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| Algebraic
Expressions |
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Factoring and expanding algebraic expressions, rules for transforming algebraic expressions |
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Factoring
algebraic
expressions - methods, the
greatest common factor |
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Factoring by grouping |
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Perfect
square trinomials - the
square of a binomial |
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The difference of
two
squares |
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Factoring
quadratic trinomials |
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The sum and difference of cubes |
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Using a variety of methods including combinations of the above to factorize
algebraic expressions |
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| Factoring algebraic
expressions |
| Factoring
algebraic expression by finding (determining) the greatest common factor |
| Examples: |
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a) 3x
- 6y
= 3 · (x
- 2y),
b) xy
- y2
= y ·
(x - y),
c) a
-
a2
= a · (1 -
a), |
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d) x3
-3x2
+
x
= x · (x2
- 3x
+1),
e) x(a
+
b)
- (a
+
b)
= (a
+
b)
· (x
-
1), |
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f)
a(x
- 3y)
- x
+
3y =
a(x -
3y) - (x
- 3y)
= (x - 3y)
· (a
- 1). |
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| Grouping
like terms, grouping and factorizing four terms |
| An addition
sign, or plus sign, in front of the brackets leaves the sign of
every term inside the brackets unchanged. |
| A minus sign
in front of the bracket indicates that, when removing the
bracket, the sign of all terms inside must be changed. |
| Examples: |
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a) ax
- bx
- a
+
b
= x(a
-
b)
- (a
-
b)
= (a
-
b)
·
(x
-
1), |
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b) a
- 1
- ab
+
b
= (a
- 1)
- b
·
(a
- 1)
= (a
- 1)
· (1 -
b), |
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c) x2
+
ax
- bx
- ab
= x(x
+
a)
- b
·
(x
+
a)
= (x +
a)
· (x
-
b), |
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d)
5ab2
- 3a3
- 10b3
+ 6a2b
= 5b2(a -
2b)
-3a2(a
- 2b)
= (a - 2b)(5b2
- 3a2). |
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| Perfect
square trinomials - the
square of a binomial |
| Examples: |
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a) 1
- 4x
+ 4x2
= 12 -
2 · 2x
+ (2x)2
= (1 - 2x)2
= (1 - 2x)
· (1 -
2x), |
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b) a5
+ 6a4b
+ 9a3b2
= a3 · (a2
+ 6ab
+ 9b2
) = a3(a
+ 3b)2
= a3(a
+ 3b)(a
+ 3b). |
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| The
difference of
two
squares |
| Examples: |
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a) 16x2
- 1
= (4x)2
- 12
= (4x
-1)
· (4x
+1), |
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b) 5y3
- 20x2y
= 5y
· (y2
- 4x2)
= 5y
[y2
- (2x)2]
= 5y(y
- 2x)(y
+ 2x), |
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c) 9x2
-
(x
+ 2)2
= [3x -
(x
+ 2)]
· [3x
+
(x
+ 2)]
= (2x
-2)
·
(4x
+ 2)
= 4(x
-1)
· (2x
+1). |
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| Factoring
quadratic trinomials |
| A
quadratic trinomial ax2
+ bx
+ c
can be factorized as |
| ax2
+ bx
+ c
= a·[x2
+ (b/a)·x
+ c/a]
= a·(x
-
x1)(x
-
x2), |
where
x1
+ x2
= b/a
and x1·
x2
= c/a |
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| That
means, to factorize a quadratic trinomial we should find such a
pair of numbers x1
and x2
whose sum equals b/a
and whose product equals c/a. |
| Therefore,
when the constant term c
is negative, then the signs of x1
and x2
will be different but when c
is positive, their signs will be the same. |
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Examples:
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a) x2
- 3x
-10
= x2
+ (-5
+
2)·x
+ (-5)·(+2)
= x2
- 5x
+
2x
-10
= |
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= x ·
(x
- 5)
+ 2 ·
(x
- 5)
= (x
- 5)
·
(x
+ 2), |
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b) 2x2
- 7x
+ 3
= 2 ·
(x2
- 7/2x
+ 3/2)
= 2(x2
- 1/2x
- 3x
+ 3/2)
= |
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= 2[x·(x
- 1/2)
- 3·
(x - 1/2)]
= 2·
(x - 1/2)·(x
- 3)
= (2x
- 1)·
(x - 3), |
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c) 3x2
- x
- 2
= 3(x2
- 1/3x
- 2/3)
= 3(x2 +
2/3x
- x
- 2/3)
= |
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= 3[x·(x
+ 2/3)
-
(x + 2/3)]
= 3·(x
+ 2/3)·(x
- 1)
= (3x + 2)·(x
- 1). |
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The sum and difference of cubes |
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Examples:
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a) x3
+ 8
= x3 + 23
= (x + 2)
· (x2
- 2x
+ 22), |
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since (x + 2)·(x2
- 2x
+ 4)
= x3
- 2x2
+ 4x
+ 2x2
- 4x
+ 8
= x3
+ 8, |
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b) 8a3
-125
=
(2a)3 -
53
= (2a - 5)·
[(2a)2
+
(2a)·5 + 52]
= (2a - 5)(4a2
+ 10a
+ 25), |
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since (2a
- 5)(4a2
+ 10a
+ 25)
= 8a3 +
20a2
+ 50a
- 20a2
- 50a
-125
= 8a3 -125. |
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| Using
a variety of methods including combinations of the above to
factorize algebraic expressions |
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Examples:
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a) x2
- 2xy
+ y2
+ 2y
- 2x
=
(x
- y)2
- 2(x
- y)
=
(x
- y)(x
- y
- 2), |
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b) x2
- y2
+ xz
- yz
=
(x
- y)(x
+ y)
+ z(x
- y)
= (x
- y)(x
+ y
+ z), |
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c) 4x2
- 4xy
+ y2
- z2
= (2x - y)2
- z2
=
(2x
- y
- z)(2x
- y
+ z), |
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| d) a3
- 7a
+ 6
= a3 - a
- 6a
+ 6
= a(a2 -1)
- 6(a
-1)
= (a -1)·[a(a
+ 1)
- 6]
= (a -1)(a2
+ a
- 6)
= |
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= (a -1)(a2
+ 3a
- 2a
- 6)
= (a -1)[a(a
+ 3)
- 2(a
+ 3)]
= (a -1)(a
+ 3)(a
- 2). |
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| Pre-calculus
contents A |
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