross multiplication’
a1c2 + c1a2
gives b. Then the factors are (a1x + c1) (a2x + c2). It soon becomes clear
whether or not this is possible with trial and error. In the current case we
soon find that
1
2
3
1
with a cross product of 1( 1) + 3(2) = 5 works. So the factors are x + 3
and 2x  1 as you can easily check
(x + 3)(2x  1) = 2x2 + 6x  x  3 = 2x2 + 5x  3
v) 2x2  8x = 2x(x  4)
2.3.11B.
Solve the quadratic equations obtained by equating the expressions in A to
zero.
Solution
With the quadratics factorised, the solution is straightforward, as we illustrate
in detail for the first of the exercises.
i)x2 + x  2 = (x  1)(x + 2) = 0 can only be true if either (x  1) = 0 or (x +
2) = 0 (because for real numbers ab = 0 implies either a or b = 0). These
two linear equations give either x = 1 or x =  2 and so the two possible
solutions are x = 1,  2.
ii) x2 + 6x + 9 = (x + 3)2 = 0 gives the solution x =  3 (twice)
iii)
x2  81 = (x  9)(x + 9) = 0 gives the solutions x =  9
iv)
1
2x2 + 5x  3 = (x + 3)(2x  1) = 0 gives the solutions x = 2 ,  3
v)
2x2  8x = 2x(x  4) = 0 gives the solutions x = 0, 4 (NB it is a
common mistake to forget the zero solution in such cases).
2.3.11C.
Complete the square
i) x2 + 2x + 2
iii)
4x2 + 4x – 8
Solution
ii)
iv)
x2 – 6x + 13
4x2 – 4x
We have to ‘add and subtract half the coefficient of x all squared’. In practice,
if you really know (x + a)2 = x2 + 2ax + a2 then this rigmarole is rarely
necessary.
i)
x2 + 2x + 2 = x2 + 2x + 1 + 1 = (x + 1)2 + 1
ii)
x2  6x + 13 = x2  6x + 9 + 4 = (x  3)2 + 22
iii)
First take out the factor of 4:
1
1
1
4x2 + 4x  8 = 4(x2 + x  2) = 4(x2 + 2  2 x + 22  22  2)
 
 
1
1
9
1
4((x + 2 )2  22  2) = 4((x + 2 )2  4 )
 
1
= 4(x + 2 )2  32
1 1
1
1
1

4x2  4x = 4(x2  x) = 4(x2  2 2 x + 2 2  2 2) = 4 x  22  4
 
 
 



iv)
1

4 x  22  1 = (2x  1)2  1


An alternative in this case is to note that
4x2  4x = 4x2  4x + 1  1 = (2x  1)2  1
which relies on knowing the general result
(ax + b)2 = a2 x2 + 2abx + b2
2.3.11D.
By completing the square determine the maximum or minimum values (as
appropriate) of the following quadratics, and the values of x at which they
occur.
x2 + 2x + 4
i)
ii)
16 – 4x – 4x2
Solution
i)
x2 + 2x + 4 = x2 + 2x + 1 + 3 = (x + 1)2 + 3
This has a minimum value of 3 when x =  1
16  4x  4x2 = 16  (4x + 4x2) = 16  (4x2 + 4x + 1  1)
ii)
= 16  ((2x + 1)2  1) = 17  (2x + 1)2
1
This has a minimum value of 17 when 2x + 1 = 0, ie at x =  2
2.3.11E.
Solve the following equations by both factorisation and formula.
i) x2 + 3x + 2 = 0
ii)
2x2 – 5x + 2 = 0
3x2 – 11x + 6 = 0
x2 + 2x – 8 = 0
iii)
v)
x2 + 10x + 16 = 0
9x2 + 6x – 3 = 0
iv)
vi)
Solution
One can always solve by formula of course, whether or not the quadratic
factorises. To get away from blind use of the formula and practice our algebra
we will solve both by factorisation and completing the square (which is
effectively deriving the formula from scratch)
i)
3
3
x2 + 3x + 2 = (x + 1)(x + 2) = x2 + 3x + 2 2  2 2 + 2
 
 
3
1

= x + 2 2  4 = 0


From the factors we get x + 1 = 0 and x + 2 = 0, ie x =  1,  2. From the
completion of the square we obtain
3
1

x + 2 2 = 4


Taking square roots of both sides we get
3
1
x+2 = 2
so
3
1
x =  2  2 =  1,  2 as above.
ii)
By inspection, the array
1
2
2
1
works to produce the factors as
1
2x2  5x + 2 = (x  2)(2x  1) = 0 giving solutions x = 2 and 2
Completing the square
5
5
5
5
2x2  5x + 2 = 2[x2  2 x + 1] = 2[x2  2 x + 4 2  4 2 + 1]
 
 
5
9
5
9


= 2[x  42  16 ] = 2x  42  8 = 0




from which
5
9

x  42 = 16


and taking square roots gives
5
3
x4 =4
from which we get the same results for x as before.
iii)
Trial and error soon shows that the array
3
2
1
3
does the job so the factors are
3x2  11x + 6 = (3x  2)(x  3) = 0
giving solutions
2
x = 3 and 3
Completing the square
11
3x2  11x + 6 = 3[x2  3 x + 2]
11
11
49
11
49

11


= 3[x  6 2   6 2 + 2] = 3[x  6 2  36 ] = 3x  6 2  12 = 0


 




from which
11
49

x  6 2 = 36


giving
11
7
x 6 = 6
reproducing the same results as before.
iv) x2 + 10x + 16 = (x + 2)(x + 8) = 0 gives x =  2 and  8
x2 + 10x + 16 = (x + 5)2  25 + 16 = (x + 5)2  9 = 0
from which
x+5=3
leading to the same results.
v) x2 + 2x  8 = (x + 4)(x  2) = 0 giving x = 2 and  4
x2 + 2x  8 = (x + 1)2  9 = (x + 1)2  32 = 0 gives x + 1 =  3, etc
1
vi) 9x2 + 6x  3 = 3(3x2 + 2x  1) = 3(x + 1)(3x  1) giving x =  1 and 3
9x2 + 6x  3 = 9x2 + 6x + 1  4 = (3x + 1)2  4 = 0, etc, etc.
2.3.11F.
Evaluate the following exactly with as little labour as possible and without a
calculator.
8(23.7)2 – 10(23.7)(45.4) + 3(45.4)2
4(23.7) – 3(45.4)
Solution
This question tests how alert you are to novel appearances of quadratic
equations. If we put x = 23.7 and y = 45.5 then the given expression is
8x2  10xy + 3y2 (4x  3y)(2x  y)
=
= 2x  y
4x  3y
4x  3y
= 2(23.7)  45.5 = 47.4  45.4 = 2
Of course, if you are still not yet fully on top of quadratics, then you may find
this approach very laborious!
2.3.11G.
For Q2.3.11E confirm your results by calculating the a) sum and b) product
of the roots.
Solution
In the general quadratic equation x2 + ax + b = 0 the sum of the roots must be
equal to  a and the product to b.
i)
For x2 + 3x + 2 = 0 we found the solutions x =  1, 
2, and indeed their sum is  3 and their product is 2, as the above result
requires.
1
ii)For 2x2  5x + 2 = 0 we found solutions x = 2 and 2. To make the
coefficient of x2 unity we divide by 2 giving the equation
5
x2  2 x + 1 = 0
5
The sum of the roots is 2 and the product is 1, as expected.
iii)3x2  11x + 6 = 0 can be rewritten as
11
x2  3 x + 2 = 0
and the solutions
2
x = 3 and 3
sum and multiply to the required results.
iv)x2 + 10x + 16 = 0 has solutions x =  2 and  8 which add to  10 and
multiply to 16, as required.
v) x2 + 2x  8 = 0 has roots x = 2 and  4 which sum to  2 and multiply
to  8
1
2
vi) 9x2 + 6x  3 = 0 gives x =  1 and 3 adding to  3 and multiplying
1
to  3

2.3.12 Powers and indices for algebraic expressions
2.3.12A.
Express in the form an
i) a2 a4
ii) a3 a2 a
iii) a a2 a–1
iv)
a7 a3/a2
vii)
a5/a–3
2
v) (a3) a–2 a3
3
vi) a–21 a2(a3)
6
2
viii) (a2) /(a3)
Solution
i)
a2a4 = a6
ii)
a3a2a = a6
iii)
aa2a 1 = a2(aa 1) = a2
iv)
a7a3 a10
10  2 = a8
a2 = a2 = a
v)
(a3)2 a 2 a3 = a6a = a7
vi)
a 21 a2(a3)6 = a 21 a2a18 = a 1
vii)
a5
= a5a3 = a8
a 3
viii)
(a2)3/(a3)2 = a6/a6 = 1
2.3.12B.
Express in the form a2n , stating the value of n.
i)
iv)
a9 a17
a27 a–3/a2
Solution
i)
a9a17 = a26 = a2(13)
So n= 13
1/4
ii) (a40)
iii)
6
1/2
a3((a3) a7 a5)
ii)
(a40)1/4 = a10 = a2(5)
So n = 5
iii)
a3((a3)6a7a5)1/2 = a3(a18a12)1/2 = a3(a30)1/2 = a3a15 = a18
So n = 9
a27a 3
27  3  2
22
a2 = a a a = a
So n = 11
iv)
2.3.12C.
Reduce to simplest form
i)
iv)
3
(a2) c12
ii)
a b2
a2 b3 c2
abc
iii)
ba12 b7 c4
(a2 b4 c6)1/2
(a3)4 c12
b–3 c10 a–1 b2
Solution
i)
a2b3c2
2
abc = ab c
ii)
(a2)3c12 a6c12
5  2 12
ab2 = ab2 = a b c
iii)
ba12b7c4
a12b8c4
11 6
=
(a2b4c6)1/2
ab2c3 = a b c
iv)
(a3)4c12
aa12c12b3b 2
=
= a13bc2
c10
b 3c10 a 1b2
2.3.12D.
Simplify the following expressions
i)
iii)
2 2 3
3
b a b ab
(x2)4 y7/2
(x6)1/3 y
ii)
t3xy
x2yt
Solution
i)
You must gather all the as and bs together
b2a2 b3a b3 = a2a b2b3 b3 = a2 + 1 b2 + 3 + 3 = a3b8
ii)
t3xy
t2
3  1x1  2 (ys cancel) = = t2x  1 =
=
t
x2yt
x
iii)
(x2)4y7/2 x8y7/2
=
= x6y3 remembering (am)n = amn and
(x6)1/3 y x2y1/2
iv)
Note that 27 = 33, 8 = 23, so
y = y1/2
81/3 27 (23)1/3 33
2
3223 = 3223 = 3  2
2.3.13 The binomial theorem
2.3.13A.
Write down the coefficients of x4 in the following expansions
i)
iv)
7
(1 + x)
(3 – 2x)8
ii)
v)
(1 + 3x)6
(3x + 2y)6
iii)
(1 – 2x)5
Solution
i)
7 C4
13x4 is the x4 term, so the coefficient is
7 C4
7!
765
= 3!4! =
3!
= 35
ii)
The required term is 6C4 13 (3x)4 so the coefficient is
6
iii)
6!
C4 34 = 2!4!  81 = 15  81 = 1215
Coefficient is 5C4 ( 2)4 = 5  16 = 80
1215
iv)
8!
Coefficient is 8C4 34( 2)4 = 4!4! 3424 = 90720
v)
Coefficient is 6C4 2234 = 1215  4 = 4860
2.3.13B.
Expand by the binomial theorem
i) (1 – 2x)7
ii)
(a + b)6
iii)
(2x
+
3y)5
(2 – 3x)6
iv)
v)
(3s – 2t)5
Solution
i)Watch out for signs and powers of 2x. It is best to treat  2x as a single entity
in the binomial expression:76
765
(1  2x)7 = 1 + 7( 2x) + 2! ( 2x)2 +
( 2x)3
3!
+
7654
7  6  5  4 3
( 2x)4 +
( 2x)5
4!
5!
+
7  6  5  4 3  2
7!
( 2x)6 +
( 2x)7
6!
7!
= 1 + 7( 2x) + 21( 2x)2 + 35( 2x)3 + 35( 2x)4 +21( 2x)5
+7( 2x)6 + ( 2x)7
(Remember that the coefficients are symmetric)
Now clear the brackets round the  2x (carefully!):
= 1  14 x + 84x2  280x3 + 560x4  672x5 + 448x6  128x7
Note that with such a finicky question it is always best to check – substituting a
suitable value will do here, say x = 1:
LHS(1) = ( 1)7 =  1
RHS(1) = 1  14 + 84  280 + 560  672 + 448  128 =  1
So we have agreement, which suggests the result is correct. This is not
bombproof of course because we have only tested the result for one particular
value of x, but it increases our confidence in the calculation.
ii)
65
654 3 3 6543 2 4
(a + b)6 = a6 + 6a5b + 2! a4b2 +
ab +
ab +
3!
4!
65432 5
ab + b6
5!
= a6 + 6a5b + 15 a4b2 + 20 a3b3 + 15 a2b4 + 6 ab5 + b6
iii)
(2x + 3y)5 =
(2x)2(3y)3 +
(2x)5 + 5 (2x)4 (3y) +
54
543
3
2
(2x)
(3y)
+
2!
3!
5432
(2x)(3y)4 + (3y)5
4!
= 32x5 + 240x4 y + 720 x3y2 + 1080 x2y3 + 810 xy4 + 243y5
iv)
65
654 3
(2  3x)6 = 26 + 6  25( 3x) + 2! 24( 3x)2 +
2 ( 3x)3 +
3!
6543 2
65432
4
2
(
3x)
+
2 ( 3x) 5 + ( 3x)6
4!
5!
= 64  6  32  3x + 15  16  9x2  20  8  27x3 + 60  4  81x 4
 12  243x5 + 729x6
= 64  576x + 2160x2  4320x3 + 19440x 4  2916x5 + 729x6
(3s  2t)5 = (3s)5 + 5 (3s)4 ( 2t) +
v)
(3s)2( 2t)3 +
54
543
3
2
2! (3s) ( 2t) +
3!
5432
(3s)(  2t)4 + ( 2t)5
4!
= 243s5  810s4 t + 1080 s3t2  720 s2t3 + 240 st4  32t5
2.3.13C.
Evaluate, without using a calculator:-
(1 +
i)
2)
3
+ (1 –
2)
3
ii)
(2 +
4
3 ) + (2 –
3)
4
Solution
The point of these questions is that on using the binomial theorem some terms
occur with opposite signs and cancel out.
i)
(1 + 2 )3 + (1  2 )3 = 1 + 3 2 + 3( 2 )2 + ( 2 )3
+ 1  3 2 + 3( 2 )2  ( 2 )3
= 2 + 6( 2 )2 = 14
ii)
(2 +
3 )4 + (2 
3 )4 = 24 + 4  23
)3 + ( 3 )4 + 24  4  23
3 + 6  22 ( 3 )2 + 4  2 ( 3
3 + 6  22 ( 3 )2  4  2 ( 3 )3 + ( 3 )4
= 2(24 + 6  22 ( 3 )2 + ( 3 )4) = 226
2.3.13D.
Use the binomial theorem to evaluate to three decimal places:
i)
(1.01)10
ii) (0.998)8
Hint: write 1.01 = 1 + 0.01 and 0.998 = 1 – 0.002
Solution
In this question we are using the fact that if one term, say b, in a binomial
expression a + b is much smaller in magnitude than the other, a, then we can
use a binomial series to express a power such as (a + b)n in terms of powers of
b, the higher ones of which can possibly be neglected to give an
approximation to a required accuracy. For example if x is much smaller than 1
then we may be able to use the approximation
(1 + x)n  1 + nx
i)
(1.01)10 = (1 + 0.01)10 = 1 + 10(0.01) +
+
10  9
2
2! (0.01)
10  9  8
10  9  8  7
(0.01)3 +
(0.01)4 + 
3!
4!
= 1 + 0.1 + 0.0045 + 0.000120 + 0.0000021 + 
= 1.105 to three decimal places
The point here is, of course, that (Provided the series converges, which
you may here assume – see Section 3.2.10) subsequent terms of the series
are not going to change the third decimal place.
ii)
Some cunning is needed here, but it is not too difficult:
(0.998)8 = (1  0.002)8 = (1  2(0.001))8
87
= 1  2  8 (0.001) + 2! (0.001)2 22 + 
= 1  0.016 + 0.000112 +  = 0.984 to 3 dp