Download Some algebra drill problems

Algebra Drill
Assembled by Katya Yurasovskaya for Math 180/184 classes and workshops.
Fall 2006
Based on Elementary Algebra by Faddeev and Sominskii 1965
Note: The most important sections are marked by ∗∗
I Rearrangement of algebraic expressions
• §1. Multiplication of powers of the same base and raising a
power to a power
Carry out the operations as indicated:
1. a3 ∗ a4 ∗ a5 ∗ a6 2. (x3 )5 3. [(c4 )3 ]2 4. [(x2 + y 2 )2 ]3
• §2. Raising a monomial to a power
Evaluate the powers:
1. (2a3 bc)4 2. (−ab2 c)5 3. ( 32 xy 2 )2 4. (−0.3a3 xy 2 )4
• §3. Multiplication of a polynomial by a monomial
Solve the following equation for x:
1. x ÷ (−3ab) = a3 + b3
• §4. Multiplication of a polynomial by a polynomial
Multiply the polynomials
1. (x2 + xy + y 2 )(2x2 − 3xy + 4y 2 )
2. (2a + b − c)(2a + b + c)
9 2
y )(14x + 9y)
3. (− 94 x2 + 27 xy − 49
2
2 2
4. (x + xy + y )
• §5. Multiplication of several polynomials
Multiply out the polynomial:
1. (x − 2y)(x + 2y)(x2 + 4y 2 )
• §6. Multiplication of polynomials containing only one letter
Multiply the polynomials:
1. (b − 1)(b5 + b4 + b3 + b2 + b + 1)
2. (x − 1)(x + 3)(x + 5)
3. (x − 1)(x − 2)(x − 3)
1
• * * §7. Rapid multiplication by formulae
(A + B)2 = A2 + 2AB + B 2
(A − B)2 = A2 − 2AB + B 2 )
(A + B)(A − B) = A2 − B 2
(A + B)3 = a3 + 3A2 B + 3AB 2 + B 3
(A − B)3 = A3 − 3A2 B + 3AB 2 − B 3
(A + B)(A2 − AB + B 2 ) = A3 + B 3
(A − B)(A2 + AB + B 2 ) = A3 − B 3
Use formulaes to do the following multiplications:
1. (3ab − x2 )(3ab + x2 ) 2. (x − 1)(x + 1)(x2 + 1)
3. (a + b)2 (a − b)2 (Solve by two methods)
4. (a2 + 3ab + b2 )(a2 − 3ab + b2 )
5. (3x − 2y)(9x2 + 6xy + 4y 2 ) 6. (2x + 1)3
7. (−a − b)2
Solve the following equations for x:
9. (a + b)x = a2 − b2
10. (3a2 − 2b)x = 9a2 − 4b2
11. (2p + 3q)x = 4p2 + 12pq + 9q 2
12. (a3 − 8b3 ) ÷ x = a − 2b
13. x ÷ (3p + q) = 9p2 − 3pq + q 2
II Factorization of Polynomials
• §1. The method of grouping
Factorize:
1. a3 + a − 2ab + 3 + 3a2 − 6b 2. a3 + a + 3a2 + a
• §2. Splitting individual terms of a polynomial into similar terms
EXAMPLE: Factorize a2 + 3ab + 2b2
Solution:
a2 + 3ab + 2b2 = a2 + ab + 2ab + 2b2 = a(a + b) + 2b(a + b) = (a + 2b)(a + b)
Factorize:
1. x2 + ax − 2a2 2. (x3 + 3bx − 4b2 x 3. x4 + 5x2 y 2 + 6y 4
2
• * *§3. The use of multiplication formulae
Factorize:
1. x4 − 16y 4 2. 4xy 3 − 4x3 y 3. a2 + 6ab + 9b2 − 4c2
4. 8x4 − 27a3 x 5. 4xy − xyz 2 6. x2 + 2x − y 2 − 2y
• * * * §4. More complicated examples
Factorize:
1. x5 + x3 − x2 − 1 2. a4 − 2a3 b − 8a2 b2 − 6ab3 − b4
3. x2 y + xy 2 + x2 z + xz 2 + y 2 z + yz 2 + 3xyz
4. x4 + 2x3 + 3x2 + 2x + 1 5. x5 − 1
III The rearrangement of fractional algebraic expressions
• §1. No ”division by zero” !
For what values of the symbols are the following algebraic expressions
meaningful?
1
1.
x−9
x2 − 64
2.
x−8
1
1
3. +
x y
4.
1
x
1
x
+
−
1
y
1
y
• §2. Division by powers of the same base
Divide:
1. a7 by a3 2. c2 by c5 3. b11 by b11
Find x in:
4. x ÷ a2 = a5 5. a3 ÷ x = a7 6. x ∗ a2 = a4
• §3. Division of monomials
Do the following divisions:
3 ∗ 2a3 bpq
1. 0.1a2 bc ÷ (−0.01ab) 2.
2 ∗ 4ab
3. 7ab2 c ÷ (5a3 b3 c3 )
• * *§4. Division of polynomial by a monomial
Examples and ideas:
a
b
c
d
a+b+c+d
=
+
+
+
m
m m m m
2. abcd + 2abc + 2abd + 2acd + 2bcd
1.
2abc 2abd 2abd 2bcd
+
+
+
)
abcd abcd
abcd abcd
2 2 2 2
= abcd(1 + + + + )
d c
b a
= abcd(1 +
3
Carry out the divisions:
1. (3x3 y − 0.1xy 2 + 7.3x2 y) ÷ (−0.1xy)
2. (x2 yz + xy 2 z + xyz 2 ) ÷ (xyz)
a3 b + a2 b2 − 3ab3
3.
2ab
xy + xz + yz
dividing term by term.
4. Rearrange
xyz
2 2
5. In the polynomial x y z − xy 2 z 2 + x2 y 2 z 2 takex2 y 2 z 2 outside brackets.
• * *§5. The application of multiplication formulae to the division
of a polynomial by a polynomial
Perform the following divisions:
x4 − 16y 2
x3 + 3x2 y + 3xy 2 + y 3
1. 2
2.
x + 4y 2
x+y
x4 + 6x2 + 9
(a + b)2 − (c + d)2
4.
3.
a+b−c−d
x2 + 3
Solve for x:
5. x(a + 2b) = a2 − 4b2 6. (a − 3b)x = a2 − 6ab + 9b2
7. (a − b − c)x = a2 − 2ab + b2 − c2
• Find some problems on long division in any algebra book and
do 3-4...
• §6. Simplification of algebraic fraction
Simplify the fractions:
x2 + 5x + 6
(x + y2)2 − (2x + y)2
1.
2.
x2 − 9
(x + 3y)2 − (3x + y)2
• §7. Addition and subtraction of algebraic fractions
Note: review finding common denominators
Add the fractions:
1
2x
3x
1.
− −
x+2 x x−2
2
1
(x + 1)2
2. 1 − + 2 −
x x
x2
• §8. Multiplication of fractions
Multiply the fractions:
a+b
a4 − b4
1
1
1
1. 2
∗
2. (1 + )(1 +
)(1 +
)
2
a +b
2ab
x
x+1
x+2
4
• * *§9. Simplification of fractions in which the numerator and
denominator are algebraic sums of fractions
Two useful examples:
1. Simplify the expression:
1
1
a + b
1
1
a + b
Solution:
1
1
a + b
1
1 =
a + b
b+a
ab
b−a
ab
=
b+a
ab
b+a
×
=
ab
b−a
b−a
2. Previous example done differently:
1
1
ab( a1 + 1b )
b+a
a + b
1 =
1 = b−a
1
1
+
−
)
ab(
a
b
a
b
3. Simplify the expression:
1
1
− x+1
2 + x−1
x+
x
x2 −1
Solution: Multiplying the numerator and denominator by x2 − 1
we get
2(x2 − 1) + (x + 1) − (x − 1)
2x2 − 2 + x + 1 − x + 1
2x2
=
=
=
x(x2 − 1) + x
x3 − x + x
x3
2
x
Simplify the fractions:
1
1
1
1
x(x+1) + (x+1)(x+2)
x+1 + x−1
2.
1. 1
1
x+2
6x
x−1 − x+1
x+2 − x
IV Powers and Roots
• §1. Properties of powers with integral indices
an ∗ am = an+m
(am )n = amn
(ab)n = an ∗ bn , (abc)n = an bn cn etc
an
a
( )n = n
b
b
5
Simplify the expressions:
a2 3
b2
) ∗ ( )4 2. (am−n )m+n for m > n 3. a ∗ (a2 )2 ∗ (a3 )3
b
a
4 Calculate (−2)6 and (−3)5
1. (
• §2. The root to any power of a number
Note: The positive
√ value of an even root is called its arithmetical value.
The notation n a for even
p n and a > 0 always means the arithmetical
value of the root. Thus (−3)2 = 3 (and not −3).
Evaluate:
p
p
√
√
√
1. 4 + 3 27 2. 3 27 + (−2)6 3. (a − 2)2
• §3. Extracting the root of products, fractions and powers
√
√
( n a)n = a, in particular ( a)2 = a)
√
n
√
n
a ∗ b ∗ ... ∗ k =
√
n
a∗
an = a
√
n
b ∗ ... ∗
√
n
k, for a, b, ...k > 0
√
n
a
a
= √
for a, b > 0
n
b
b
√
√
nk
n
a=
ak for a > 0
r
n
Rearrange:
q
√
√
1. 2 × 3 3 × 5 16
√
√
2× 33
√
3. √
3× 32
√
3
16
2. √
8
• §4. Raising a root to a power and extracting the root of a root
(
q
√
n
n
a)k = ak
q
k
√
n
a=
√
nk
a
p√
√
√
Rearrange: 1. ( 2)4 2. ( 3 5)2 3.
25
6
• §5. Removing a rational factor from under the root sign.
Putting a rational factor under the root sign
If a and b are√positive
numbers then
√
n
n
a b = an b
r
n
√
n
b
b
=
n
a
a
Put all factors under the root
rsign:
q
√
a
−
b
a+b
1. 2 3 2. 3 3 13 3.
a+b a−b
• * *§6 Rationalizing a denominator
A fractional expression containing radicals in the denominator can be
converted into a fractional expression without radicals in the denominator.
This process is called rationalizing the denominator.
1. The denominator is a radical, i.e. the fraction is of the form
A
√
n
b
In this case a factor must be chosen which, when multiplied by the
base, gives a perfect√nth power
√
5
532
532
5√
3
√
Example: √
= √
= √
=
2
3
3
3
3
2
4
4 2
8
2. Expression is of form
c
c
√ or
√
a+ b
a− b √
√
Then multiply numerator and denominator by a − b or a + b respectively. Then we get:
√
a
c(a − b)
c(a − sqrtb)
√ =
√
√ =
a2 − b2
a+ b
(a + b)(a − b)
3. The denominator is sum or difference of two square roots, i.e. the
fraction is of form
c
c
√ or √
√
√
a+ b
a− b
Use same method as in the previous case.
Rationalize the
√ denominators
√
√ of:
a
2+ 3
2−1
√
1. √
2.
3. √
5
6
2+1
a3
7
a3 − x3
√
4. √
a− x
V Quadratic equations and equations reducing to quadratics
• §1. The reduced quadratic equation
Reduced quadratic equation x2 + px + q = 0
1. ( 12 p)2 − q < 0
Equation has no solution
2. ( 12 p)2 − q = 0
Equation has only one solution, x = − 12 p
3. ( 12 p)2 − q > 0
Equation has two solutions:
r
p
p
x = − ± ( )2 − q
2
2
or equivalently
p
−p ± p2 − 4q
x=
2
Solve:
1. x3 − x − 96 = (x − 1)(x − 2)(x − 3) 2. (x + 1)2 = 2x2 − 7
• * * * §2. The general quadratic equation
General quadratic equation: ax+ bx + c = 0
To find roots use quadratic formula
√
−b ± b2 − 4ac
x=
2a
Solve:
1. 5x2 + 3x − 1 = 0 2. (a + b)x2 + (4a − 2b) + b − 5a = 0
3. (x + 1)(x + 2)(x + 3) = x3 + 52
• §3. Factorizing a quadratic trinomial
1. Given reduced quadratic trinomial x2 + px + q with real roots x1 , x2 ,
the factorized form is
x2 + px + q = (x − x1 )(x − x2 )
2. Given a quadratic trinomial ax2 + bx + c with real roots x1 , x2 , the
factorized form is
ax2 + bx + c = a(x − x1 )(x − x2 )
Factorize:
1. 3x2 − 10x + 3 2. x2 + 2x − 5
8
Answers
Section I
§1
1
1
2
2
• 1. a 8 2. x 5 3. c 4 4. (x + y 2 )6
§2
1
4 4
• 1. 16a 2b c
5 1
5
2. −a b 0c
3.
4 2 2
9x y
4. 0.0081a1 2x4 y 8
§3
4
4
• 1. x = −3a b − 3ab
§4
4
3
2 2
3
• 1. 2x − x y + 3x y + xy + 4y 4 2. 4a2 + 4ab + b2 − c2
81
3
4. x4 + 2x3 y + 3x2 y 2 + 2xy 3 + y 4
3. − 56
9 x − 49
§5
• 1. x4 − 16y 4
§6
6
3
2
• 1. b − 1 2. x + 7x + 7x − 15 3. x3 − 6x2 + 11x − 6
§7
• 1. 9a2 b2 − x4 2. x4 − 1 3. a4 − 2a2 b2 + b4 4. a4 − 7a2 b2 + b4 5.
27x3 −8y 3 6. 8x3 +12x2 +6x+1 7. a2 +2ab+b2 8. x4 +4x3 +2x2 −4x+1
9. x = a − b 10. x = 3a2 + 2b 11. x = 2p + 3q 12. x = a2 + 2ab + 4b2
13. x = 27p3 + q 3
9
Section II
§1
2
• 1. (a + 3)(a + 1 − 2b) 2. (a + 3)(a2 + 1)
§2
• 1. (x + 2a)(x − a) 2. x(x + 4b)(x − b) 3. (x2 + 2y 2 )(x2 + 3y 2
§3
2
2
• 1. (x + 2y)(x − 2y)(x + 4y ) 2. 4xy(y + x)(y − x)
3. (a + 3b + 2c)(a + 3b − 2c) 4. x(2x − 3a)(4x2 + 6ax + 9a2 )
5. xy(2 + z)(2 − z) 6. (x − y)(x + y + 2)
Section III
§1
• 1. x 6= 9 2. x 6= 8 3. x 6= 0 and y 6= 0 4. x 6= 0 and y 6= 0 and x 6= y
§2
• 1. a4 2.
1
c3
3. 1 4. x = a7 5. x =
1
a4
6. x = a2
§3
• 1. −10ac 2.
4 2
3 a pq
3.
7
5a2 bc2
§4
• 1. −30x2 + y − 73x 2. x + y + z 3.
4. x1 + y1 + z1 5. x2 y 2 z 2 ( z1 − x1 + y1 )
a2 +ab−3b2
2
§5
2
2
2
2
• 1. x − 4y 2. x + 2xy + y 3. a + b + c + d 4. x2 + 3
5. x = a − 2b 6. x = a − 3b 7. x = a − b + c
§6
• 1.
x+2
x−3
• 1.
x3 −11x2 +4
x(x2 −4)
• 1.
(a+b)2 (a−b)
2ab
2.
3
8
§7
2.
− x4
§8
2.
2x(x2 +x+1)
x+1
§9
• 1. x 2.
2
5x2 −4x−4
10
Section IV
§1
2 5
• 1. a b
m2 −n2
1
3. a 4 4. x6 + 4x5 − 2x4 − 20x3 − 7x2 + 24x + 16
2. a
§2
• 1. 5 2. 11 3. a − 2 for a ≥ 2; 2 − a for a < 2
§3
• 1.
√
30
q
29 × 34 2.
• 1. 4 2.
√
3
25 3.
1
2
6
√
3.
q
6
2
3
§4
5
§5
• 1.
• 1.
√
12 2.
√
5
a2
2.
√
3
9 3.
q
√
√
2 3+3 2
6
a−b
a+b
§6
√
√
√
3. 3 − 2 2 4. (a2 + ax + x2 )( a + x)
Section V
§1
• 1. −3, 5 2. −2, 4
§2
• 1.
√
−3± 29
10
2. 1,
• 1. 3(x − 3)(x −
1
3)
b−5a
a+b
3. 2, − 23
6
§3
√
√
2. (x + 1 − 6)(x + 1 + 6)
11