CHAPTER 2 SOLUTIONS TO REINFORCEMENT EXERCISES IN ALGEBRA
2.3.1Multiplication of linear expressions
2.3.1A.
Identify which of the following are algebraic expressions in the variables involved
i)14ii)x – 3
iii)2t + 1 = 0iv)2x + y
v)x3 – 2x2 + x – 1vi)ex + e–x
vii)x2 – 3x + 2 = 0viii)2– x2
ix)x)sin x – 3 cos x
xi)s– 2s2 + s = 0 xii)
xiii)2 cos x = 1xiv)u2 + 2uv – 3v
xv)= 0xvi)x ln x + x2 = 0
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 equationthat equates the linear algebraic expression 2t + 1 to zero. It tells us that t = .
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 ex = 1/ex are algebraic expressions, so neither is ex + ex. 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)2 x2 = 2x x2 is an algebraic expression roots are allowed.
ix) 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)s 2s2 + s = 0 is an algebraic equation in s obtained by equating the algebraic expression s 2s2 + s to zero. This is not so easy to solve for s!
xii) is rational (algebraic) function.
xiii)2 cos x = 1 is an equation, equating the transcendental expression 2 cos x (not algebraic) to 1. It tells us that cos x = .
xiv)u2 + 2uv 3v is algebraic in both u and v.
xv) = 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+da+bc+bdviii) a+bc+bd a+bc+b2d
ii) a+bc+dac+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
v) a–bc–d ac–bc–d xii) x2–3 x+4 x3–3x+4
vi) a–bc+d ac–bc+d
vii) x2–3x+4 x2–3x–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)= ac + bc + ad + bd
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)= 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)2(x + 2)ii)3(x – 1) – (x – 4)
iii)3t(t – 1)iv)(s– t)(s + 2t)
v)a2(a – 3)vi)(x2 + 2x – 1)(x – 1)
vii)– 2u(u2 + 3)viii)9(x2 – 3) – 2(x + 4)
ix)(a2 – 1)(a + 2) – 3(a – 3)x)x(x – 1)(x + 2) – 3x2
xi)– (x – x2)(x – 2)xii)– [(x2 – 1)(x – 2) – (x – 3)(x + 2)]
xiii)(1 – t)(1 – s)(1 – u)xiv)(a – 2b)2 – (a + 2b)2
xv)(x – y)2 + (x + y)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)2(x + 2) = 2x + 2 2 = 2x + 4
ii)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)
= [IE1]
= 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.
2.3.2Polynomials
2.3.2A.
Which of the following are polynomials? For those that are give the degree and list the coefficients.
i)t2 – t + 4ii)0iii)
iv)7t3 – 2t + 1v)4x4 – 2x3 + 3x – vi)27x4 – 3x2 + 1
vii)viii)x + ix)3x2 + t3
x)x2y +
Solution
i)t2 t + 4 is a polynomialof degree 2 (a quadratic) in t, with coefficients 1, 1, 4. (Note that the minus sign is included).
ii)0 is a polynomialof degree 0 (a constant) with coefficient 0.
iii) is not a polynomial, it is a rational function in u.
iv)7t3 2t + 1 is a third degree polynomial with coefficients 7, 0, 2, 1 (note the zero coefficient).
v)4x4 2x3 + 3x is not a polynomial because of the part.
vi)27x4 3x2 + 1 is a polynomial of degree 4 with coefficients 27, 0, 3, 0, 1.
vii) is actually a rational function, not a polynomial, even though we might be tempted to cancel the x to get
= x2 + 2
we can only do this if x 0. So exists only provided x 0 while the polynomial x2 + 2 exists for all x.
viii)The means that x + is not a polynomial, as = 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 + 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)
iii)(x – 1)(x + 1)2iv)(x – 2)(x – 3)(x + 1)(x + 2) v) (u – 1)2(u + 1)2 vi) (x – 1)3 (x + 2)
vii)(t + 1)(t – 2)(t + 2)viii)(u – 2)(u + 3)(u – 3)
ix)(s – 2)4x)(x – 1)(x + 2)(x – 3)(x + 4) xi) (x + 2)2(x – 3)2 xii) (2t + 1)(3t – 4) xiii) (3s – 1)(s + 2)(4s + 3) xiv) (3x + 2)2(2x – 1)(x + 2)
xv)(3x + 1)(3x – 1)(x + 3)
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
2.3.3Factorisation of polynomials by inspection
2.3.3A.
Factorise the following, retaining only real coefficients
i) x2 + xii)3x3 – 2x2iii)– 7x2 + 42x4
iv)t4 – 3t3 + t2v)u2 – 9vi)t2 – 121
vii)s24 – 16s22viii)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 (x 1) (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)
i)t2 + 5t + 6ii)t3 + t2 – 4t – 4iii) y4 – 2y3 – 7y2 + 8y + 12
iv)9x3 + 27x2 – x – 3
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.
2.3.4Simultaneous equations
2.3.4A.
Solve the following systems of linear equations, verifying your solution by back substitution in each case.
i)x – y = 1ii)A + B = 0iii)s + 3t = 1
x + 2y = 03A – B = 1s – 2t = 1
iv)3x + 2y = 2v)u + 4v = 1vi)7x1 – 2x2 = 1
– 2x + 3y = 1u – v = 23x1 – 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 (1)
x + 2y = 0 (2)
Subtracting (1) from (2) removes x to give
3y = 1
so y =
Then from (2)
x = 2y =
Substituting these values in the equations gives
x y =
x + 2y = = 0
as required.
ii)A + B = 0 (1)
3A B = 1(2)
(1) + (2) gives 4A = 1 so A =
(1) gives B = A =
Then A + B = = 0
and 3A B = 3 () = 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(1)
2x + 3y = 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
x =
From the original equation (1) we now obtain
2y = 2 3x = 2
soy =
Checking :-
3x + 2y = = 2
2x + 3y = = 1
v)u + 4v = 1(1)
u v = 2(2)
(1) (2) gives 5v = 1 so v =
Then
u = 2 + v = 2
Checking:
u + 4v = + 4 = 1
u v = = 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)
4x1 = 1, so x1 =
Then (2) gives
x2 =
7x1 2x2 = 7 2 = 1
3x1 2x2 = 3 2 = 0
2.3.4B.
Comment on the following systems of equations
i)x + y = 1ii)2x – y = 3iii)x + y = 0
3x + 3y = 34x – 2y = 1x – y = 0
iv)2A + B = 1v)u + v = – 1vi)x + y = 0
4A + 2B = – 1 3u + 3v = – 3 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
2x y =
Such a system has no solutions, since it would require 3 = . 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) , v = s s arbitrary.
vi)x + y = 0 (1)
x2 y2 = 1(2)
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.
2.3.5Equalities and identities
2.3.5A.
Determine the real values of A, B, C, D in the following identities
i)(s – 1)(s + 2) As2 + Bs + C
ii)(x – 1)3 Ax3 + 2Bx2 – 3Cx + D
iii)(x – A)(x + B) x2 – 4
iv)(x + 2)(x – 3)(2x – 1) A(x – 1)2 + Bx + Cx3 – Dx2
v)A(x – 1)+ B(x + 2) x – 3
vi)(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
SoA = 1
2B = 3 or B =
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 B =
x = 2 knocks out B to give
3A = 5 so A =
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 or A = 1
and
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 =
Put x = a so
B(a b) = a2 + b
and
B =
Second method
Rearrange and equate coefficients:-
A(x A) + B(x b) (A + B) x (aA + bB)
ax + b
So
A + B = a(1)
aA + bB = b(2)
(2) b(1) gives
(a b)A = b ab
or
A =
as before, and
B = a A = a
=
=
=
as before.
2.3.6Roots and factors of a polynomial
Use the factor theorem to factorize the following polynomials
i)u3 – u2 – 4u + 4ii)x3 + 4x2 + x – 6 iii) t4 – t3 – 7t2 + t + 6 iv) x3 – x2 – x + 1 v) x3 + 3x2 – 10x – 24
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)
2.3.7Rational functions
2.3.7A.
Identify the rational functions and state a) the denominator, b) the numerator.
i)ii)iii)
iv)v)
Solution
i)Rational function a) x 1 b) x + 1.
ii)Rational function (despite the deliberately misleading ) a) x2 1 b) x + .
iii)Not a rational function because the numerator is not a polynomial.
iv)Rational function a) x4 2x + 3 b) x3 + 2x 1
v)Rational function a) x4 + 1 b) 2x4 1.
2.3.7B.
For what values of x does – exist?
Solution
It may appear that
= 0
but of course this assumes the existence of both terms in the difference.
exists for all values of x except for x = 1
So exists (and = 0) for all values of x except x = 1.
We should strictly write
= 0 x 1
2.3.8Algebra 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)– ii)–
iii)– iv)+
v)+ + vi)– –
vii)+ viii)x – +
ix)+ x)2x – 1 + –
Solution
i)
=
=
Check: x = 0 on LHS gives 1 + 3 = 4 and on the RHS = 4
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:-
=
etc
ii) =
=
LHS (0) = RHS (0) =
iii) =
=
LHS (0) = RHS (0) =
iv)
=
LHS (0) = 1 1 = 2 RHS (0) = = 2
v)
Here the common denominator is (x 1)(x 2)(x 3), so we convert each fraction to have this denominator.
= + +
=
(With practice you will soon be able to jump direct to the last line)
=
=
LHS (0) = 1 RHS (0) =
vi) =
=
= Note care needed with signs and brackets.
=
LHS (0) = RHS (0) =
vii)
In this case care is needed – the common denominator obtained by cross multiplication is not the last word:-
=
Noticing that 2x2 x 1 (2x + 1)(x 1), this becomes
= =
What has happened here is that in fact the LCD is not (x 1)(x2 1) but (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:-
=
=
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) x
=
=
Notice how the numerator is of degree greater than the denominator i.e. the fraction is improper.
LHS (0) = 1 + 1 = 0 RHS (0) = 0
ix) =
=
The x2 makes no difference to the methodology here.
LHS (0) = 3 + 1 = 4 RHS (0) = = 4
x)2x 1 + =
=
=
=
LHS (0) = 1 2 RHS (0) =
2.3.8B.
Put over a common denominator
i)+ ii)–
iii)+ iv)–
v) 1 +
Solution
i) =
=
ii)
iii) =
=
=
iv) =
=
=
v)1 + =
2.3.9Division and the remainder theorem
2.3.9A.
Perform the following divisions, whenever permissible:
i)ii)
iii)iv)
v)vi)
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:-
= 1 but only if x 0
ii) =
=
= 1 + provided x 1
iii) =
=
= x3 +
= x3 + x2 +
= x3 + x2 + 3x +
= x3 + x2 + 3x +
= x3 + x2 + 3x + 13 + (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) =
=
= x2
(Notice how it is best to pull the sign out in such cases.)
= x2
= 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) =
= 2 x 1
2.3.9B.
Find the remainders when the following polynomials are divided by:-
a) x – 1b)x + 2c)x
i)3x3 + 2x – 1ii)x5 – 2x2 + 2x – 1
iii)x4 – x2iv)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)Remainder = p(0) = 1
ii)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)p(0) = 1
iii)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) p(1) = 2 3 + 4 2 + 1 = 2
b)p( 2) = 2( 2)7 3( 2)5 + 4 ( 2)3 2( 2)2 + 1
= 199
c)p(0) = 1
2.3.10Partial 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
+
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
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:
+
+ as above
ii)From now on we will use the cover-up rule where convenient.
+
+
as in 2.3.8A ii).
iii) +
as in 2.3.8A iii)
iv) + +
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
+ +
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