MODULE II
FOR MID – TERM PERIOD
Learning Objectives
At the end of Module II, the students are expected to be able to:
1. Enumerate and discuss the laws of exponents;
2. Simplify algebraic expressions with exponents;
3. Multiply polynomials;
4. Divide polynomials using the long method;
5. Divide polynomials using synthetic division;
6. Enumerate and explain the different expressions under special products;
and,
7. Simplify polynomials under special products.
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
MODULE II
2.1 LAWS OF EXPONENTS
To be able to multiply algebraic expression effectively, it is necessary that
understanding about exponents and their operations must be given primary
consideration. In the expression where a and n represent real numbers, a is called the
base while n is called the exponent. If n is a positive integer, then as a definition
An = a • a • a • . . . •a • a (n times).
A number raised to an exponent means the number is to be multiplied by itself
n – times. For example 26 = 2 • 2 • 2 • 2 • 2 • 2 = 64. This definition can be
extended to values of n equal to zero or a negative integer, provided a ≠ 0 for a 0 =
1. The expression 00 is undefined. For negative integers and a ≠ 0, define
1
a- n =
𝑎ⁿ
.
Illustrations:
1. 30 = 1
2. (- 5) -2 =
1
(−5)²
=
1
25
For polynomials in two or more variables, the degree is the largest sum of
exponents of the variables in any one term. The degree of
2x2y3 + x2y2z2
is 6 .
Laws of exponents
A. Multiplication Law: aman = a m+n
When multiplying terms with the same base, the exponents are
added and the base is retained.
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
Illustrations:
1. 23 • 24 = 2 3+4 = 27
2. x2 • x5 = x 2+5 = x7
3. n • n5 • n3 = n 1+ 5 + 3 = n 9
4. (s + t) k (s + t )4 = (s + t) k + 4
B. Power of a Power: (am) n = a mn
When a term raised to an exponent is raised to another exponent,
the exponents are multiplied and the base is retained.
Illustrations:
1. (23)4 = 23x4 = 212
2. (x4)2 = x4x2 = x8
3. (a5)m = a5m
C. Power of a Product: (ab)n = an bn
When a product is being raised to an exponent, each of the factors is
raised to that exponent. That is, the exponent of a product may be
distributed to each factor.
Illustrations:
1. (2 • 3)4 = 24 • 34 = 16 • 81 = 1, 296
2. (2m2n5)3 = 23 (m2)3 (n5)3 = 8m6n15
3. (5x2ynz)3 = 53 (x2)3 (yn)3 z3 = 125x 2•3 y n• 3 z3 = 125x6 y3n z3
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
D. Division Law:
𝑎ⁿ
= an-m if n ≥ m or
𝑎ᵐ
1
𝑎ᵐ ‾ ⁿ
if n ˂ m, and a ≠ 0
When dividing terms with the same base, if the exponent of the numerator is
greater than the exponent of the denominator, the base is retained in the numerator;
otherwise the base is retained in the denominator. The new exponent is equal to the
difference of the exponents of the numerator and denominator.
Illustrations:
1.
2.
3.
3⁵
3²
𝑎⁴
𝑎¹²
𝑥ᵏ
𝑥
= 3 5-2 = 33 = 27
=
1
𝑎¹²‾ ⁴
=
1
𝑎⁸
= x k-1
𝑎
𝑎ⁿ
E. Power of a Quotient: ( 𝑏 )ⁿ =
where b ≠ 0
𝑏ⁿ
When quotient is raised to an exponent, both numerator and
denominator
are raised to that exponent. That is, the exponent of a quotient may be
distributed to the numerator and denominator.
5
5⁴
3
3⁴
1. ( )⁴ =
2. (
2𝑥 2
𝑦3
)³ =
𝑎2 𝑏3
=
625
81
(2𝑥² )³
3. (
)ᵐ =
𝑐ⁿ 𝑑 4
(𝑦 3 )³
=
2³ (𝑥 2 )³
𝑦⁹
(𝑎2 )ᵐ (𝑏3 )ᵐ
(𝑐ⁿ)ᵐ (𝑑 4
)ᵐ
=
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
=
8𝑥⁶
𝑦⁹
𝑎²ᵐ 𝑏³ᵐ
𝑐ᵐⁿ 𝑑⁴ᵐ
Algebraic expressions are also considered simplified when there are
nonnegative exponents. Zero and negative exponents can be eliminated in any
algebraic expression using the laws of exponents.
Illustrations:
Simplify the following expressions:
1
1. – 52 x0 y -3 = - 25 • 1 • 𝑦³ = −
25
𝑦³
2. (a3b-2) -2 = (a3) -2 (b-2) -2 = a - 6 • b4 =
3.
𝑥‾ ³(𝑦𝑧 3 )‾ ²
=
𝑥‾ ²𝑦‾ ⁵ 𝑧
=
𝑥‾ ³𝑦‾ ²(𝑧 3 )‾ ²
(𝑥‾2 )3 (𝑦‾⁵)3 𝑧
x -3-(-6) • y -2-(-15) •
=
1
𝑧¹‾⁽‾⁶⁾
Simplify the following:
1.
2.
3‾²
𝑚‾²
𝑚²𝑛⁰
+
3²
𝑎‾²𝑏⁰𝑐‾¹
𝑎‾¹𝑐³
3. (
𝑥 2 𝑧‾3
𝑦‾²
)‾¹
4. (x-2 y3) (
20 𝑦
𝑥𝑦‾¹
)
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
1
𝑎⁶
𝑥‾ ³𝑦‾ ²𝑧‾ ⁶
𝑥‾ ⁶𝑦‾ ¹⁵𝑧
𝑏⁴
𝑎⁶
• b4 =
=
𝑥‾ ³
𝑥‾ ⁶
= x3 • y¹³ = •
•
1
𝑧⁷
𝑦‾ ²
𝑦‾ ¹⁵
=
•
𝑧‾ ⁶
𝑧
𝑥³𝑦¹³
𝑧⁷
5. a2 • a7
11. (a5)4
6. 3x4 • 2x3
12. (3b)5
7. 3x3 • x4 • x5
13. (5c)3
8. (-3)4 • (-3)5
14. (3a)4
9. y13 • y11
15. ((2a3)5
10. (23)4
16. (ar • as)t
2.2 MULTIPLICATION OF POLYNOMIALS
In multiplying polynomials, we have to consider the associative, distributive,
as well as the laws of exponents discussed previously.
I. Multiplying Monomial over Binomial (Distributive property over
addition)
x(ax + a) = ax2 + ax
Examples:
Find the product of
a. x ( x + 3) = x2 + 3x
b. a (a2 – 2b2) = a3 - 2ab2
c. – 3x2y (5x2y3 – 6) = - 15x4y4 + 18x2y
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
II. Multiplying monomials over Trinomials (Distributive property over
addition)
x(ax2 + ax + a) = ax3 + ax2 + ax
Examples:
Find the product of
a. x(x2 + 3x + 4) = x3 + 3x2 + 4x
b. 2x2(3x2 - 4x + 2) = 6x4 - 8x3 + 4x2
c. – a2m3x(4a3b2 - 6m2n + 7x2y) = - 4a5b2m3x + 6a2m5nx –
7a2m3x3y (Distributive)
III. Multiplying Monomials over Polynomials of Four terms
x(x3 + 2x2 + 5x – 1) = x4 + 2x3 + 5x2 - x
Examples:
Find the product of
a. -3bc (2b3c3 + 3b2c2 - bc -1) = -6b4c4 – 9b3c3 + 3b2c2 + 3bc
b. 2a2 (3a3 - 4a2 + 5a) = 6a5 - 8a4 + 10a3
c. –x (- 5a + 4b - 3c) = 5ax – 4bx + 3cx (Distributive)
IV. Multiplying Two Binomials
Using FOIL Method (for first term, outer term, inner term, and last term)
(ax + b) (cx + d) = acx2 + adx + bcx + bd
= acx2 + (ad + bc) x + bd
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
Examples:
Find the product of
a. ( 2x + 3) (3x + 5) = (2x)(3x) + (2x)(5) + (3)(3x) + (3)(5)
= 6x2 + 10x + 9x + 15 (combining
similar terms.)
= 6x2 + 19x + 15
This method is also called horizontal solution and combining like
terms. It
can also be solved vertically, as shown below
2x + 3
3x + 5
= 6x2 + 10x + 9x + 15
= 6x2 + 19x + 15
b. (4x – 2) (7x + 3) = (4x)(7x) + (4x)(3) - (2)(7x) - (2)(3)
= 28x2 + 12x - 14x - 6
= 28x2 - 2x – 6
c. (x – 6) (2x – 7) = (x)(2x) – (x)(7) - (6)(2x) + (6)(7)
= 2x2 – 7x - 12x + 42
= 2x2 - 19x + 42
V. Multiplying Binomial over Trinomial
(ax +b) (cx2 + dx + e) = (ax)(cx2) + (ax)(dx) + (ax)(e) + (b)(cx2) + (b)(dx) +
(b)(e)
= acx3 + adx2 + aex + bcx2 + bdx + be
(No like terms.)
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
Examples:
Find the product of the following:
a. (2x – 4) (x2 – 6x + 7) = (2x)(x2) – (2x)(6x) + (2x)(7) –
(4)(x2) + (4)(6x) – (4)(7)
= 2x3 – 12x2 + 14x – 4x2 + 24x – 28
= 2x3 – 12x2 – 4x2 + 38x – 28
= 2x3 – 16x2 + 38x – 28
b. (3x – 5) (2x2 – 4x – 1) = (3x)(2x2) – (3x)(4x) – (3x)(1) –
(5)(2x2) + (5)(4x) + (5)(1)
= 6x3 – 12x2 – 3x – 10x2 + 20x + 5
= 6x3 – 22x2 – 17x + 5
VI. MULTIPLYING TRINOMIAL OVER TRINOMIAL
(mx2 + nx + p) (ax2 + bx + c) = (mx2)(ax2) + (mx2)(bx) + (mx2)(c) +
(nx)(ax2) + (nx)(bx) + (nx)(c) +
(p)(ax2) + (p)(bx) + (p)(c)
= amx4 + bmx3 + cmx2 + anx3 + bnx2 +
cnx + apx2 + bpx + cp (No like
terms.)
Examples:
Find the product of the following:
1. (x2 + 2x – 1) (2x2 – 3x + 4) = (x2)(2x2) - (x2)(3x) + (x2)(4)
+
(2x)(2x2) – (2x)(3x)
+ (2x)(4) – (1)(2x2) + (1)(3x)
– (1)(4)
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
= 2x4 – 3x3 + 4x2 + 4x3 – 6x2 +
8x – 2x2 + 3x – 4
= 2x4 + x3 – 4x2 + 11x – 4
2. (3x2 – 4x – 5) (6x2 + 5x – 2) = (3x2)(6x2) + (3x2)(5x) –
(3x2)(2) – (4x)(6x2) –
(4x)(5x) + (4x)(2) –
(5)(6x2) – (5)(5x) + (5)(2)
= 18x4 + 15x3 – 6x2 – 24x3 –
20x2 + 8x
– 30x2 – 25x + 10
= 18x4 – 9x3 – 56x2 – 17x +
10
Solve the following:
1. (xa – yb) (xm + yn)
2. (x + 1 + x2) (x3 – 2x + 1)
3. x7 (x6 – 7x4 + 5x2 – 20)
4. (x2 – 2x + 3) (x2 – 2x + 3)
5. (x + 1) (x-1) (x + 1)
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
6. 2ax (3ax2 – bx + 4c)
7. (y + 3) (y2 + 3y – 2)
9. (5m2n – 2ab + 3) (2m3n2 + 3a2b2 – 7)
10. (3u2v + 4y2z) ( 2u3v2 – y2z2 – 8)
2.3 DIVISION OF POLYNOMIALS
We have learned that in division the following terms are used: dividend which
is the expression being divided; divisor is the number or expression that divides the
dividend; and, the quotient which is the result of division. In the example below
12 ÷ 4 = 3 or in fraction form
12
4
= 3,
The number 12 is the dividend, number 4 is the divisor, and the 3 is the quotient.
In dividing polynomials, the nature of the divisor will determine the method in
carrying out the division. In all the illustrations, the divisors are assumed to be not
equal to zero – since division by zero is undefined. If the divisor is a monomial, write
the division in fraction form, then divide each term of the numerator by the
denominator, that is, distribute the denominator to each term in the numerator and
simplify each resulting fraction using the law of exponents.
Illustration:
1. 6x ÷ 3 = 2x, or
6𝑥
3
= 2x
2. (24x – 18y) ÷ 6xy =
24𝑥
18𝑦
= 4y – 3x
6𝑥𝑦 6𝑥𝑦
3. (– 24a3b3 + 32a2b4 – 16a5b3) ÷ 8a2b2 =
− 24𝑎³𝑏³
8𝑎²𝑏²
+
32𝑎²𝑏⁴
8𝑎²𝑏²
-
= -3ab + 4b2 – 2a3b
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
16𝑎⁵𝑏³
8𝑎²𝑏²
A. Divide the following:
1.
2.
3.
21x³ − 35x
7x
2𝑥−4
2
x(x+3)−2(x+3)
x+3
Long division for polynomials works in much the same way. The method as
well as the form is similar to the numerical long division. The following are the steps
in long division:
1. Arrange the terms of the dividend and divisor in descending powers of one
of the variables. If some powers are missing, write zero as its coefficients.
2. Divide the first term of the dividend by the first term of the divisor. This is
the first term of the quotient.
3. Multiply the entire divisor by the first term of the quotient and subtract the
product from the entire dividend. The difference obtained is now the new
dividend.
4. Apply steps 1 to 3 to this new dividend.
5. Continue doing step 4 until the difference in step 3 is zero or has degree
less than the degree of the divisor. The non – zero difference with degree
less than the divisor is called the remainder.
Illustrations:
1. Divide x2 – 9x – 10 by x + 1
First, set up the division:
For the moment, ignore the other terms and look
just at the leading x of the divisor and the
leading x2of the dividend.
If the leading x2 inside is divided by the leading x in
front, the quotient x appears. Write the x on top:
Now multiply the top x with the leading x. The
ISHRM 2014
Learning product
Material is x². Write this product underneath:
College Algebra
MT 111/MT 211 Module II
Then multiply the x (on top) by the 1 (on the "side"),
and carry the 1x underneath:
next draw the "equals" bar, so I can do the
subtraction.
To subtract the polynomials, change all the signs in
the second line...
...and then add down. The first term (the x2) will
cancel out:
remember to carry down that last term, the
"subtract ten", from the dividend:
Now look at the x from the divisor and the new
leading term, the –10x, in the bottom line of the
division. Divide the –10x by the x, to get – 10 and
write it on top:
Multiply the –10 (on top) by the leading x(on the
"side"), and carry the –10x to the bottom:
...and I'll multiply the –10 (on top) by the 1 (on the
"side"), and carry the –10 to the bottom:
Draw the equals bar, and change the signs on all
the terms in the bottom row:
Then, finally, add down:
𝒙²+𝟗𝒙+𝟒
2. Simplify
𝒙+𝟕
This can be done in either of two ways: Factor the quadratic and then cancel
the common factor, like this:
Also by the use long division:
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
B. Divide the following:
1. (2x2 – 5x - 6) ÷ (2x – 1)
2. (x3 – y3) ÷ (x – y)
3. (x2 + 2xy + y2) ÷ (x + y)
4. (3x4 + 2x2 – 6x + 1) ÷ (x+1)
5. (x3 – 6x2 + 5) ÷ (x2 + 3x – 2)
6. (x3 + 3x2 – 2x + 1) ÷ (x – 2)
2.4 SYNTHETIC DIVISION.
In dividing polynomials in one variable, say x, a more convenient method
called synthetic division can be used when the divisor of the form x – a, where a is a
constant. Suppose 5x3 – 11x2 – 14x – 10 is divided by x – 3. Using long division the
solution is:
5x2 + 4x
- 2
X – 3 5x3 - 11x2 - 14x - 10
5x3 - 15x2
4x2 - 14x
4x2 - 12x
-2x - 10
-2x + 6
-16
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
If in the solution the powers of x are removed and only the numerical
coefficients
are retained, then the solution becomes
1–3
5 4
5 - 11
5 - 15
4
4
-2
- 14 - 10
-
14
12
2 - 10
2
6
- 16
Moreover, since the coefficient of the first term of the divisor is 1, then the
coefficient of the first term of each dividend is equal to the coefficient of the
succeeding term of the quotient. Also, the coefficient of the first term of the next
partial product is equal to the coefficient of the first term of each new dividend.
Hence, removing first terms will result to the above solution.
Moving the numbers up so that the solution part is arranged in three rows and
writing the coefficient of the quotient on the bottom row together with the remainder
and placing the constant of the divisor to the right will result to
Dividend coefficient
5
5
- 11
- 14
- 10
- 15
- 12
+6
4
- 2
Quotient coefficients
| -3
- 16
Remainder
Observe that the numbers in the second row are the products of the previous
terms
in the third row and -3, while the numbers in the third row are the differences of the
first and second row. If instead of multiplying by -3, multiply by 3, the third row can
be obtained by adding the first two rows. Thus, the computation will be
Dividend coefficient
5
5
- 11
- 14
-10
+ 15
+ 12
- 6
4
- 2
Quotient coefficient
|+3
- 16
Remainder
The method of solution shown above is referred to as synthetic division. It is
simpler and less prone to errors.
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
Steps in using synthetic division of a polynomial in x by x - a.
1. Write the terms of the polynomial according to descending powers of x,
supplying each missing power with zero as corresponding coefficient.
2. Write the coefficients of the polynomial in a horizontal row. This will be
the first row.
3. Bring down the first coefficient to the bottom row.
4. Multiply the third row by a and write the product in the second row of the
next column.
5. Add the first and the second row entries of the column and write the sum at
the third row of the same columns.
6. Repeat steps 4 and 5 up to the last column.
7. The last number in the third row is the remainder. The preceding numbers
are the coefficients of the successive terms of the quotient which is a
polynomial of degree one less than the dividend.
Illustrations:
Divide the following using synthetic division:
1. (x3 – 4x2 – 5x – 6) + (x + 3)
Solution:
1
1
-4
-5
-6
-3
21
- 48
-7
16
- 54
|-3
Hence, the quotient is x2 – 7x + 16 and the remainder is – 54.
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
2. (x4 – 2x3 + 14x2 – 30x – 63) ÷ (x – 3)
3. (3x4 – 8x2 – 11x + 1) ÷ (x – 2)
1
4. (2x3 – 7x2 – 5x + 4) ÷ ( 𝑥 − 2 )
MT 111/MT 211 Module II
WORK PROJECT / LEARNING ACTIVITY NO. 1, MODULE II
Name: ___________________________________ Yr/Sec: ____________ Date:
____________
I. OPERATION ON THE LAWS OF EXPONENTS
A. Simplify/evaluate the following.
𝑥⁴𝑦⁶𝑧⁷
1. (xy)3
6.
2. (2x)4
7.
3. (-3y)2
8. (x2y3)6 •
4
4. ( ) ³
9.
5. (x3)⁴
10.
𝑥
𝑥¹⁰𝑦³𝑧⁸
(
𝑎3 𝑏2
𝑐5
)² (
−64𝑥⁵𝑦⁷
8𝑥⁷𝑦⁵
𝑐8
𝑏
)
𝑚⁴𝑛⁴
𝑥¹²𝑦¹⁸
•
𝑥²
𝑦²
(20𝑟 2 𝑠 3 𝑡 4 )(2𝑟 2 𝑠 2 𝑡)
(−4𝑟𝑠𝑡)(3𝑟𝑠 4 𝑡 2 )
B. Simplify the following (no zero or negative exponent in the final answers):
1.
2.
𝑎‾²𝑏⁰𝑐‾¹
𝑎‾¹𝑐³
(
𝑥 2 𝑧‾3
𝑦‾2
) ‾¹
3. (x-2y3) (
4.
(
5.
(
𝑢‾3
𝑣4
𝑥𝑦‾1
)
)² (w10)0
𝑎2𝑏0
𝑐5
20 𝑦
)²(
𝑐3
𝑏
)³
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
WORK PROJECT / LEARNING ACTIVITY NO. 2, MODULE II
Name: ___________________________________ Yr/Sec: ____________ Date:
____________
II. MULTIPLICATION AND DIVISION OF POLYNOMIALS
Perform the indicated operations and simplify the final answer.
1. (-x + 1) (x + 3)
2. (2x + y) (x – 2y)
3. (5w + 2) (-w2 + 2w – 3)
4. (a + 4b) (2b – a) (3a – b)
5. (m – 2n + 5) (m + 2n + 5)
6. (5a3b2c + 20a2b – 10ac) ÷ 5abc
7. (4x3 – 4x2 – 7x + 6) ÷ (2x – 3)
8. (x5 + 2x4 – 6x3 + x2 – 5x + 1) ÷ (3x + 1)
9. (2x5 + 5x4 – 4x3 + 8x2 + 1) (2x2 – x + 1)
10. (2x3 – 11x2y + 13xy2 – 4y3) ÷ (x – 4y)
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
WORK PROJECT / LEARNING ACTIVITY NO. 3, MODULE II
Name: ___________________________________ Yr/Sec: ____________ Date:
____________
III. SYNTHETIC DIVISION
Divide the following using synthetic division: (Indicate your quotient and
remainder.)
1. (2x3 – 5x2 – 15x + 4) ÷ (x – 4)
2. (2x4 + 5x3 + 10x + 3) ÷ (x + 1)
3. (x5 + 10x2 + 8) ÷ (x + 2)
4. (x4 – x2 + 1) ÷ (x – 1)
1
5. (2x3 + 5x2 – 5x + 1) (x - 2 )
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
2.5 SPECIAL PRODUCTS
There are instances when the expression to be multiplied appears to have
certain similarities or pattern. They are called special products. Mastery of these
products leads to a faster way of getting the product of polynomials which can be
verified by performing the actual multiplication. The following are some special
product formulas:
SP-A. Product of the Sum and Difference of two Binomials
Where: (x + y) (x - y) = x2 - y2 (Use distributive to check
solution.)
Rule: Square of the first terms minus the square of the
second terms
Simplify the following:
1. (3x – 7) (3x + 7) = 9x2 – 49
2. (5a2 + 2b) (5a2 – 2b) = 20a4 – 4b2
3. (3u + v2) (3u – v2)
4. (4m2n) – (7x2y) (4m2n + 7x2y)
5. (3s – 2t) (3s + 2t)
6. (7x2y2 + 5y3z3) (7x2y2 – 5y3z3)
7. (10ab – 9bc) (10ab + 9bc)
8. (5x3y2z – 7ab2c3) (5x3y2z – 7ab2c3)
SP-B. The Square of the Binomial
Where: (a + b)2 = (a + b) (a + b) = a2 + ab + ab + b2 (add like terms)
= a2 + 2ab + b2 (Perfect Trinomial Square)
Rule: Square of the first term, twice the product of the first and
the second terms, and square of the second term.
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
Simplify the following:
1. (m + n)2 = m2 + 2[(m)(n)] + n2 = m2 + 2mn + n2
2. (3x + 2y)2 = 9x2 + 2[(3x)(2y)] + 4y2 = 9x2 + 2(6xy) + 4y2
= 9x2 + 12xy + 4y2
3. (m - n)2 = m2 - 2[(m)(n)] + n2 = m2 – 2mn + n2
4. (2m – 4n)2 = 4m2 – 2[(2m)(4n)] + 16n2 = 4m2 – 2(8mn) +
16n2
= 4m2 – 16mn + 16n2
5. (m2n + x2y)2
6. (5a2 – 3b)2
7. (3yz2 – 5mn2)2
8. (9x2 – 7y2)2
9. (6a3b2 + 5c2d)2
1
10. ( 2 x2 +
1
4
𝑦²)²
SP-C. The Cube of a Binomial
Where: (m + n)3 = (m + n)(m + n)(m + n)
= [(m2) + 2[(m)(n)] + (n2)][(m + n)]
= m3 + 2m2n + mn2 + m2n + 2mn2 + n3
(Combine like terms.)
= m3 + 3m2n + 3mn2 + n3
(m – n)3 = (m – n)(m – n)(m – n) = (m2 – mn – mn + n2)(m
– n)
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
= (m2 – 2mn + n2) (m – n)
= m3 – 2m2n + mn2 – m2n + 2mn2 – n3
(Combine like terms.)
= m3 – 3m2n + 3mn2 – n3
Rule: Cube of the first term, thrice the product of the first term
squared and the second term, thrice the product of the
first term and square of the second term, and cube of the
second term.
Simplify the following:
1. (2m – 3n)3 = (2m)3 – 3(2m)2(3n) + 3(2m)(3n)2 – (3n)3
= 8m3 – 3(4m2)(3n) + 3(2m)(6n2) – 9n3
= 8m3 – 3(12m2n) + 3(12mn2) – 9n3
= 8m3 – 36m2n + 36mn2 – 9n3
2. (3x + 2y)3
3. ( 4a + 6b)3
4. (5 – 2a)3
5. (2x2 – 3)3
6. (3m2 + 5n2)3
7. (4y + 2z)3
8. (3x – 2a)3
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
SP-D. The Square of a Trinomial
Where: (m + n + p)2 = (m + n + p) (m + n + p)
= m2 + n2 + p2 + 2mn + 2mp + 2np
Rule: Square of the first term, square of the second term,
square of the third term, then twice the product of the
first and second terms, twice the product of the first and
the third terms, and twice the product of the second and
the third terms.
Simplify the following:
1. (3a + 2b + c)2 = (3a)2 + (2b)2 + (c)2 + 2(3a)(2b) + 2(3a)(c)
+ 2(2b)(c)
= 9a2 + 4b2 + c2 + 2(6ab) + 2(3ac) +
2(2bc)
= 9a2 + 4b2 + c2 + 12ab + 6ac + 4bc
2. (2x + 2y – 2z)2
3. (4m – 3n + 2p)2
4. (b + 2c + 2d)2
5. (3x – 2y – z)2
6. (u + v + 2w)2
7. (b – 3c + 2d)2
8. (3s + 2t + 2u)2
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
SP-E. The Cube of a Trinomial
Where: (m + n + p)3 = (m + n + p)2 (m + n + p)
= (m2 + mn + mp+ mn + n2 + np + mp + np + p2)(m + n+ p)
= (m2 + n2 + p2 + 2mn + 2mp+ 2np) (m + n + p)
= m3 + mn2 + mp2 + 2m2n + 2m2p +2mnp + m2n + n3 + np2 +
2mn2 + 2mnp + 2n2p + m2p + n2p + p3 + 2mnp + 2mp2 + 2np2
= m3 + n3 + p3 + 3m2n + 3m2p + 3mn2 + 3n2p + 3mp2 + 3np2 +
6mnp
Rule: Cube of the first, plus second and third terms, thrice the
product of the first term squared and the second term,
plus thrice the product of the first term squared and the
third term, plus thrice the product of the first term and
the square of the second term, plus thrice the product of
the second term squared and the third term, plus thrice
the product of the first term and square of the third term,
plus thrice the product of the second term and the
square of the third term, and plus the sixth times the
product of the first, the second, and the third term.
Simplify the following
1. (2x + y - 2z)3
2. (3x - 4y + 5z)3
3. (3x – 5y + 2)3
4. (2m + 2n + 2p)3
5. (4a – 3b – 2c)3
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
SP-F. Product Resulting to the Sum and Difference of Two Cubes
Where: (a + b) (a2 – ab + b2) = a3 + b3
and: (a – b) (a2 + ab + b2) = a3 + b3
Rule: Cube of the first terms and cube of the last terms.
Simplify the following:
1. (2a + 2b) (4a2 – 4ab + 4b2) = 8a3 + 8b3
Check:
4a2 - 4ab + 4b2
2a + 2b
8a3 - 8a2b + 8ab2
8a2b - 8ab2 + 8b3
8a3
2. (3x – 4y) ( 9x2 + 12xy + 16y2)
3. (4m + 2n) (16m2 - 8mn + 4n2)
4. (2a + 5b) (4a2 – 10ab + 25b2)
5. (5 – b) (25 + 5b + b2)
6. (3a + 2) (9a2 – 6a + 4)
7. (4x – 5y) (16x2 + 20xy + 25y2)
8. (6s – 7t) (36s2 + 42st + 49t2)
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
+ 8b3
= 8a3 + 8b3
WORK PROJECT / LEARNING ACTIVITY NO. 4, MODULE II
Name: ___________________________________ Yr/Sec: ____________ Date:
____________
IV. SPECIAL PRODUCT.
Use special product to simplify the following:
1. (4u + 2v2) (4u – 2v2)
2. (3m2n) – (5x2y) (3m2n + 5x2y)
3. (7s – 4t) (7s + 4t)
4. (5x2y2 + 6y3z3) (5x2y2 – 6y3z3)
5. (2x3y2z – 7ab2c3) (2x3y2z – 7ab2c3)
6. (2yz2 – 6mn2)2
7. (7x2 – 5y2)2
8. (3a3b2 + 7c2d)2
1
9. ( 4 𝑎2 +
3
4
𝑏²)²
10. (2x + 5y)3
11. ( 5a + 3b)3
12. (8 – 3a)3
13. (5x2 – 2)3
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II
WORK PROJECT / LEARNING ACTIVITY NO. 5, MODULE II
Name: ___________________________________ Yr/Sec: ____________ Date:
____________
IV. SPECIAL PRODUCT.
Use special product to simplify the following:
1. (2b + 2c + 2d)2
2. (4x – 2y – 3z)2
3. (u + 3v + 2w)2
5. (2b – 3c + 2d)2
6. (4s + 2t + 2u)2
7. (3b + c + 2d)2
8. (3x – 3y – z)2
9. (x + 3y - z)3
10. (5x - 4y + 3z)3
11. (2x – 3y + 2)3
12. (3m + 3n + 3p)3
13. (5a – 4b – 3c)3
14. (4 – c) (16 + 4c + c2)
15. (2a + 3) (4a2 – 6a + 9)
16. (5x – 6y) (25x2 + 30xy + 36y2)
17. (5s – 9t) (25s2 + 45st + 81t2)
ISHRM 2014
Learning Material
College Algebra
MT 111/MT 211 Module II