CHAPTER 2 SOLUTIONS TO REINFORCEMENT EXERCISES IN ALGEBRA

2.3.1 Multiplication of linear expressions2.3.1A.Identify which of the following are algebraic expressions in the variables involved

i) 14 ii) x – 3iii) 2t + 1 = 0 iv) 2x + yv) x3 – 2x2 + x – 1 vi) ex + e–x

vii) x2 – 3x + 2 = 0 viii) 2– x2

ix) x) sin x – 3 cos xxi) s– 2s2 + s = 0 xii)

xiii) 2 cos x = 1 xiv)u2 + 2uv – 3vxv) = 0 xvi) x ln x + x2 = 0

Solutioni) 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 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.

SolutionAs 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 v) a–bc–d ac–bc–d xii) x2–3 x+4 x3–3x+4 vi) a–bc+d ac–bc+dvii) x2–3x+4 x2–3x–12

SolutionThis 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 + bdii) (a + b)c + d = ac + bc + diii) (a + b)(c + d) = ac + bc + ad + bdiv) a b(c + d) = a bc bdv) (a b)c d = ac bc dvi) (a b)c + d = ac bc + d

vii) x2 3(x + 4) = x2 3x 12viii) a + b(c + bd) = a + bc + b2d

ix) (a + bc + b)d = ad + bcd + bdx) (a b)(c d) = ac bc ad + bdxi) 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

SolutionThis ‘removing the brackets’ is essentially repeated application of the distributive rule.

a(b + c) = ab + acJust be careful with signs and squares.

i) 2(x + 2) = 2x + 2 2 = 2x + 4ii) 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 6uviii) 9(x2 3) 2(x + 4) = 9x2 27 2x 8

= 9x2 2x 35ix) (a2 1)(a + 2) 3(a 3) = a3 + 2a2 a 2 3a + 9

= a3 + 2a2 4a + 7x) x(x 1) (x + 2) 3x2 = x(x2 + x 2) 3x2

= x3 + x2 2x 3x2

= x3 2x2 2xxi) (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)

= = 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) = 8abxv) 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.

2.3.1EFactorise each of your answers to Question 2.3.1D as far as

possible.SolutionThis 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 equationx3 + 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 expression1 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 Polynomials2.3.2A.Which of the following are polynomials? For those that are give the degree and list the coefficients.

i) t2 – t + 4 ii) 0 iii)iv) 7t3 – 2t + 1 v) 4x4 – 2x3 + 3x – vi) 27x4 – 3x2 + 1vii) viii) x + ix) 3x2 + t3

x) x2y +

Solutioni) 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) 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)2 iv) (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)4 x) (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.

SolutionThis 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 + 12LHS = 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 = 1vi) (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 8xv) (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 inspection2.3.3A.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 – 121vii) s24 – 16s22 viii) 4x12 – 64x8

Solutioni) 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 + 6 ii) t3 + t2 – 4t – 4 iii) y4 – 2y3 – 7y2 + 8y + 12

iv) 9x3 + 27x2 – x – 3

SolutionThis 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 that9x3 + 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 equations2.3.4A.Solve the following systems of linear equations, verifying your solution by back substitution in each case.

i) x – y = 1 ii) A + B = 0 iii) s + 3t = 1x + 2y = 0 3A – B = 1 s – 2t = 1

iv) 3x + 2y = 2 v) u + 4v = 1 vi) 7x1 – 2x2 = 1– 2x + 3y = 1 u – v = 2 3x1 – 2x2 = 0

SolutionIn 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 = 1so 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 = = 0and 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 = 1Substituting back into the equations, we haves + 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 = 62 (2) 4x + 6y = 2

Subtract these equations to get13x = 4

sox =

From the original equation (1) we now obtain2y = 2 3x = 2

so y = 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 = 1u 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 = 13x1 2x2 = 3 2 = 0

2.3.4B. Comment on the following systems of equations

i) x + y = 1 ii) 2x – y = 3 iii) x + y = 03x + 3y = 3 4x – 2y = 1 x – y = 0

iv) 2A + B = 1 v) u + v = – 1 vi) x + y = 04A + 2B = – 1 3u + 3v = – 3

x2 – y2 = 1

SolutionHere 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 = 1This 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 system2x y = 32x 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 = 14A + 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) = 1From (1) this gives 0 = 1, which is not possible.

2.3.5 Equalities and identities2.3.5A.Determine the real values of A, B, C, D in the following identities

i) (s – 1)(s + 2) As2 + Bs + Cii) (x – 1)3 Ax3 + 2Bx2 – 3Cx + D

iii) (x – A)(x + B) x2 – 4iv) (x + 2)(x – 3)(2x – 1) A(x – 1)2 + Bx + Cx3 – Dx2

v) A(x – 1)+ B(x + 2) x – 3vi) (x + A)2 + B2 x2 – 2x + 5

SolutionIn 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 + DSo A = 1

2B = 3 or B = 3C = 3 or C = 1D = 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 = 54or

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

soA = 1 and B = 2

Alternatively expand both sides and equate coefficients:-

(x + A)2 + B2 x2 + 2Ax + A2 + B2

x2 2x + 5so

2A = 2 or A = 1and

A2 + B2 = 5so

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

SolutionThis 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

soA =

Put x = a so

B(a b) = a2 + band

B =

Second methodRearrange and equate coefficients:-

A(x A) + B(x b) (A + B) x (aA + bB) ax + b

SoA + B = a (1)aA + bB = b (2)

(2) b(1) gives

(a b)A = b ab

orA =

as before, and

B = a A = a

=

=

=

as before.

2.3.6 Roots and factors of a polynomialUse the factor theorem to factorize the following polynomials

i) u3 – u2 – 4u + 4 ii) x3 + 4x2 + x – 6iii) t4 – t3 – 7t2 + t + 6 iv) x3 – x2 – x + 1v) x3 + 3x2 – 10x – 24

SolutionHere 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 = 113 12 4 + 4 = 0

u = 223 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 = 0so 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 = 0so (t 1) is a factor

p( 1) = 1 + 1 7 1 + 6 = 0so (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 = 0so (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) 0p( 2) = 0 so (x + 2) is a factorp(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) = 0confirms this. Thus

x3 + 3x2 10x 24 (x + 2)(x 3)(x + 4)

2.3.7 Rational functions2.3.7A.Identify the rational functions and state a) the denominator, b)

the numerator.

i) ii) iii)

iv) v)

Solutioni) 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?

SolutionIt may appear that

= 0but of course this assumes the existence of both terms in the difference.exists for all values of x except for x = 1So exists (and = 0) for all values of x except x = 1.We should strictly write

= 0 x 1

2.3.8 Algebra of rational functions2.3.8A.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 + –

Solutioni)

= =

Check: x = 0 on LHS gives 1 + 3 = 4 and on the RHS = 4Notes: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 + =

2.3.9 Division and the remainder theorem2.3.9A.Perform the following divisions, whenever permissible:

i) ii)

iii) iv)

v) vi)

SolutionThe 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 – 1 b) x + 2 c) x

i) 3x3 + 2x – 1 ii) x5 – 2x2 + 2x – 1

iii) x4 – x2 iv) 2x7 – 3x5 + 4x3 – 2x2 + 1

SolutionBy the remainder theorem the remainder when the polynomial p(x) is divided by x a is p(a).

i) p(x) = 3x3 + 2x 1a) Remainder when divided by x 1 is p(1) = 3 + 2 1 = 4b) Remainder = p( 2) (note the sign)

= 3( 2)3 + 2(2) 1= 24 4 1 = 29

c) Remainder = p(0) = 1ii) p(x) = x5 2x2 + 2x 1

a) p(1) = 1 2 + 2 1 = 0So (x 1) actually divides, i.e. is a factor of, x5 2x2 +

2x 1b) p( 2) = ( 2)5 2( 2)2 + 2( 2) 1

= 32 8 4 1 = 45c) 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 = 28c) 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

2.3.10 Partial fractions2.3.10A.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.

SolutionWe 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

+ + + +

as in 2.3.8A v).

vi) + +

as in 2.3.8A vi).

vii) In this case we will not get the original result. Direct cover-up rule gives

+ which you may think looks nothing like the original:

+ However, let’s write this as

+ + + + using the cover-up rule

+ 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.

x + x +

x + + x +

as in RE 2.3.8A vii)

ix) 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

+ as 2.3.8A ix).

x) Once we have divided out we obtain a simple partial fractions to do

2x 1 + 2x 1 +

as 2.3.8A x).

xi) + +

as 2.3.8B i).

xii) Nothing much you can do with !

xiii) In the case of we could use the decomposition+ +

and determine the coefficients. However, we will take the hint from the original form given in RE 2.3.8B iii), and assume

+

SoA(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 Constants4A C = 13, so C = 4A 13 = 1So

+ as 2.3.8B iii).

xiv) We assume the form +

We can get A from the cover-up rule:A = = 4

So

So (Bx + C)(x 2) 4(x2 + 1) 2x2 5x 2

Equating coefficients gives:

x2

B 4 = 2, so B = 2Constant 2C 4 = 2, so C = 1

So

as in RE 2.3.8B iv).

xv) This is just a matter of dividing out 1 +

as in 2.3.8B v).

2.3.10B.Split into partial fractions

i) ii) iii)

iv) v) vi)

vii)

Solutioni) + +

ii)

iii) = + = iv) + + Note that C as well as A can be found using the cover-up rule

A = = 3C = = 1

So + +

We can get B easily by putting x = 0, giving= 1 B

So B = 1 and therefore + +

v) + + by the cover up rule

Equating coefficients in the numerators now givesx2

+ B = 0 and so B = Constant+ C = 3 and so C = Hence

vi) Write + +

We can get D from the cover-up ruleD = =

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 = CConstant 16 = 16 B 32C + 12 or 4B 8C = 7x20 = 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 4B = 5 so B =

Substituting in (2) gives C = then A = So finally

+ +

vii) The cover-up rule works well here + + +

2.3.11 Properties of quadratic expressions and equations2.3.11A.Factorise the quadratics

i) x2 + x – 2 ii) x2 + 6x + 9

iii) x2 – 81 iv) 2x2 + 5x – 3

v) 2x2 – 8x

Solutioni) 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 2ii)(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 32 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.

SolutionWith 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 = 9iv) 2x2 + 5x 3 = (x + 3)(2x 1) = 0 gives the solutions x = ,

3v) 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 ii) x2 – 6x + 13iii) 4x2 + 4x – 8 iv) 4x2 – 4x

SolutionWe 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 + 1ii) x2 6x + 13 = x2 6x + 9 + 4 = (x 3)2 + 22

iii)First take out the factor of 4:4x2 + 4x 8 = 4(x2 + x 2) = 4(x2 + 2 x + 2)

4((x + )2 2) = 4((x + )2 )= 4(x + )2 32

iv)4x2 4x = 4(x2 x) = 4(x2 2 x + 2 2) = 4

4 = (2x 1)2 1

An alternative in this case is to note that4x2 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.

i) x2 + 2x + 4 ii) 16 – 4x – 4x2

Solutioni) x2 + 2x + 4 = x2 + 2x + 1 + 3 = (x + 1)2 + 3 This has a minimum value of 3 when x = 1

ii) 16 4x 4x2 = 16 (4x + 4x2) = 16 (4x2 + 4x + 1 1) = 16 ((2x + 1)2  1) = 17 (2x + 1)2 

This has a minimum value of 17 when 2x + 1 = 0, ie at x =

2.3.11E.Solve the following equations by both factorisation and formula.

i) x2 + 3x + 2 = 0 ii) 2x2 – 5x + 2 = 0iii) 3x2 – 11x + 6 = 0 iv) x2 + 10x + 16 = 0v) x2 + 2x – 8 = 0 vi) 9x2 + 6x – 3 = 0

SolutionOne 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) x2 + 3x + 2 = (x + 1)(x + 2) = x2 + 3x + 2 2 + 2 = 2 = 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

2 = Taking square roots of both sides we get

=

so = 1, 2 as above.

ii) By inspection, the array1 22 1

works to produce the factors as2x2 5x + 2 = (x 2)(2x 1) = 0 giving solutions x = and 2

Completing the square2x2 5x + 2 = 2[x2 + 1] = 2[x2 + 2 2 + 1]

= 2[ ] = 2 = 0 from which

= and taking square roots gives

= from which we get the same results for x as before.

iii) Trial and error soon shows that the array3 21 3

does the job so the factors are3x2 11x + 6 = (3x 2)(x 3) = 0

giving solutionsx = and 3

Completing the square 3x2 11x + 6 = 3[x2 + 2]

= 3[ + 2] = 3[ ] = 3 = 0 from which

= giving

= reproducing the same results as before.

iv) x2 + 10x + 16 = (x + 2)(x + 8) = 0 gives x = 2 and 8x2 + 10x + 16 = (x + 5)2 25 + 16 = (x + 5)2 9 = 0 from which

x + 5 = 3leading to the same results.

v) x2 + 2x 8 = (x + 4)(x 2) = 0 giving x = 2 and 4x2 + 2x 8 = (x + 1)2 9 = (x + 1)2 32 = 0 gives x + 1 = 3, etc

vi) 9x2 + 6x 3 = 3(3x2 + 2x 1) = 3(x + 1)(3x 1) giving x = 1 and 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.

SolutionThis 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

= = 2x y = 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.ii)For 2x2 5x + 2 = 0 we found solutions x = and 2. To make the coefficient of x2 unity we divide by 2 giving the equation

x2 x + 1 = 0The sum of the roots is and the product is 1, as expected.

iii)3x2 11x + 6 = 0 can be rewritten as x2 x + 2 = 0

and the solutionsx = 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

vi) 9x2 + 6x 3 = 0 gives x = 1 and adding to and multiplying to

2.3.12 Powers and indices for algebraic expressions2.3.12A.Express in the form an

i) a2 a4 ii) a3 a2 a iii) a a2 a–1

iv) a7 a3/a2 v) (a3)2 a–2 a3 vi) a–21 a2(a3)6

vii) a5/a–3 viii) (a2)3/(a3)2

Solutioni) a2a4 = a6

ii) a3a2a = a6

iii) aa2a 1 = a2(aa 1) = a2

iv) = = a10 2 = a8 v) (a3)2 a 2 a3 = a6a = a7

vi) a 21 a2(a3)6 = a 21 a2a18 = a 1 vii) = a5a3 = a8 viii) (a2)3/(a3)2 = a6/a6 = 1

2.3.12B.Express in the form a2n , stating the value of n.

i) a9 a17 ii) (a40) iii) a3((a3)6 a7 a5)

iv) a27 a–3/a2

Solutioni) a9a17 = a26 = a2(13)

So n= 13

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

iv) = a27 a 3 a 2 = a22 So n = 11

2.3.12C.Reduce to simplest form

i) ii) iii)

iv)

Solutioni) = ab2c

ii) = = a5b 2c12 

iii) = = a11b6c

iv) = = a13bc2

2.3.12D.Simplify the following expressions

i) b2 a2 b3 ab3 ii)

iii)

Solutioni) You must gather all the as and bs together

b2a2 b3a b3 = a2a b2b3 b3 = a2 + 1 b2 + 3 + 3 = a3b8 ii) = t3 1x1 2 (ys cancel) = = t2x 1 = iii) = = x6y3 remembering (am)n = amn and = y1/2 iv) Note that 27 = 33, 8 = 23, so

= = 3 2 2

2.3.13 The binomial theorem2.3.13A.Write down the coefficients of x4 in the following expansions

i) (1 + x)7

ii) (1 + 3x)6 iii) (1 – 2x)5

iv) (3 – 2x)8 v) (3x + 2y)6

Solutioni) 7C4 13x4 is the x4 term, so the coefficient is 7C4 = =

= 35ii) The required term is 6C4 13 (3x)4 so the coefficient is

6C4 34 = 81 = 15 81 = 1215

iii) Coefficient is 5C4 ( 2)4 = 5 16 = 80

1215

iv) Coefficient is 8C4 34( 2)4 = 3424 = 90720

v) Coefficient is 6C4 2234 = 1215 4 = 4860

2.3.13B.

Expand by the binomial theoremi) (1 – 2x)7 ii) (a + b)6 iii) (2x + 3y)5

iv) (2 – 3x)6 v) (3s – 2t)5

Solutioni) Watch out for signs and powers of 2x. It is best to treat 2x as a single

entity in the binomial expression:-

(1 2x)7 = 1 + 7( 2x) + ( 2x)2 + ( 2x)3

+ ( 2x)4 + ( 2x)5

+ ( 2x)6 + ( 2x)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) (a + b)6 = a6 + 6a5b + a4b2 + a3b3 + a2b4 + ab5 + b6

= a6 + 6a5b + 15 a4b2 + 20 a3b3 + 15 a2b4 + 6 ab5 + b6

iii) (2x + 3y)5 = (2x)5 + 5 (2x)4 (3y) + (2x)3(3y)2 + (2x)2(3y)3 + (2x)(3y)4 +

(3y)5

= 32x5 + 240x4 y + 720 x3y2 + 1080 x2y3 + 810 xy4 + 243y5

iv) (2 3x)6 = 26 + 6 25( 3x) + 24( 3x)2 + 23( 3x)3 + 22( 3x)4 + 2 ( 3x) 5 + ( 3x)6

= 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

v) (3s 2t)5 = (3s)5 + 5 (3s)4 ( 2t) + (3s)3( 2t)2 + (3s)2( 2t)3 + (3s)(

2t)4 + ( 2t)5

= 243s5 810s4 t + 1080 s3t2 720 s2t3 + 240 st4 32t5

2.3.13C.Evaluate, without using a calculator:-

i) 3 + 3 ii) 4 + 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 + )3 + (1 )3 = 1 + 3 + 3()2 + ()3

+ 1 3 + 3()2 ()3

= 2 + 6()2 = 14

ii) (2 + )4 + (2 )4 = 24 + 4 23 + 6 22 ()2 + 4 2 ()3 + ()4 + 24 4 23 + 6 22 ()2 4 2 ()3 + ()4

= 2(24 + 6 22 ()2 + ()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

SolutionIn 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 + nxi) (1.01)10 = (1 + 0.01)10 = 1 + 10(0.01) + (0.01)2

+ (0.01)3 + (0.01)4 + = 1 + 0.1 + 0.0045 + 0.000120 + 0.0000021 +

= 1.105 to three decimal placesThe 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 = 1 2 8 (0.001) + (0.001)2 22 +

= 1 0.016 + 0.000112 + = 0.984 to 3 dp