Download The task is available in PDF-format here

Transformations of Expressions∗
1
Introduction
Quite often for the solution of various problems in algebra it turns out to be helpful to
factorise polynomials into products of simpler ones, to simplify rational expressions or to
convert expressions containing radicals. All of these are examples of transformations of
expressions into equivalent expressions. This short course is aimed to help you to learn
how to make such transformations correctly.
For successful manipulation with transformations it is important:
1) To learn a series of practical examples which can help one to perform transformations. The more examples you have in your arsenal and the more actively you use them,
the more difficult problems you will be able to solve.
2) To have a clear understanding of what you do when you perform transformations,
what objects you are working with and which seemingly equivalent expressions are not
equivalent. Without an understanding of these matters you risk making mistakes, e.g. to
lose or to gain extra roots of equations, to change domains of the definition of functions,
etc.
Let us now recall some definitions that you need to remember and understand properly
when you are solving any problem.
1.1
Expressions.
An expression is a sequence of numbers and variables connected by symbols for algebraic
operations subject to some rules. The rigorous definition is boring and tiresome; for us it
is only important to understand what is behind it. We will just give you some examples.
a) The simplest expressions are numbers and variables:
Y;
x;
a;
b;
1;
2;
3.14;
0.999...
are examples of elementary expressions. Variables are letters (usually) of the Latin alphabet. By numbers we understand sequences of digits written in the decimal system.
∗
Please, send comments, corrections and queries at [email protected].
1
2
http : //extramaths4fun.com/
Strictly speaking, the decimal expression and the number itself are different creatures.
The number is an abstract notion. Every number admits a decimal expression though we
cannot always write it all down (but we allow periodic decimal expressions in writing).
Different expressions may denote the same real number, e.g. both 0.999... and 1 denote
the same number called one.
b) Starting from the simplest expressions with help of algebraic symbols we can construct more complicated expressions. Symbols of algebraic operations are “+” — addition,
“−” — subtraction, “·” or “×” — multiplication, “:” — division (usually division is de2 +1
, but here we often write a : b for ab ), another symbol of
noted by “−”, for example, xy−1
√
algebraic operation is radical “ ”.
As an example we give some more complex constructions of expressions:
a + b,
a + Y,
2
,
x
5 · z,
√
2,
√
0.999.
Finally the following are also expressions:
a
(2x )
,
x
a ,
2
y+1
,
³√
0.2
´2y2 −0.999...
,
each is formed from two simpler expressions, one of which is the base and the other one
the exponent of the power.
1.2
Numerical expressions.
√
Numerical expressions are formed without variables. For example, 0.999...; 5 + 3 2 ; 2; 00
are numerical expressions, while x + 1, a − a are not.
√
a) Numerical expressions which do not make sense. 00 , −1, (−1)3/4 are numerical
expressions which do not make sense. A numerical expression does not make sense if and
only if the algebraic operations, whose symbols are used in the expression, are not defined
everywhere, e.g. 00 does not make sense, as the division operation is defined if and only if
√
the divisor is non zero, −1 does not make sense because the square root is defined only
for non-negative numbers etc.
b) It is important to understand that a numerical expression which makes sense, necessarily denotes a number. For instance, the decimal expression of a number is (the simplest)
numerical expression which denotes the given number. Different numerical expressions
may denote the same number. You can see it on the level of decimal representations: for
example, 0.999... and 1 are different expressions for the same number denoted for the
sake of convenience by 1, equally 0.333... and 1/3 denote the same number.
Test Question 1 Describe all the cases when distinct decimal expressions denote the
same number.
3
http : //extramaths4fun.com/
If we consider not just the simplest numerical expressions, like decimals,
√ then the set
of expressions denoting the same number becomes larger. For example: 4, 2, 1.999...,
3 − 1 are distinct expressions for the same number 2.
c) An equality or identity is of the form A = B, where A and B are numerical expressions. If any of the expressions A and B does not make sense, then we say that the
equality itself does not make sense. If an equality makes sense and in addition both of
A and B denote the same number, then the equality is called true and the expressions A
and B are equal. If an equality makes sense but A and B denote distinct numbers then
the equality is false. To prove equality A = B means that we are required to show that
A = B is a true equality.
d) Simplification. By a simplification of a numerical expression we understand a change
of it into another expression, which denotes the same number, but is itself a simpler
expression. In every concrete example it is assumed to be clear what means “simpler”. If
the original expression does not make sense, then we include under simplification of the
expression a proof that the expression does not make sense.
e) Calculation. By a calculation of the value of a numerical expression we understand
a process of finding a decimal expression for the number which is denoted by the original
expression. In that case the calculation is the final achievable simplification.
Example 1.1 Calculate: 0.999... + 0.999... − 2.
Solution. 0.999... + 0.999... − 2 = 1 + 1 − 2 = 0.
Example 1.2 Calculate:
√
√
2+1
√
− 3 − 2.
2−1
Solution.
√
√
2+1
√
−3− 2=
2−1
√
√
√
√
√
√
( 2 + 1)( 2 + 1)
2+2 2+1
√
= √
−3− 2=
− 3 − 2 = 2.
2−1
( 2 − 1)( 2 + 1)
√
We cannot continue the calculations as 2 is an irrational number and does not admit
a representation
in terms of a periodic decimal fraction. Therefore, the answer to the
√
question is 2 because it is the simplest expression with the same value as the original
expression.
Test Question 2 What is necessary and sufficient condition that a numerical expression
cannot be written as an equal decimal expression?
http : //extramaths4fun.com/
1.3
4
Expressions and functions.
Just as every numerical expression denotes a number, every expression (not necessarily
numerical) denotes a function.
Test Question 3 Give the definition of a function.
The expression x+1 denotes a function which associates to every number x the number
exceeding the original one by 1. Among numerical expressions, we have seen some that do
not make sense and, therefore, they do not denote any number. In the case of expressions
with variables we obtain the notion of the set of admissible values of the variables for
which the expression makes sense. This set of admissible values is called the domain of
definition of the expression.
a) The set of values of variables for which the expression makes sense. Consider for
the sake of simplicity an expression which contains only one variable x. Then the set of
values of x for which a given expression makes sense is the set of numbers such that after
substituting them for x in the given expression we obtain a numerical expression that
makes sense.
Example 1.3 xx makes sense for x 6= 0, i.e. the set of values of the variable x for which
the expression makes sense is the set of all non-zero numbers.
Example 1.4
√
x is an expression that makes sense for x ≥ 0.
b) Functions
associated to expressions. Let V (x) be an expression with variable x, for
√
example, x. We associate to it a function with the domain of definition consisting of
those x for which V (x) makes sense. In our example it is {x : x ≥ 0}. The function
associates to every value r from the domain of definition a number given by a numerical
expression,
obtained from V (x) after substituting the number r for the variable x. There√
√
fore, x denotes a function with domain {x : x ≥ 0} and the correspondence r → r for
every r ≥ 0.
If for all values of a variable the corresponding numerical expression does not make
sense, then we say, that the original expression does not make sense (and, thus, it does
not define any function).
Example 1.5
x−x
x−x
is an expression which does not make sense.
c) Equivalent transformations. As in the case of numerical expressions, expressions of
the same variables may be equal. This happens when they denote the same function.
For this in particular, it is necessary that they have the same domain of definition of the
variable for which they make sense. Such expressions are called equal, and a change of
one to another is called an equivalent transformation.
5
http : //extramaths4fun.com/
Example 1.6 x2 and
√
x4 are equal expressions.
Test Question 4 Are the expressions
x2
x
and x equal or not?
Test Question 5 Explain what it means to say that two expressions are equal on some
set of values of a variable.
2
Hint: the expressions xx and x are equal, for example, when x > 0; the expressions x and
√
x2 are equal, for example, when x ≥ 2.17.
In the same vein as with numerical expressions one defines simplifications. By a
problem of simplification we mean the problem of finding a change of an expression for
an identically equal one with a simpler formula. Finally a question to prove an identity
A ≡ B, where A and B are expressions, means that we have to prove the identity of
expressions A and B.
d) Conditional identities and equalities.
Example 1.7 Calculate:
µ
¶ µ
¶
1
1
1
1
√
+√
: √
−√
x−1
x−1
x+1
x+1
if x =
a2 +b2
,
2ab
a > 0, b > 0, a 6= b.
In this example we are asked to do the following. Let us substitute into the expression
¶ µ
¶
µ
1
1
1
1
√
: √
+√
−√
x−1
x−1
x+1
x+1
instead of x its expression in terms of a and b. We obtain an expression V (a, b). We are
then expected to simplify V (a, b) to a simpler expression α where V (a, b) = α.
This problem is an exercise on transformation of expressions under some additional
condition.
Example 1.8 Eliminate irrationality from the denominator of the fraction
α2 − 3α − 1
,
α2 + 2α + 1
where α3 + α2 + 3α + 4 = 0.
In this example we are dealing with a conditional equivalent transformation, as we are
required to change
α2 − 3α − 1
,
α2 + 2α + 1
to an equal one which does not contain irrationals in the denominator. The transformation
should be done under the condition that α3 + α2 + 3α + 4 = 0.
6
http : //extramaths4fun.com/
Example 1.9 Calculate
2.
√
√
√
√
25 − x2 + 15 − x2 if it is known that 25 − x2 − 15 − x2 =
√
√
2 +
It is implied here that we have to prove the identity
25
−
x
15 − x2 = α, where α
√
√
is the unknown number such that the identity 25 − x2 − 15 − x2 = 2 holds for all x.
We suggest that you only try to understand the objects you will be working with
rather then learn them by heart. Later on you will come across various problems with
combinations of questions on simplification and calculation. Nevertheless, if you were able
to understand the formulations of problems given in this section, it should not be difficult
for you to understand what is required in every particular situation.
And finally, when you are solving inequalities, equations or systems of equations it is
inadmissible to lose or to gain roots. In this case the problem will be counted as solved
incorrectly. To save yourself from this you have to understand the content of this section
because every intermediate step of any problem in algebra involves understanding transformations between equivalent expressions or, more generally, transformations between
expressions equivalent under some given condition.
2
2.1
Division of polynomials. The Remainder Theorem.
Definition
Let P (x) and Q(x) be two polynomials of degree n and m respectively:
P (x) = a0 xn + a1 xn−1 + · · · + an ,
(a0 6= 0),
Q(x) = b0 xm + b1 xm−1 + · · · + bm ,
(b0 6= 0).
To divide a polynomial P (x) by Q(x) with a remainder means to rewrite P (x) in the
form P (x) = Q(x)T (x) + R(x), where the quotient T (x) is a polynomial, the remainder
R(x) is either 0 or a polynomial of degree less than the degree of Q(x). Division of a
polynomial P (x) by Q(x) is always possible unless Q(x) is identically zero. The process
of division consists in finding T (x) and R(x) such that the following identity holds P (x) =
Q(x)T (x) + R(x). We say that a polynomial P (x) is divisible by the polynomial Q(x) if
R(x) ≡ 0.
√
√
Test Question 6 Let P (x) = 2; Q(x) = 3. Divide P (x) by Q(x).
2.2
Long division.
This algorithm is analogous to the long division of numbers.
7
http : //extramaths4fun.com/
Example 2.1 Divide a polynomial P (x) by a polynomial Q(x) with a remainder, where
P (x) = 2x4 + x3 + 7x − 3; Q(x) = x2 − 2x + 4.
Step 1. We start by comparing the degrees of the polynomials P (x) and Q(x).
Step 2. Find a monomial (it is 2x2 ) such that after multiplying it by Q(x) we get a
polynomial D1 (x) with the highest term equal to the highest term of P (x):
¡ 2¢ ¡ 2
¢
2x · x − 2x + 4 = 2x4 − 4x3 + 8x2 = D1 (x).
Step 3. Subtract from P (x) the polynomial D1 (x). We obtain a polynomial P1 (x) =
5x3 − 8x2 + 7x − 3.
These calculations can be written in the following way:
2x4 + x3
+
7x− 3
x2 − 2x + 4
2x2
2x4 − 4x3 + 8x2
5x3 − 8x2 + 7x− 3
Step 4. Compare the degrees of Q(x) and P1 (x). Because deg P1 (x) ≥ deg Q(x) (by
deg P (x) we denote the degree of P (x)), we can continue the process of division which
will consist of repeating the previous steps but replacing P (x) by P1 (x).
Step 5. Find a monomial (it is 5x) such that after multiplying it by Q(x) we obtain a
polynomial D2 (x) with the highest term equal to the highest term of P1 (x):
(5x) · (x2 − 2x + 4) = 5x3 − 10x2 + 20x = D2 (x).
Add this monomial 5x to the previous one 2x2
2x4 + x3
+
7x− 3
2x4 − 4x3 + 8x2
x2 − 2x + 4
2x2 + 5x
5x3 − 8x2 + 7x
5x3 − 10x2 + 20x
2x2 − 13x−3
Step 6. Subtract D2 (x) from P1 (x), we obtain a polynomial
2x2 − 13x − 3 = P2 (x).
Step 7. Compare the degrees of Q(x) and P2 (x). Because deg P2 (x) ≥ Q(x) (2 ≥ 2), the
process can be continued.
8
http : //extramaths4fun.com/
Step 8. Find a monomial (it is 2) such that after multiplying it by Q(x) we get a polynomial D3 (x) with the highest term equal to the highest term of P2 (x):
¡
¢
2 · x2 − 2x + 4 = 2x2 − 4x + 8.
Add this monomial to the previously constructed 2x2 + 5x:
2x4 + x3
+
7x− 3
2x4 − 4x3 + 8x2
x2 − 2x + 4
2x2 + 5x + 2
5x3 − 8x2 + 7x
5x3 − 10x2 + 20x
2x2 − 13x−3
2x2 − 4x+ 8
−9x− 11
Step 9. Subtract from P2 (x) the polynomial D3 (x). We obtain a polynomial P3 (x) =
−9x − 11.
2x4 + x3
+
7x− 3
2x4 − 4x3 + 8x2
x2 − 2x + 4
2x2 + 5x + 2 = T (x)
5x3 − 8x2 + 7x
5x3 − 10x2 + 20x
2x2 − 13x−3
2x2 − 4x+ 8
−9x− 11 = R(x)
Step 10. Compare degrees: deg P3 (x) < deg Q(x), (1 < 2). The division process terminates here, P3 (x) = R(x) is the remainder.
Example 2.2 Divide a polynomial P (x) = x5 + 5x3 − 2x2 + 6x − 4 by a polynomial
Q(x) = x3 + 3x − 2 with a remainder.
We give a non-detailed description of all steps using the following picture:
x5 + 5x3 − 2x2 + 6x− 4
x5 + 3x3 − 2x2
2x3 +
6x− 4
2x3 +
6x− 4
x3 + 3x − 2
x2 + 2 = T (x)
0 = R(x)
9
http : //extramaths4fun.com/
In this example P (x) is divisible by Q(x).
Example 2.3 Divide with a remainder a polynomial P (x) = 3x4 − 10ax3 + 22a2 x2 −
24a3 x + 10a4 by a polynomial Q(x) = x2 − 2ax + 3a2 .
Again we give a non-detailed description of all steps:
3x4 − 10ax3 + 22a2 x2 − 24a3 x+ 10a4 x2 − 2ax + 3a2
3x4 − 6ax3 + 9a2 x2
−4ax3 + 13a2 x2 − 24a3 x
3x2 − 4ax + 5a2 = T (x)
−4ax3 + 8a2 x2 − 12a3 x
5a2 x2 − 12a3 x+ 10a4
5a2 x2 − 10a3 x+ 15a4
−2a3 x− 5a4 = R(x)
Test Question 7 In Example 2.3 we considered a as a parameter and the polynomial as
a polynomial in x. But we could have considered all the polynomials as polynomials in
a and set x to be a parameter. If we divide now one polynomial by another we obtain a
remainder R1 (a). Without division show that R1 (a) 6= 0.
2.3
The Remainder Theorem
Theorem 2.1 A polynomial P (x) is divisible by x − a if and only if a is a root of P (x)
(i.e. when P (a)=0).
Proof. See Problem 24.
Let us illustrate some applications of this theorem.
Example 2.4 Find all values a for which the polynomial x4 + a2 x3 − 5a + 5 is divisible
by x − 1.
Solution. From the Remainder Theorem the values a are those for which 1 is a root of
x4 + a2 x3 − 5a + 5. Substituting x = 1 into this polynomial, we obtain 1 + a2 − 5a + 5 = 0.
From here a = 2 or a = 3.
The Remainder Theorem can be successfully applied to the problem of factorisation
of polynomials:
Example 2.5 Factorise the expression
(a + b + c)3 − (a + b)3 − (a + c)3 − (b + c)3 + a3 + b3 + c3 .
10
http : //extramaths4fun.com/
Solution. Let us consider this expression as a polynomial in a. Then 0 is a root of this
polynomial. Indeed,
(b + c)3 − b3 − c3 − (b + c)3 + b3 + c3 = 0,
i.e. this expression is divisible by a. In the same vein we prove that the original expression
is divisible by b and c. Because our expression is a polynomial of degree 3 with respect
to a, b, c, it can differ from abc only by a scalar multiple (why?), i.e.
(a + b + c)3 − (a + b)3 − (a + c)3 − (b + c)3 + a3 + b3 + c3 = λabc,
To find λ it is enough to consider some explicit values for a, b, c, for example, a = b = c = 1.
Then
33 − 23 − 23 − 23 + 1 + 1 + 1 = 6 = λ · 1 · 1 · 1.
From that we deduce λ = 6.
Answer:
(a + b + c)3 − (a + b)3 − (a + c)3 − (b + c)3 + a3 + b3 + c3 = 6abc,
Remark 2.1 Let P (x) 6= 0 be a polynomial. To factorise this polynomial means to
represent it as P (x) = P1 (x)·P2 (x) · · · Ps (x), where Pi (x), deg Pi (x) < deg P (x), i = 1, ...s,
are polynomials which do not admit further factorisation. A polynomial P (x) does not
admit a factorisation if it can not be represented in the form:
P (x) = P1 (x) · P2 (x) · · · · · Ps (x),
deg Pi (x) < deg P (x),
i = 1, ...s.
Test Question 8 Formulate what it means to factorise a polynomial of three variables
x, y, z.
2.4
Identically zero polynomials.
Example 2.6 A polynomial of degree less than or equal to 1 in a single variable which
takes the value zero at two distinct points is identically zero.
Proof. Let us write our polynomial in the form ax + b, and set x1 and x2 to be the points
where the polynomial takes zero values: x1 6= x2 , i.e.
½
ax1 + b = 0
ax2 + b = 0.
Then a(x1 − x2 ) = 0. Because x1 6= x2 we deduce that a = 0 and, therefore, b = 0.
A more general theorem (Problem 25 from the problem sheet) is valid. A polynomial
of degree less than or equal to n in one variable which takes the value zero at n+1 distinct
points is identically zero.
Let us consider some applications of this theorem.
11
http : //extramaths4fun.com/
Example 2.7 Simplify the expression:
a − b b − c c − a (a − b)(b − c)(c − a)
+
+
+
.
a + b b + c c + a (a + b)(b + c)(c + a)
Solution. This expression makes sense whenever a 6= −b, b 6= −c and c 6= −a. Therefore,
we assume that a 6= −b, b 6= −c, c 6= −a. Let us multiply the expression by (a + b)(b +
c)(c + a). We get in the numerator
(a − b)(b + c)(c + a) + (b − c)(a + b)(c + a) + (c − a)(a + b)(b + c) + (a − b)(b − c)(c − a).
Consider this expression as a polynomial of degree 2 with respect to the variable a. It
takes the value 0 for a = 0, a = b, a = c. If the numbers 0, b, c are pairwise distinct,
then due to the theorem mentioned above, applied to the polynomial of degree 2, the
polynomial is identically zero. If b = 0 (similarly for c = 0) or b = c we show that
the polynomial is identically zero by explicit verification. We conclude that the original
expression is identically zero when a 6= −b, b 6= −c, c 6= −a. If a = −b, or a = −c, or
b = −c, then this expression is not defined.
3
Factorisation of polynomials based on properties of
roots of quadratic equations.
Let α, β be the roots of the quadratic equation Ax2 + Bx + C = 0. The trinomial
Ax2 + Bx + C can be factorised as:
Ax2 + Bx + C = A(x − α)(x − β).
In the very same way one can factorise another trinomial Ax4 + Bx2 + C. If α, β are the
roots of the quadratic equation Az 2 + Bz + C = 0, then
Ax4 + Bx2 + C = A(x2 − α)(x2 − β).
The trinomial Ax4 +Bx2 +C can be factorised into a product of two quadratic polynomials
in another way. We set
Ax4 + Bx2 + C = A(x2 + px + q)(x2 − px + q)
Performing the multiplication on the right hand side and comparing the coefficients of
the same powers of x we obtain
B
= 2q − p2 ,
A
C
= q2.
A
From which we get
p=
s
B
− ±2
A
r
C
,
A
q=±
r
C
.
A
12
http : //extramaths4fun.com/
Example 3.1 Factorise x4 + 4.
In this situation A = 1, B = 0, C = 4 and by the formula from the previous section we
obtain:
s
r
C
B
= 2,
− +2
A
A
q
therefore, p = 2, q = CA . Thus,
x4 + 4 = (x2 + 2x + 2)(x2 − 2x + 2).
Example 3.2 Factorise x4 + x2 + 1.
In this example A = 1, B = 1, C = 1.
s
r
C
B
− +2
= 1,
A
A
r
C
= 1.
A
Therefore, p = 1, q = 1. The factorisation we are looking for has the form
x4 + x2 + 1 = (x2 + x + 1)(x2 − x + 1).
Let us consider two cases of factorisations of polynomials of two variables:
a) Consider a polynomial:
Ax2 + Bxy + Cy 2 + Dx + Ey + F.
It can be rewritten in the form
Ax2 + (By + D)x + (Cy 2 + Ey + F ).
This polynomial can be factorised as A(x−α)(x−β), where α, β are roots of the equation
Ax2 + (By + D)x + (Cy 2 + Ey + F ) = 0.
Example 3.3 Factorise the polynomial
2x2 + xy − 6y 2 − 7x + 7y + 3.
13
http : //extramaths4fun.com/
The roots of the equation
2x2 + (y − 7)x − 6y 2 + 7y + 3 = 0
can be expressed by the formula
p
−y + 7 ± (y − 7)2 + 48y 2 − 56y − 24
−y + 7 ± (7y − 5)
=
.
x1,2 =
4
4
From this we get
1
x2 = −2y + 3.
x1 = (3y + 1),
2
The factorisation required has the following form
2x2 + xy − 6y 2 − 7x + 7y + 3 = 2(x − x1 )(x − x2 ) = (2x − 3y − 1)(x + 2y − 3).
b) Consider a polynomial
Ax2 y 2 + Bx2 + Cxy + Dy 2 + E;
it can be factorised as (Ay 2 + B)(x − α)(x − β), where α, β are the roots of the quadratic
polynomial
(Ay 2 + B)x2 + Cyx + (Dy 2 + E) = 0.
Example 3.4 Factorise
x2 y 2 − x2 + 4xy − y 2 + 1.
Solution. This polynomial can be factorised as
(y 2 − 1)(x − α)(x − β),
where α, β are the roots of
(y 2 − 1)x2 + 4xy − (y 2 − 1) = 0.
Solving this equation we find
x1,2 =
−2y ±
p
4y 2 + (y 2 − 1)2
−2y ± (y 2 + 1)
=
.
y2 − 1
y2 − 1
From here
y−1
y+1
,
β=−
.
y+1
y−1
The factorisation we are looking for has the form
α=
x2 y 2 − x + 4xy − y 2 + 1 = (xy + x − y + 1)(xy − x + y + 1).
14
http : //extramaths4fun.com/
Example 3.5 Factorise
a2 (b + c) + b2 (c + a) + c2 (a + b) + 3abc.
Solution. Taking a as the leading variable we can rewrite the polynomial in the form
(b + c)a2 + (b2 + c2 + 3bc)a + bc(b + c),
which factors as (b + c)(a − α)(a − β), where α, β are the roots of the equation
(b + c)x2 + (b2 + c2 + 3bc)x + bc(b + c) = 0.
Solving this equation we obtain
x1,2 =
−(b2 + c2 + 3bc) ±
=
p
(b2 + c2 + 3bc)2 − 4bc(b + c)2
=
2(b + c)
−(b2 + c2 + 3bc) ± (b2 + bc + c2 )
.
2(b + c)
From here
α=−
bc
,
b+c
β = −(b + c).
The factorisation has the form
a2 (b + c) + b2 (c + a) + c2 (a + b) + 3abc =
= (ab + bc + ca)(a + b + c).
Example 3.6 Factorise
a4 + b4 + c4 − 2a2 b2 − 2a2 c2 − 2b2 c2 .
Solution. Taking a as the leading symbol we factorise this polynomial as
(a2 − α)(a2 − β),
where α, β are the roots of the polynomial
x2 − 2(b2 + c2 )x + b4 + c4 − 2b2 c2 = 0.
Solving this equation we find the roots
α = b2 + c2 + 2bc,
β = b2 + c2 − 2bc.
The factorisation has the form
a4 + b4 + c4 − 2a2 b2 − 2a2 c2 − 2b2 c2 =
= (a2 − b2 − c2 − 2bc)(a2 − b2 − c2 + 2bc).
15
http : //extramaths4fun.com/
Example 3.7 Factorise
(a2 + d2 )bc + (b2 + c2 )ad.
Solution. Taking a as the leading symbol we find that the polynomial is factorised as
bc(a − α)(a − β), where α, β are the roots of
bcx2 + (b2 + c2 )dx + bcd2 = 0.
Solving this equation we find the roots
bd
cd
,
β=− .
b
c
The expression we are looking for has the form
α=−
(a2 + d2 )bc + (b2 + c2 )ad = (ab + cd)(ac + bd).
4
Transformations of expressions with radicals
Sometimes for the solution of special problems it is helpful to use the following identities
s
s
√
√
q
2
√
A+ A −B
A − A2 − B
+
(1)
A+ B =
2
2
s
s
√
√
q
2
√
A+ A −B
A − A2 − B
−
(2)
A− B =
2
2
These identities are true for those A and B for which both sides are defined. Let us
show applications of these formulas.
Example 4.1 Simplify the expression
√
√
2− 3
2+ 3
p
p
√ +√
√
√ = A.
2+ 2+ 3
2− 2− 3
q
q
q
q
p
p
√
√
2+1
2−1
2+1
+
, due to (2): 2 − 3 =
− 2−1
,
Solution. Due to (1): 2 + 3 =
2
2
2
2
√
√
2+ 3
2− 3
p
p
p
p
A= √
+√
,
2 + 3/2 + 1/2
2 − 3/2 + 1/2
√
√
A
2+ 3 2− 3
1
1
√ =
√ +
√ =2−
√ −
√ =
2
3+ 3 3− 3
3+ 3 3− 3
√
√
√
3+ 3+3− 3
√
=2−
= 1; A = 2.
32 − ( 3)2
In some situations it is convenient in the process of transformations of expressions
with radicals to introduce letters to stand for more complex expressions.
16
http : //extramaths4fun.com/
Example 4.2 Prove the identity:
q
3
Solution. We denote
√
3
√
3
2−1=
r
3
1
−
9
r
3
2
+
9
r
3
4
.
9
2 by α. Then the identity is equivalent to
9(α − 1) = (1 − α + α2 )3 ,
where α3 = 2.
Let us prove this relation. Because α3 = 2, α4 = 2α we have
(1 − α + α2 )2 = α4 − 2α3 + 3α2 − 2α + 1 = 3α2 − 3.
Therefore,
(1 − α + α2 )3 = 3(α2 − 1)(1 − α + α2 ) =
= 3(α4 − α3 + α − 1) = 9α − 9,
q.e.d.
5
Eliminating irrationality in the denominators of
fractions
An irrational number — is a number which cannot be represented in the form p/q,
√ where
p is an integer, q is a natural number. Irrational numbers exist.
For
example,
2. Let
√
us prove that this is an irrational number. Assume that 2 is rational. Then it is
representable
as a fraction p/q, where p is an integer and q is a natural number. We
√
have 2 = p/q (we assume that p and q have no common factors, as in the opposite
case we could√cancel in the fraction the largest common divisor of p and q). Consider the
identity p = 2q. Let us square both sides of this identity. We obtain p2 = 2q 2 , from
which it follows that p is even, i.e. p = 2p1 . Thus, 4p21 = 2q 2 , i.e 2p21 = q 2 and q is √
even.
Therefore, p and q have common factor 2 which contradicts our assumption. Thus, 2 is
an irrational number.
Let α be an irrational number which is a root of a polynomial P (x) with integer
1 (α)
coefficients. To eliminate irrationality in the denominator of a fraction Q
, where Q1 (x)
Q2 (α)
and Q2 (x) are polynomials with integer coefficients, means to find a polynomial G(x) with
Q1 (α)
= G(α).
rational coefficients such that Q
2 (α)
Example 5.1 Eliminate irrationality in the denominator of a fraction
root of the equation α2 + 2α − 2 = 0.
1
α+2
where α is a
17
http : //extramaths4fun.com/
Solution. From the equation α2 + 2α − 2 = 0 it follows that α + 2 = α2 . Substituting the
expression for α + 2 into the original fraction we obtain
1
α
1
=
= .
α+2
2/α
2
Answer: α/2.
Let us discuss a general method of eliminating irrationality in the denominator. We
shall demonstrate it in an example containing radicals of degree three, but it is applicable
for eliminating radicals of any degree.
√
√
√
√
3
3
Consider an expression A + B 3 x + C x2 . Multiplying it by P + Q 3 x + R x2 , where
P, Q, R are, for the moment, unknown parameters, we find
³
´³
´
√
√
√
√
3
3
A + B 3 x + C x2 P + Q 3 x + R x2 =
(3)
√
√
3
= AP + CQx + BRx + 3 x (P B + AQ + CRx) + x2 (P C + AR + BQ)
If we require that
½
P B + AQ + CRx = 0,
P C + AR + BQ = 0
(4)
then the expression (3) becomes AP + CQx + BRx. We define P, Q, R in such a way that
they satisfy (4). From (4) it follows that
A2
P
R
Q
= 2
= 2
.
− BCx
B − AC
C x − AB
(5)
This means that P, Q, R are proportional to the denominators of the corresponding fractions in (5). Taking the arbitrary coefficient of proportionality to be equal to 1 we deduce
P = A2 − BCx,
R = B 2 − AC,
Q = C 2 x − AB.
Substituting the values obtained for P, Q, R into the expression AP + CQx + BRx we get
A3 + B 3 x + C 3 x2 − 3ABCx
(6)
√
√
3
+ C x2 , then
It is clear that if we have a fraction with a denominator A √
+ B 3 x√
3
multiplying both the numerator and the denominator by P + Q 3 x + C x2 , we would
transform the denominator into a rational expression of the form (6), i.e. we would
eliminate irrationality in the denominator.
Example 5.2 Eliminate irrationality in the denominator of a fraction
root of the equation α3 − 3α + 1 = 0.
α
,
α+1
where α is a
18
http : //extramaths4fun.com/
Solution. We shall try to construct the solution in the form of a polynomial in α with
unknown coefficients. Because powers of α can be expressed as polynomials of the second
degree in α:
α3 = 3α − 1,
α4 = α(α3 ) = 3α2 − α,
α5 = α(α4 ) = α(3α2 − α) = 3α3 − α2 = −α2 + 9α − 3,
etc.,
it is sufficient to consider polynomials of second degree only a0 +a1 α+a2 α2 with unknown
coefficients a0 , a1 , a2 .
α
From the identity α+1
= a0 + a1 α + a2 α2 it follows that
α = a0 + α(a0 + a1 ) + α2 (a1 + a2 ) + α3 a2 .
But α3 = 3α − 1, therefore
α = (a0 − a2 ) + α(a0 + a1 + 3a2 ) + α2 (a1 + a2 ).
On the left and the right hand sides of this equality we have polynomials in α. Two
polynomials are equal if and only if the coefficients of the same terms coincide. Therefore,
from the last equality we obtain a system of equations

 0 = a 0 − a2
1 = a0 + a1 + 3a2

0 = a 1 + a2
Solving this system we deduce that a0 = 1/3, a1 = −1/3, a2 = 1/3.
Thus,
1 1
1
α
= − α + α2 .
α+1
3 3
3
The same method is also applicable to the case of several irrationalities.
1
, where α,
Example 5.3 Eliminate irrationalities in the denominator of the fraction α+β
2
2
β are roots of the equations α − α − 1 = 0 and β + 2β − 6 = 0 respectively.
Solution. We shall try to find a solution as a polynomial in α and β. As in the previous
example it is possible to show that we need work only with a polynomial of the form
a00 + a01 α + a10 β + a11 αβ with unknown coefficients a00 , a01 , a10 , a11 . From the equality
1
= a00 + a01 α + a10 β + a11 αβ it follows that 1 = (a00 + a01 α + a10 β + a11 αβ)(α + β).
α+β
Expanding parentheses and taking into account that
a11 α2 β = a11 (α + 1)β,
a10 β 2 = a10 (6 − 2β),
a01 α2 = a01 (α + 1),
a11 αβ 2 = a11 α(6 − 2β),
19
http : //extramaths4fun.com/
we deduce that
1 = (a01 + 6a10 ) + α(a00 + a01 + 6a11 ) + β(a00 − 2a10 + a11 ) + αβ(a01 + a10 − a11 ).
To find the unknown coefficients we investigate the system

a01 + 6a10 = 1



a00 + a01 + 6a11 = 0
a

00 − 2a10 + a11 = 0


a01 + a10 − a11 = 0
7
; a01 = − 29
; a10 =
Solving this system we obtain a00 = 13
29
7
6
1
1
= 13
−
α
+
β
−
αβ.
Answer: α+β
29
29
29
29
6
;
29
1
a11 = − 29
.
Let us discuss now how to apply the method of elimination of radicals to the solution
of equations.
Example 5.4 Solve the equation:
p
√
√
1 − x + 2 3 1 − x − 3 3 (1 − x)2 = 0.
√
√
Solution. Let 1 − x = u, 3 1 − x = z. Then u + 2z − 3z 2 = 0. Eliminate the radical z
following (3)-(6). We obtain
u3 + 8z 3 − 27z 6 + 18uz = 0.
Substituting instead of u and z their values we get
√
√
(1 − x) 1 − x + 8(1 − x) + 18(1 − x) 1 − x − 27(1 − x)2 = 0,
or, in another form,
£ √
¤
(1 − x) 19 1 − x + 8 − 27(1 − x) = 0.
√
From here it follows that either x = 1 or 19 1 − x = 19 − 27x. From the last equation
665
.
we find x = 0, x = 729
Testing shows that x = 0, x = 1 satisfy our equation and x = 665
is an irrelevant root
729
(where did it come from?).
Answer: x = 0, x = 1.
20
http : //extramaths4fun.com/
6
Problem Sheet.
Problem 1 Answer the test questions inside the text.
Problem 2 Simplify the expressions:
a)
x2 + x1
;
x + x1 − 1
b)
x4 + 1
√
.
x2 − 2x + 1
√
√
√
√
Problem 3 Calculate 25 − x2 + 15 − x2 if 25 − x2 − 15 − x2 = 2.
Problem 4 Calculate
¢
0.666... + 31 : 0.25
+ 12.5 · 0.64.
0.12333... : 0.0925
¡
Problem 5 Divide the polynomial P (x) by the polynomial Q(x) with a remainder:
a) P (x) = 5x7 + 8x5 + 9x4 + 11x3 + 6, Q(x) = x2 − 2x + 1.
b) P (x) = x6 + 7x5 + 10x4 + 13x3 + 33x2 + 14x, Q(x) = x2 + 2x.
Problem 6 Prove the identities:
a) an − bn = (a − b) (an−1 + an−2 b + · · · + abn−2 + bn−1 ) for all natural numbers n.
b) an + bn = (a + b) (an−1 − an−2 b + · · · − abn−2 + bn−1 ) for all odd n.
Please, prove this by long division of an − bn by a − b and of an + bn by a + b.
Problem 7 Prove that it is always possible to perform division of any polynomial P (x)
by Q(x) 6= 0 with a remainder.
Problem 8 Show that the polynomials
a) x2 + 1,
b) x2 + x
√+ 1
2
c) x + 2x + 1
are not factorisable.
Problem 9 Factorise x3 + y 3 + z 3 − 3xyz into factors of a smaller degree.
Problem 10 Prove the identity
(a + b + c + d)4 − (a + b + c − d)4 − (a + b − c + d)4 − (a − b + c + d)4 −
(−a + b + c + d)4 + (a + b − c − d)4 + (a − b + c − d)4 + (a − b − c + d)4 = 192abcd.
21
http : //extramaths4fun.com/
Problem 11 Prove that if a + b + c = 0 then
a3 + b3 + c3 = 3abc.
Problem 12 Prove that if a + b + c = 0, then
2(a5 + b5 + c5 ) = 5abc(a2 + b2 + c2 ).
Problem 13 Find all values a for which the polynomial
a3 x5 + (1 − a)x4 + (1 + a3 )x2 + (1 − 3a)x − a3
is divisible by x − 1.
Problem 14 For which values a and b is the polynomial (a + b)x5 + abx2 + 1 divisible by
x2 − 3x + 2?
Problem 15 Simplify the expression
(a + b + c)5 − a5 − b5 − c5
(a + b + c)3 − a3 − b3 − c3
Problem 16 Calculate
µ
if x =
a2 +b2
2ab
1
1
√
+√
x−1
x+1
¶ µ
¶
1
1
−√
: √
x−1
x+1
with a > 0, b > 0, a 6= b.
Problem 17 Find the largest set of values of a, b, c, for which the following identity holds:
a2 b 3 − a 3 b 2 + b 2 c 3 − b 3 c 2 + c 2 a3 − c 3 a2
=
ab3 − a3 b + bc3 − b3 c + ca3 − c3 a
ab + ac + bc
a+b+c
(Hint: when a = b both the numerator and the denominator of the left hand side take zero
value).
=
Problem 18 Eliminate irrationality in the denominator:
Ã
!3
1
1
p
p
.
√
√
√ −√
4+ 7+ 2
2− 4− 7
22
http : //extramaths4fun.com/
Problem
the identities
r 19 Prove
q
p
√
√
a) 17 + 6 4 − 9 + 4 2 − 2 = 3,
p
p
√
√
3
3
b) 20 + 14 2 + 20 − 14 2 = 4.
Problem 20 Eliminate irrationality in the denominator:
2
, where α3 + α2 + 3α + 4 = 0;
a) αα2 −3α−1
+2α+1
2
+α+1
, where α4 + 2α3 + 2α2 + α + 1 = 0;
b) αα2 +2α+2
2 +β 2
c) αα+β
, where α2 + α + 1 = 0, β 2 + β − 5 = 0;
1
, where α3 + α + 1 = β, α2 + β 2 = 2;
d) α+β
1
√
e) √
3 a+ 3 b+ √
3 c;
f)
√ 1√ .
2+ 3
Problem
√ 21 Prove that
a) √ 3,
b) 8
are irrational.
Problem 22 Prove that
p
p
√
√
3
3
7 + 50 − 7 − 5 2 is rational.
Problem 23 Prove the identities (1), (2) from Section 4.
Problem 24 Prove The Remainder Theorem.
Problem 25 Prove that polynomial P (x) = an xn + · · · + a1 x + a0 , an 6= 0, has no more
than n distinct roots.
Problem 26 Factorise
x3 + 3xy + y 3 − 1.
Problem 27 Factorise
a8 + a 6 b 2 + a 4 b 4 + a 2 b 6 + b 8 .
Problem 28 Prove the identity
q
2+
√
3·
r
2+
q
2+
√
3·
s
2+
r
2+
q
2+
√
3·
s
2−
r
2+
q
2+
√
Problem 29 Prove the identity
(x + 1)(x + 3)(x + 5)(x + 7) + 15 = (x + 6)(x + 2)(x2 + 8x + 10).
3 = 1.
http : //extramaths4fun.com/
23
Problem 30 Prove the identity
q
√
√
√
√
√
√
10 + 24 + 40 + 60 = 2 + 3 + 5.
Acknowledgements. Various materials from “Kvant”, All Union Mathematical
Olympiads, Malyi MehMat have been used in the preparation of this task. We express
a deep gratitude to all people involved in these kinds of activities and especially wish to
thank Andrei Nikolaevich Kolmogorov who initiated most of them in USSR many years
ago.
The materials are prepared and can be used for non commercial purposes only.