104 Chapter 6
678 Problem
Solve for x 2
È
x a
+ 3
È
a x
= b
a +
6a b
.
679 Problem Solve
x − 7x − 3x + 5x + 1 = 1680 .
680 Problem Solve
x
4
+ x
3
− 4x
2
+ x + 1 = 0 .
681 Problem Solve the equation
2
sin2 x
+ 5 · 2
cos2 x
= 7 .
682 Problem If the equation
q
x +
È
x +
p
x + √
··· = 2 made sense, what would be the value of x?
683 Problem How many real solutions are there to
sin x = log
e
x?
684 Problem Solve the equation
|x + 1| − |x| + 3|x − 1| − 2|x − 2| = x + 2.
685 Problem Find the real roots of
p
x + 3 − 4 √
x − 1 +
p
x + 8 − 6 √
x − 1 = 1 .
686 Problem
Solve the equation 6x
4
− 25x
3
+ 12x
2
+ 25x + 6 = 0 .
687 Problem Solve the equation
x2x + 1x − 22x − 3 = 63 .
688 Problem
Find the value of √
30 · 31 · 32 · 33 + 1.
689 Problem Solve
x + √
x
2
− 1 x −
√ x
2
− 1 +
x − √
x
2
− 1 x +
√ x
2
− 1 = 98
.
690 Problem
Find a real solution to x
2
− 9x − 1
10
+ 99x
10
= 10x
9
x
2
− 1 .
Hint: Write this equation as x
2
− 9x − 1
10
− 10x
9
x
2
− 9x − 1 + 9x
10
= 0 .
691 Problem
Find the real solutions to
É
x + 2
q
x + 2
È
x + ··· + 2
p
x + 2 √
3x
| {z
}
n radicals
= x .
692 Problem Solve the equation
1 1 +
1 1 +
1 1 +
. .
. 1 +
1 x
= x .
where the fraction is repeated n times.
693 Problem Solve for x
p
x + √
x + 11 +
p
x + √
x − 11 = 4 .
6.2 Systems of Equations
694 Example Solve the system of equations
x + y + u =
4 ,
y + u + v =
−5 ,
u + v + x =
, v + x + y
= −8
. Solution: Adding all the equations and dividing by 3,
x + y + u + v = −3 .
This implies 4 + v
= −3
, −5 + x
= −3
, 0 + y
= −3
, −8 + u
= −3
,
Practice 105
whence x = 2 , y = −3, u = 5, v = −7.
695 Example
Solve the system x + yx + z = 30
, y + zy + x = 15
, z + xz + y = 18
. Solution: Put u = y + z
, v = z + x, w = x + y. The system becomes vw = 30
, wu = 15, uv = 18. 6.7
Multiplying all of these equations we obtain u
2
v
2
w
2
= 8100, that is, uvw = ±90. Dividing each of the equations in 7, we
gather u = 3 , v = 6, w = 5, or u = −3, v = −6, w = −5. This yields
y + z =
3 , or y + z
= −3
, z + x
= 6
, or z + x =
−6 ,
x + y =
5 , or x + y =
−5 ,
whence x = 4 , y = 1, z = 2 or x = −4, y = −1, z = −2..
Practice
696 Problem Let a,b, c be real constants, abc
6= 0. Solve x
2
− y − z
2
= a
2
, y
2
− z − x
2
= b
2
, z
2
− x − y
2
= c
2
.
697 Problem Solve
x
3
+ 3x
2
y + y
3
= 8 ,
2x
3
− 2x
2
y + xy
2
= 1 .
698 Problem Solve the system
x + 2 + y + 3 +
p
x + 2y + 3 = 39 ,
x + 2
2
+ y + 3
2
+ x + 2y + 3 = 741 .
699 Problem Solve the system
x
4
+ y
4
= 82 ,
x − y = 2 .
700 Problem
Solve the system x
1
x
2
= 1 , x
2
x
3
= 2 , . . . , x
100
x
101
= 100 , x
101
x
1
= 101 .
701 Problem Solve the system
x
2
− yz = 3 ,
y
2
− zx = 4 ,
z
2
− xy = 5 .
702 Problem Solve the system
2x + y + z + u = −1 x + 2y + z + u = 12
x + y + 2z + u = 5 x + y + z + 2u = −1
703 Problem Solve the system
x
2
+ x + y = 8 ,
y
2
+ 2xy + z = 168 ,
z
2
+ 2yz + 2xz = 12480 .
6.3 Remainder and Factor Theorems
The Division Algorithm for polynomials states that if the polynomial px is divided by ax then there exist polynomials qx
, rx with px = axqx + rx
6.8 and 0
≤ degree rx degree ax. For example, if x
5
+ x
4
+ 1 is divided by x
2
+ 1 we obtain x
5
+ x
4
+ 1 = x
3
+ x
2
− x − 1x
2
+ 1 + x + 2 ,
and so the quotient is qx = x
3
+ x
2
− x − 1 and the remainder is rx = x + 2 .
106 Chapter 6
704 Example
Find the remainder when x + 3
5
+ x + 2
8
+ 5x + 9
1997
is divided by x + 2 .
Solution: As we are dividing by a polynomial of degree 1, the remainder is a polynomial of degree 0, that is, a constant. Therefore, there is a polynomial qx and a constant r with
x + 3
5
+ x + 2
8
+ 5x + 9
1997
= qxx + 2 + r Letting x = −2 we obtain
−2 + 3
5
+ −2 + 2
8
+ 5−2 + 9
1997
= q−2−2 + 2 + r = r .
As the sinistral side is 0 we deduce that the remainder r = 0.
705 Example A polynomial leaves remainder −2 upon division by x − 1 and remainder −4 upon division by x + 2
. Find the remainder when this polynomial is divided by x
2
+ x − 2. Solution: From the given information, there exist polynomials q
1
x , q
2
x with px = q
1
xx − 1 − 2 and px = q
2
xx + 2 − 4 . Thus p1 = −2 and p−2 = −4. As x
2
+ x − 2 = x − 1x + 2 is a polynomial of degree 2 the remainder rx upon dividing px by x
2
+ x − 1 is of degree 1 or less, that is rx = ax + b for some constants a , b which we
must determine. By the Division Algorithm, px = qxx
2
+ x − 1 + ax + b .
Hence −2 = p1 = a + b
and −4 = p−2 = −2a + b
. From these equations we deduce that a = 2
3, b = −83. The remainder sought is rx = 2x3 − 83.
706 Example Let f x = x
4
+ x
3
+ x
2
+ x + 1 . Find the remainder when f x
5
is divided by f x .
Solution: Observe that f xx − 1 = x
5
− 1 and f x
5
= x
20
+ x
15
+ x
10
+ x
5
+ 1 = x
20
− 1 + x
15
− 1 + x
10
− 1 + x
5
− 1 + 5 .
Each of the summands in parentheses is divisible by x
5
− 1 and, a fortiori, by f x . The remainder sought is thus 5.
Using the Division Algorithm we may derive the following theorem.
707 Theorem Factor Theorem The polynomial px is divisible by x − a if and only if pa = 0
.
Proof As x − a is a polynomial of degree 1, the remainder after diving px by x − a is a polynomial of degree 0, es that is, a constant. Therefore
px = qxx − a + r .
From this we gather that pa = qaa − a + r = r , from where the theorem easily follows.
708 Example If px is a cubic polynomial with p1 = 1
, p2 = 2, p3 = 3, p4 = 5, find p6. Solution: Put gx = px − x. Observe that gx is a polynomial of degree 3 and that g1 = g2 = g3 = 0
. Thus gx = cx − 1x − 2x − 3 for some constant c that we must determine. Now, g4 = c4 − 14 − 24 − 3 = 6c and
g4 = p4 − 4 = 1 , whence c = 16. Finally
p6 = g6 + 6 = 6 − 16 − 26 − 3
6 + 6 = 16
.
Practice 107
709 Example
The polynomial px has integral coefficients and px = 7 for four different values of x. Shew that px never equals 14.
Solution: The polynomial gx = px − 7 vanishes at the 4 different integer values a , b, c, d. In virtue of the Factor Theorem,
gx = x − ax − bx − cx − dqx ,
where qx is a polynomial with integral coefficients. Suppose that pt = 14 for some integer t. Then gt = pt − 7 = 14 − 7 = 7
. It follows that 7 = gt = t − at − bt − ct − dqt
, that is, we have factorised 7 as the product of at least 4 different factors, which is impossible since 7 can be factorised as
7−11, the product of at most 3 distinct integral factors. From this contradiction we deduce that such an integer t does not exist.
Practice
710 Problem If px is a polynomial of degree n such that
pk = 1 k, k = 1, 2, . . . ,n + 1, find pn + 2.
711 Problem The polynomial px satisfies p−x = −px
. When px is divided by x − 3 the remainder is 6. Find the remainder when px is divided by x
2
− 9.
6.4 Viète’s Formulae
