Identities and Factorization

1
Identities and Factorization
4
Case Study
4.1 Meaning of Identity
4.2 Difference of Two Squares
4.3 Perfect Square
4.4 Factorization by Taking out Common Factors
4.5 Factorization by Grouping Terms
Chapter Summary
2
Case Study
Rachael wants to calculate the area of a square.
She measures the length of the square is 103 cm.
Without using a calculator, Rachael can find the
area by the use of the perfect square.
We have the following formula for the perfect
square (a ? b)2 ? a2 ? 2ab ? b2
We can consider 103 ? 100 ? 3, that is a ? 100
and b ? 3.
Thus, the area ? 1032 cm2 ? (100 ? 3)2 cm2
? 1002 ? 2(100)(3) ? 32 cm2
? (10000 ? 600 ? 9) cm2 ? 10609 cm2
3
4.1 Meaning of Identity
Consider an equation x ? 2 ? 2x
.............................. (1)
L.H.S. ? x ? 2
R.H.S. ? 2x
L.H.S. ? R.H.S.?
Consider an equation x ? 2 ?(2 ? x)
..................... (2)
L.H.S. ? x ? 2
R.H.S. ? ?(2 ? x)
L.H.S. ? R.H.S.?
Only one value of x can satisfy equation (1).
However, all values of x satisfy equation (2).
4
4.1 Meaning of Identity
Consider the equations x ? 2 ? 2x
.............................. (1) x ? 2 ?(2
? x) ..................... (2) If we substitute
any value of x into the equation (2), the L.H.S.
is always equal to the R.H.S. We call such an
equation an identity.
If any value of the unknown(s) can satisfy an
equation, then the equation is called an identity.
However, to prove that an equation is an
identity, it is impossible for us to check with
all values of unknown(s).
Thus, we can simplify the algebraic expressions
on both sides and match up the like terms.
If all the like terms and the corresponding
coefficients on both sides of the equation are
equal, then the given equation is an identity.
5
4.1 Meaning of Identity
Example 4.1T
Prove that the following equations are
identities. (a) 6x ? 7 ? 3(2x ? 3) ? 2 (b) (5x
? 3) ? 2(x ? 4) ? 3(x ? 3) ? 2
Solution
(a) R.H.S. ? 3(2x ? 3) ? 2
? 6x ? 9 ? 2
? 6x ? 7
? L.H.S.
? 6x ? 7 ? 3(2x ? 3) ? 2 is an identity.
(b) L.H.S. ? (5x ? 3) ? 2(x ? 4)
R.H.S. ? 3(x ? 3) ? 2
? 5x ? 3 ? 2x ? 8
? 3x ? 9 ? 2
? 3x ? 11
? 3x ? 11
? L.H.S. ? R.H.S.
? (5x ? 3) ? 2(x ? 4) ? 3(x ? 3) ? 2 is an
identity.
6
4.1 Meaning of Identity
Example 4.2T
Solution
(a) L.H.S. ? (x ? 2)2
? (x ? 2)(x ? 2)
? x(x ? 2) ? 2(x ? 2)
? x2 ? 2x ? 2x ? 4
? x2 ? 4x ? 4
? R.H.S.
? (x ? 2)2 ? x2 ? 4x ? 4 is an identity.
7
4.1 Meaning of Identity
Example 4.2T
Solution
(b) L.H.S. ? (2x ? 1)(x ? 2) ? x(x ? 4)
? 2x(x ? 2) ? (x ? 2) ? x(x ? 4)
? 2x2 ? 4x ? x ? 2 ? x2 ? 4x
? x2 ? x ? 2
R.H.S. ? (x ? 2)(x ? 1)
? x(x ? 1) ? 2(x ? 1)
? x2 ? x ? 2x ? 2
? x2 ? x ? 2
? L.H.S. ? R.H.S.
? (2x ? 1)(x ? 2) ? x(x ? 4) ? (x ? 2)(x ? 1) is
an identity.
8
4.1 Meaning of Identity
Example 4.2T
Solution
? x ? 4 ? x ? 2
? ?6
? R.H.S.
9
4.1 Meaning of Identity
Example 4.3T
Prove that 3(2x ? 1) ? 4(4x ? 3) ? ?5(x ? 3) is
not an identity.
Solution
L.H.S. ? 3(2x ? 1) ? 4(4x ? 3)
? 6x ? 3 ? 16x ? 12
? ?10x ? 15
R.H.S. ? ?5(x ? 3)
? ?5x ? 15
? L.H.S. ? R.H.S.
? 3(2x ? 1) ? 4(4x ? 3) ? ?5(x ? 3) is not an
identity.
Alternative Solution
When x ? 0,
L.H.S. ? 32(0) ? 1 ? 44(0) ? 3
? 15
R.H.S. ? ?5(0) ? 3
? ?15
? L.H.S. ? R.H.S.
? 3(2x ? 1) ? 4(4x ? 3) ? ?5(x ? 3) is not an
identity.
10
4.1 Meaning of Identity
By using the fact that the like terms on both
sides of an identity are equal, we can find the
value of the unknown coefficients or constants in
an identity.
For example, consider Ax ? B ? 2x ? 3.
Ax and 2x are like terms. B and 3 are also like
terms.
By comparing the coefficients of the like terms,
we have A ? 2 and B ? 3.
11
4.1 Meaning of Identity
Example 4.4T
If Ax2 ? Bx ? C ? (3x ? 2)(3x ? 2), find the
values of A, B and C.
Solution
R.H.S. ? (3x ? 2)(3x ? 2)
? 3x(3x ? 2) ? 2(3x ? 2)
? 9x2 ? 6x ? 6x ? 4
? 9x2 ? 4
? Ax2 ? Bx ? C ? 9x2 ? 4
12
4.1 Meaning of Identity
Example 4.5T
If 2x2 ? 5x ? C ? 2(x ? 2)(Ax ? 1) ? Bx, find the
values of A, B and C.
Solution
R.H.S. ? 2(x ? 2)(Ax ? 1) ? Bx
? (2x ? 4)(Ax ? 1) ? Bx
? 2x(Ax ? 1) ? 4(Ax ? 1) ? Bx
? 2Ax2 ? 2x ? 4Ax ? 4 ? Bx
? 2Ax2 ? (2 ? 4A ? B)x ? 4
? 2x2 ? 5x ? C ? 2Ax2 ? (2 ? 4A ? B)x ? 4
By comparing the coefficients of like terms, we
have 2A ? 2
2 ? 4A ? B ? 5
?C ? ?4
2 ? 4(1) ? B ? 5
13
4.1 Meaning of Identity
Example 4.5T
If 2x2 ? 5x ? C ? 2(x ? 2)(Ax ? 1) ? Bx, find the
values of A, B and C.
Alternative Solution
Substitute different values of x into the
identity to find the unknowns.
When x ? 0, 2(0)2 ? 5(0) ? C ? 2(0 ? 2)A(0) ?
1 ? B(0)
C ? 2(2)(1) 0
When x ? 2, 2(2)2 ? 5(2) ? C ? 2(2 ? 2)A(2) ?
1 ? B(2)
8 ? 10 ? 4 ? 2B
When x ? 1, 2(1)2 ? 5(1) ? C ? 2(1 ? 2)A(1) ?
1 ? B(1)
2 ? 5 ? 4 ? 2(?1)(A ? 1) ? 7
?4 ? ?2(A ? 1)
14
4.2 Difference of Two Squares
For any values of a and b,
(a ? b)(a ? b) ? a2 ? b2
The above identity is called the difference of
two squares.
We can prove the above identity algebraically.
L.H.S. ? (a ? b)(a ? b)
? a(a ? b) ? b(a ? b)
? a2 ? ab ? ba ? b2
? a2 ? b2
? R.H.S.
? (a ? b)(a ? b) ? a2 ? b2
15
4.2 Difference of Two Squares
Example 4.6T
Expand the following expressions. (a) (5x ? 2)(5x
? 2) (b) (?4 ? 3x)(?4 ? 3x) (c) (?2x ? 7y)(?2x ?
7y)
Solution
(a) (5x ? 2)(5x ? 2) ? (5x)2 ? 22
(b) (?4 ? 3x)(?4 ? 3x) ? (?4)2 ? (3x)2
(c) (?2x ? 7y)(?2x ? 7y) ? (?2x)2 ? (7y)2
16
4.2 Difference of Two Squares
Example 4.7T
Solution
17
4.2 Difference of Two Squares
Example 4.8T
Evaluate the following without using a
calculator. (a) 1252 ? 252 (b) 100.5 ? 99.5
Solution
(a) 1252 ? 252 ? (125 ? 25)(125 ? 25)
? 150 ? 100
(b) 100.5 ? 99.5 ? (100 ? 0.5)(100 ? 0.5)
? 1002 ? 0.52
? 10 000 ? 0.25
18
4.3 Perfect Square
For any values of a and b,
(a ? b)2 ? a2 ? 2ab ? b2
(a ? b)2 ? a2 ? 2ab ? b2
and
The above identities are called the perfect
square.
We can prove the above identities algebraically.
L.H.S. ? (a ? b)2
L.H.S. ? (a ? b)2
? (a ? b)(a ? b)
? (a ? b)(a ? b)
? a(a ? b) ? b(a ? b)
? a(a ? b) ? b(a ? b)
? a2 ? ab ? ba ? b2
? a2 ? ab ? ba ? b2
? a2 ? 2ab ? b2
? a2 ? 2ab ? b2
? R.H.S.
? R.H.S.
? (a ? b)2 ? a2 ? 2ab ? b2
? (a ? b)2 ? a2 ? 2ab ? b2
19
4.3 Perfect Square
Example 4.9T
Expand the following expressions. (a) (3x ?
2y)2 (b) (2 ? 5x)2
Solution
(a) (3x ? 2y)2 ? (3x)2 ? 2(3x)(2y) ? (2y)2
(b) (2 ? 5x)2 ? 22 ? 2(2)(5x) ? (5x)2
20
4.3 Perfect Square
Example 4.10T
Solution
(a) 3(?2x ? y)2 ? 3(?2x)2 ? 2(?2x)(y) ? y2
? 3(4x2 ? 4xy ? y2)
21
4.3 Perfect Square
Example 4.11T
Evaluate the following without using a
calculator. (a) 9952 (b) 1052
Solution
(a) 9952 ? (1000 ? 5)2
? 10002 ? 2(1000)(5) ? 52
? 1 000 000 ? 10 000 ? 25
(b) 1052 ? (100 ? 5)2
? 1002 ? 2(100)(5) ? 52
? 10 000 ? 1000 ? 25
22
4.4 Factorization by Taking out Common
Factors
We already know how to expand the polynomial a(b
? c) into ab ? ac and the process is called
expansion.
The process of expressing ab ? ac into a(b ? c)
is called factorization. Factorization is the
reverse process of expansion.
Consider the polynomial 2xy ? xz. x is the common
factor for the terms 2xy and xz. We can take this
common factor out 2xy ? xz ? x(2y ? z)
This method is called factorization by taking out
common factors.
If we want to know whether the results of
factorization are correct, then we can expand the
expressions to check the results.
23
4.4 Factorization by Taking out Common
Factors
Example 4.12T
Factorize the following expressions. (a) 5a2x ?
15a2x2 (b) 45pq ? 60pqr (c) 18x3y ? 24x2y2 ? 30xy3
Solution
(a) 5a2x ? 15a2x2 ? 5a2x(1) ? 5a2x(3x)
(b) 45pq ? 60pqr ? 15pq(3) ? 15pq(4r)
(c) 18x3y ? 24x2y2 ? 30xy3 ? 6xy(3x2) ? 6xy(4xy)
? 6xy(5y2)
24
4.4 Factorization by Taking out Common
Factors
Example 4.13T
Factorize the following expressions. (a) (2a ?
b)c ? (2a b)d (b) 2m(x ? 2y) ? 4n(2y ?
x) (c) ?5rt ? 2t(3r ? 4s) (d) 18m2n(p ? q)2 ?
27mn2(q ? p)
Solution
(b) 2m(x ? 2y) ? 4n(2y ? x) ? 2m(x ? 2y) ? 4n(x ?
2y)
(c) ?5rt ? 2t(3r ? 4s) ? ?t5r ? 2(3r ? 4s)
? ?t(5r ? 6r ? 8s)
which is equivalent to t(8s ? 11r)
(d) 18m2n(p ? q)2 ? 27mn2(q ? p) ? 18m2n(p ? q)2
? 27mn2(p ? q)
? 9mn(p ? q)2m(p ? q) ? 3n
25
4.5 Factorization by Grouping Terms
In the previous section, we learnt how to
factorize polynomials by taking out common
factors.
However, for some expressions like ac ? ad ? bc ?
bd, we may be unable to find any common factors
for all terms.
In such a situation, we can group the terms first
and then take out the common factors of each
group to factorize the expression.
ac ? ad ? bc ? bd ? (ac ? ad) ? (bc ? bd)
? a(c ? d) ? b(c ? d)
? (a ? b)(c ? d)
This method is called factorization by grouping
terms.
26
4.5 Factorization by Grouping Terms
Example 4.14T
Factorize the following expressions. (a) 1 ? y ?
5xy ? 5x (b) 6xyz ? 6wyz ? 5w ? 5x
Solution
(a) 1 ? y ? 5xy ? 5x ? (1 ? y) ? (5xy ? 5x)
OR
? (1 ? y) ? 5x(y ? 1)
? 5x(y ? 1) ? (1 ? y)
? (1 ? y) ? 5x(1 ? y)
? 5x(y ? 1) ? (y ? 1)
Alternative Solution
(a) 1 ? y ? 5xy ? 5x ? (1 ? 5x) ? (5xy ? y)
? (1 ? 5x) ? y(5x ? 1)
? (1 ? 5x) ? y(1 ? 5x)
27
4.5 Factorization by Grouping Terms
Example 4.14T
Factorize the following expressions. (a) 1 ? y ?
5xy ? 5x (b) 6xyz ? 6wyz ? 5w ? 5x
Solution
(b) 6xyz ? 6wyz ? 5w ? 5x ? (6xyz ? 6wyz) ?
(5x ? 5w)
? 6yz(x ? w) ? 5(x ? w)
Alternative Solution
(b) 6xyz ? 6wyz ? 5w ? 5x ? (6xyz ? 5x) ?
(6wyz ? 5w)
? x(6yz ? 5) ? w(6yz ? 5)
28
4.5 Factorization by Grouping Terms
Example 4.15T
Factorize the following expressions. (a) 3az ?
6bz ? 12b ? 6a (b) x2y2 ? y2 ? x2 ? 1
Solution
(a) 3az ? 6bz ? 12b ? 6a ? 3z(a ? 2b) ? 6(2b ? a)
? 3z(a ? 2b) ? 6(a ? 2b)
(b) x2y2 ? y2 ? x2 ? 1 ? (x2y2 ? y2) ? (x2 ? 1)
? y2(x2 ? 1) ? (x2 ? 1)
Alternative Solution
(b) x2y2 ? y2 ? x2 ? 1 ? (x2y2 ? x2) ? (y2 ? 1)
? x2(y2 ? 1) ? (y2 ? 1)
29
Chapter Summary
4.1 Meaning of Identity
1. If any value of the unknown(s) can satisfy an
equation, then the equation is called an identity.
2. If all the like terms and their corresponding
coefficients on both sides of the equation are
equal, then the given equation is an identity.
3. The like terms on the two sides of an identity
are always equal.
30
4.2 Difference of Two Squares
Chapter Summary
For any values of a and b, (a ? b)(a ? b) ? a2
? b2
31
4.3 Perfect Square
Chapter Summary
For any values of a and b, (a ? b)2 ? a2 ? 2ab
? b2 and (a ? b)2 ? a2 ? 2ab ? b2
32
4.4 Factorization by Taking out Common Factors
Chapter Summary
1. The process of rewriting a polynomial as a
product of its factors is called factorization.
2. Factorization is the reverse process of
expansion.
3. When each term of a polynomial has one or more
common factors, we can factorize the polynomial
by taking out common factors.
33
4.5 Factorization by Grouping Terms
Chapter Summary
Divide a polynomial into several groups first,
then take out the common factors of each group to
factorize the polynomial.
34
4.1 Meaning of Identity
Follow-up 4.1
Prove that the following equations are
identities. (a) 3(3x ? 9) ? ?9(3 ? x) (b) 5 ?
(3x ? 13) ? 3(6 ? x)
Solution
(a) L.H.S. ? 3(3x ? 9)
R.H.S. ? ?9(3 ? x)
? 9x ? 27
? ?27 ? 9x
? 9x ? 27
? L.H.S. ? R.H.S.
? 3(3x ? 9) ? ?9(3 ? x) is an identity.
(b) L.H.S. ? 5 ? (3x ? 13)
R.H.S. ? 3(6 ? x)
? 5 ? 3x ? 13
? 18 ? 3x
? ?3x ? 18
? ?3x ? 18
? L.H.S. ? R.H.S.
? 5 ? (3x ? 13) ? 3(6 ? x) is an identity.
35
4.1 Meaning of Identity
Follow-up 4.2
Solution
(a) L.H.S. ? (x ? 4)(x ? 1)
? x(x ? 1) ? 4(x ? 1)
? x2 ? x ? 4x ? 4
? x2 ? 3x ? 4
? R.H.S.
? (x ? 4)(x ? 1) ? x2 ? 3x ? 4 is an identity.
36
4.1 Meaning of Identity
Follow-up 4.2
Solution
(b) L.H.S. ? (x ? 2)(x ? 3) ? 4x
R.H.S. ? (x ? 6)(x ? 1)
? x(x ? 3) ? 2(x ? 3) ? 4x
? x(x ? 1) ? 6(x ? 1)
? x2 ? 3x ? 2x ? 6 ? 4x
? x2 ? x ? 6x ? 6
? x2 ? 5x ? 6
? x2 ? 5x ? 6
? L.H.S. ? R.H.S.
? (x ? 2)(x ? 3) ? 4x ? (x ? 6)(x ? 1) is an
identity.
37
4.1 Meaning of Identity
Follow-up 4.2
Solution
? 2x ? 5 ? 2x ? 3
? 4x ? 2
R.H.S. ? 2(2x ? 1)
? 4x ? 2
? L.H.S. ? R.H.S.
38
4.1 Meaning of Identity
Follow-up 4.3
Prove that 4x ? 8(x ? 2) ? 12(x ? 1) is not an
identity.
Solution
L.H.S. ? 4x ? 8(x ? 2)
? 4x ? 8x ? 16
? 12x ? 16
R.H.S. ? 12(x ? 1)
? 12x ? 12
? L.H.S. ? R.H.S.
? 4x ? 8(x ? 2) ? 12(x ? 1) is not an identity.
Alternative Solution
When x ? 0,
L.H.S. ? 4(0) ? 8(0 ? 2)
? 16
R.H.S. ? 12(0) ? 1
? 12
? L.H.S. ? R.H.S.
? 4x ? 8(x ? 2) ? 12(x ? 1) is not an identity.
39
4.1 Meaning of Identity
Follow-up 4.4
If Ax2 ? Bx ? C ? (2x ? 1)(x ? 4), find the
values of A, B and C.
Solution
R.H.S. ? (2x ? 1)(x ? 4)
? 2x(x ? 4) ? (x ? 4)
? 2x2 ? 8x ? x ? 4
? 2x2 ? 7x ? 4
? Ax2 ? Bx ? C ? 2x2 ? 7x ? 4
40
4.1 Meaning of Identity
Follow-up 4.5
If 9x2 ? Bx ? 10 ? (Ax ? 1)(x ? 2) ? C, find the
values of A, B and C.
Solution
R.H.S. ? (Ax ? 1)(x ? 2) ? C
? Ax(x ? 2) ? (x ? 2) ? C
? Ax2 ? 2Ax ? x ? 2 ? C
? Ax2 ? (2A ? 1)x ? (2 ? C)
? 9x2 ? Bx ? 10 ? Ax2 ? (2A ? 1)x ? (2 ? C)
By comparing the coefficients of like terms, we
have
B ? 2A ? 1
2 ? C ? ?10
? 2(9) ? 1
41
4.2 Difference of Two Squares
Follow-up 4.6
Expand the following expressions. (a) (3x ? 1)(3x
? 1) (b) (?2x ? 5)(?2x ? 5) (c) (6x ? 5y)(6x ? 5y)
Solution
(a) (3x ? 1)(3x ? 1) ? (3x)2 ? 12
(b) (?2x ? 5)(?2x ? 5) ? (?2x)2 ? 52
(c) (6x ? 5y)(6x ? 5y) ? (6x)2 ? (5y)2
42
4.2 Difference of Two Squares
Follow-up 4.7
Solution
43
4.2 Difference of Two Squares
Follow-up 4.8
Evaluate the following without using a
calculator. (a) 9952 ? 52 (b) 302 ? 298
Solution
(a) 9952 ? 52 ? (995 ? 5)(995 ? 5)
? 1000 ? 990
(b) 302 ? 298 ? (300 ? 2)(300 ? 2)
? 3002 ? 22
? 90 000 ? 4
44
4.3 Perfect Square
Follow-up 4.9
Expand the following expressions. (a) (3x ?
4y)2 (b) (2x ? y)2
Solution
(a) (3x ? 4y)2 ? (3x)2 ? 2(3x)(4y) ? (4y)2
(b) (2x ? y)2 ? (2x)2 ? 2(2x)(y) ? y2
45
4.3 Perfect Square
Follow-up 4.10
Solution
(a) 4(?1 ? 4x)2 ? 4(?1)2 ? 2(?1)(4x) ? (4x)2
? 4(1 ? 8x ? 16x2)
46
4.3 Perfect Square
Follow-up 4.11
Evaluate the following without using a
calculator. (a) 1022 (b) 892
Solution
(a) 1022 ? (100 ? 2)2
? 1002 ? 2(100)(2) ? 22
? 10 000 ? 400 ? 4
(b) 892 ? (90 ? 1)2
? 902 ? 2(90)(1) ? 12
? 8100 ? 180 ? 1
47
4.4 Factorization by Taking out Common
Factors
Follow-up 4.12
Factorize the following expressions. (a) 3a2x ?
12ax2 (b) 25a2c ? 40abc2 (c) 14m3n ? 28m2n2 ?
21mn3
Solution
(a) 3a2x ? 12ax2 ? 3ax(a) ? 3ax(4x)
(b) 25a2c ? 40abc2 ? 5ac(5a) ? 5ac(8bc)
(c) 14m3n ? 28m2n2 ? 21mn3 ? 7mn(2m2) ? 7mn(4mn)
? 7mn (3n2)
48
4.4 Factorization by Taking out Common
Factors
Follow-up 4.13
Factorize the following expressions. (a) (a ? c)b
? (a c)d (b) 5m(p ? q) ? n(q ? p) (c) (4x ?
3y)z ? 5yz (d) 21p2q(a ? b) ? 28pq2(a ? b)2
Solution
(b) 5m(p ? q) ? n(q ? p) ? 5m(p ? q) ? n(p ? q)
(c) (4x ? 3y)z ? 5yz ? z(4x ? 3y) ? 5y
? z(4x ? 2y)
(d) 21p2q(a ? b) ? 28pq2(a ? b)2 ? 7pq(a ? b)3p
? 4q(a ? b)
49
4.5 Factorization by Grouping Terms
Follow-up 4.14
Factorize the following expressions. (a) 6a ?
6b ? ac ? bc (b) 4rst ? 6rsu ? 2t ? 3u
Solution
(a) 6a ? 6b ? ac ? bc ? (6a ? 6b) ? (ac ? bc)
? 6(a ? b) ? c(a ? b)
(b) 4rst ? 6rsu ? 2t ? 3u ? (4rst ? 6rsu) ? (2t ?
3u)
? 2rs(2t ? 3u) ? (2t ? 3u)
50
4.5 Factorization by Grouping Terms
Follow-up 4.15
Factorize the following expressions. (a) ?2x2y ?
6xz ? 12xy ? 36z (b) ab ? 1 ? b ? a
Solution
(a) ?2x2y ? 6xz ? 12xy ? 36z ? ?2(x2y ? 3xz ?
6xy ? 18z)
? ?2(x2y ? 3xz) ? (6xy ? 18z)
? ?2x(xy ? 3z) ? 6(xy ? 3z)
(b) ab ? 1 ? b ? a ? (ab ? a) ? (1 ? b)
? a(b ? 1) ? (b ? 1)