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![CHAPTER IV: Algebra
Special Products Formula, If a, b, c, d are constants, x, y, and z are variables then,
1. (x – y)(x + y) = x2
– y2
Ex. (2x – 3y)(2x + 3y) = (2x)2
– (3y)2
= 4x2
– 9y2
2. (x + y)2
= x2
+ 2xy + y2
Ex. (2x + 3y)2
= (2x)2
+ 2(2x)(3y) + (3y)2
= 4x2
+ 12xy +
9y2
3. (x – y)2
= x2
– 2xy + y2
Ex. (2x - 3y)2
= (2x)2
- 2(2x)(3y) + (3y)2
= 4x2
- 12xy + 9y2
4. (x + y)(x2
– xy + y2
) = x3
+ y3
Ex. (2x + 3y)(4x2
- 6xy + 9y2
) = (2x)3
+ (3y)3
= 8x3
+ 27y3
5. (x – y)(x2
+ xy + y2
) = x3
– y3
Ex. (2x - 3y)(4x2
+ 6xy + 9y2
) = (2x)3
- (3y)3
= 8x3
- 27y3
6. (x + y)3
= x3
+ 3x2
y + 3xy2
+ y3
Ex. (2x + 3y)3
= (2x)3
+ 3(2x)2
(3y) + 3(2x)(3y)2
+ (3y)3
= 8x3
+ 36x2
y + 54xy2
+ 27x3
7. (x – y)3
= x3
– 3x2
y + 3xy2
– y3
Ex. (2x - 3y)3
= (2x)3
- 3(2x)2
(3y) + 3(2x)(3y)2
- (3y)3
= 8x3
- 36x2
y + 54xy2
- 27x3
8. (x + y + z)2
= x2
+ y2
+ z2
+ 2(xy + xz + yz)
Ex. (2x - 3y + 4z)2
= (2x)2
+ (-3y)2
+ (4z)2
+ 2[(2x)(-3y) + (2x)(4z) –(3y)(4z)
= (2x)2
+ (-3y)2
+ (4z)2
+ 2(-6xy + 8xz – 12yz)
= 4x2
+ 9y2
+ 16z2
+ 2(-6xy + 8xz – 12yz)
9. (ax+by)(cx+dy)=acx2
+(ad+bc)xy+bdy2
Ex. (2x + 3y)( 4x - 5y), a = 2, b = 3, c = 4, d = -5
= (2)(4)x2
+[(2)(-5)+(3)(4)]xy + (3)(-5)y2
= 8x2
+ (-10 + 12)xy – 15y2
= 8x2
+ 2xy – 15y2](https://image.slidesharecdn.com/c4l1ppt-250308050036-5223c4b4/75/Chapter-4-Learning-Outcome-1_Mathematics-for-Technologists-14-2048.jpg)
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- 2. Algebra Chapter IV 21st Century Mathematics forthe Industrial Technology Learners: Automotive Technology
- 3. CHAPTER IV: Algebra Learning Outcome1: Conceptualized thorough understanding of the essentials of Algebra;
- 4. The Number Line CHAPTERIV: Algebra
- 5. Algebra begins witha systematic study of the operations and rules of arithmetic. The operations of addition, subtraction, multiplication and division serves as a basis for all arithmetic calculations. In order to achieve generality, letters of the alphabet are used in algebra to represent numbers. A letter such as x, y, z, a, b or c can stand for a particular number (known or unknown), or it can stand for any number at all. A letter that represents an arbitrary number is called a variable and constant otherwise. The sum of a and b denoted by a + b; difference, a – b; product, a x b or a•b and quotient, a ÷ y or a/y for simpler notations. CHAPTER IV: Algebra
- 6. Rules in Operationswith signed numbers: 1. To add numbers of the same sign, add their absolute values and attach the common sign. Example 1. 10 + 2 = 12 Example 2. (-10) + (-2) = -12 2. To add unlike signed numbers, subtract the smaller number from the bigger number then attach the sign of the larger number. Example 3. 10 + (-2) = 8 Example 4. (-10) + 2 = -8 3. To subtract two signed numbers, changed the sign of the subtrahend then proceed to addition. (1 and 2) Example 5. 10 - 2 Example 6. 10 – (- 2) 10 + (-2) = 8 10 + 2 = 12 CHAPTER IV: Algebra
- 7. Rules in Operationswith signed numbers: 4. To multiply/divide liked signed numbers, multiply/divide their absolute value and attach a positive sign. Example 7. 10 x 2 = 20 Example 8. (-10) x (-2) = 20 Example 9. 10/2 = 5 Example 10. (-10)/(-2) = 5 5. To multiply/divide unlike signed numbers. Multiply/divide their absolute value and attach a negative sign. Example 11. (-10) x 2 = -20 Example 12. 10 x (-2) = -20 Example 13. 10/(-2) = -5 Example 14. (-10)/2 = -5 CHAPTER IV: Algebra
- 8. ORDER OF OPERATIONS ON REALNUMBER 1. First, perform the operations within each pair of grouping symbols (parentheses, brackets and braces) beginning with the innermost pair. 2. Next, perform multiplication and division as they occur from left to right. 3. Last, perform addition and subtraction as they occur from left to right. Note: PEMDAS (Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction) CHAPTER IV: Algebra Example: Solve 58÷ (4 x 5) + 32 Solution: 58 ÷ (4 x 5) + 32 As per the PEMDAS* rule, first, we must perform the operation which is in the parentheses. = 58 ÷ 20 + 32 Now perform the exponent/power operation = 58 ÷ 20 + 9 The division should be performed. = 2.9 + 9 And the last, addition. = 11.9 Therefore, 58 ÷ (4 x 5) + 32 = 11.9.
- 9. CHAPTER IV: Algebra Letus take a look at x + x + x + x + x, this can be written as 5x. The use of exponents provides an economy of notation for products; like, x∙x = x2 , and x⋅x⋅x = x3 . In general, if n is a positive integer, xn = x⋅x⋅x⋯x⋅x⋅x⋯x (n times) and x–n = 1/xn . In using the exponential notation xn , we refer to x as a base and n as the exponent, or the power to which the base is raised. When the exponent is negative, we must assume that the base is nonzero to avoid a zero in the denominator.
- 10. Laws of Exponents,for all x, y in the real number and not equal to zero n, m are integers. CHAPTER IV: Algebra
- 11. CHAPTER IV: Algebra Analgebraic expression is an expression formed from any combination of numbers and variables by using the operations of addition, subtraction, multiplication, division, exponentiation (raising to powers), or extraction of roots. A polynomial is an algebraic sum in which no variables appear in denominators or under radical signs, and all variables that do appear are raised only to positive-integer powers. The degree of a term in a polynomial is the sum of all the exponents of the literal coefficient in the term unless stated otherwise. For example, 3xy2 has degree 3 but of degree one in x and of degree 2 in y. The constant term has degree 0. The highest degree of all terms that appear with nonzero coefficients in a polynomial is called the degree of the polynomial. Let us look with this example, 4x6 – 2x2 – x – 3 where the degree of the polynomial is 6, the first term (4x6) has degree 6, the second term (– 2x2) is 2, the third term(-x) is one and the last term (-3) is of degree zero since it is a constant.
- 12. CHAPTER IV: Algebra Additionand Subtraction of Polynomials To find the sum of two or more polynomials, we use the associative and commutative properties of addition to group like terms together, and we combine the like terms by using the distributive property then arranged in ascending or descending order. Example 1. Horizontal: Addition Simplify (x3 + 5x2 – 2x - 8) + (2x3 + 3x – 6) + (–2x2 + x – 2) + (5x4 -3x2 + 4). Collect similar terms: 5x4 + (x3 + 2x3 ) + (5x2 -2x2 -3x2 ) + (-2x + 3x + x) + ( -8 -6 -2 + 4) Apply properties of real: 5x4 + 3x3 + 0 + 2x + ( -12) = 5x4 + 3x3 + 2x -12 qed. Example 2. Vertical: Subtraction Simplify (x3 + 5x2 – 2x – 8) – (-2x2 + x – 2) Solution: Simplify (x3 + 5x2 – 2x – 8) – (-2x2 + x – 2) → x3 + 5x2 – 2x – 8 → x3 + 5x2 – 2x – 8 –2x2 + x – 2 + + 2x2 – x + 2 x3 + 7x2 – 3x – 6 qed.
- 13. CHAPTER IV: Algebra Multiplicationof Polynomials To multiply a polynomial by a monomial, we use the distributive property. Thus, we multiply each term of the polynomial by the monomial and then simplify the resulting products by using the properties of exponents. Certain products of polynomials occur so often that it is useful to know the expanded forms by heart. The following list contains some of these special products formula. Note that, factoring is just the inverse of the special product. monomial by monomial: monomial by binomial: 3x3 yz(-5x2 yz4 )=3(-5)x3 x2 yyzz4 3xy(-5x2 y + 3xy)=3xy(-5x2 y)+3xy(3xy) = -15x5 y2 z5 =-15x3 y2 + 9x2 y2 monomial by trinomial: 3xy(-5x2 + 3y - z) = -15x3 y + 9xy2 – 3xyz binomial by binomial: (x – 4)(5 + y) = xy + 5x – 4 y – 20
- 14. CHAPTER IV: Algebra SpecialProducts Formula, If a, b, c, d are constants, x, y, and z are variables then, 1. (x – y)(x + y) = x2 – y2 Ex. (2x – 3y)(2x + 3y) = (2x)2 – (3y)2 = 4x2 – 9y2 2. (x + y)2 = x2 + 2xy + y2 Ex. (2x + 3y)2 = (2x)2 + 2(2x)(3y) + (3y)2 = 4x2 + 12xy + 9y2 3. (x – y)2 = x2 – 2xy + y2 Ex. (2x - 3y)2 = (2x)2 - 2(2x)(3y) + (3y)2 = 4x2 - 12xy + 9y2 4. (x + y)(x2 – xy + y2 ) = x3 + y3 Ex. (2x + 3y)(4x2 - 6xy + 9y2 ) = (2x)3 + (3y)3 = 8x3 + 27y3 5. (x – y)(x2 + xy + y2 ) = x3 – y3 Ex. (2x - 3y)(4x2 + 6xy + 9y2 ) = (2x)3 - (3y)3 = 8x3 - 27y3 6. (x + y)3 = x3 + 3x2 y + 3xy2 + y3 Ex. (2x + 3y)3 = (2x)3 + 3(2x)2 (3y) + 3(2x)(3y)2 + (3y)3 = 8x3 + 36x2 y + 54xy2 + 27x3 7. (x – y)3 = x3 – 3x2 y + 3xy2 – y3 Ex. (2x - 3y)3 = (2x)3 - 3(2x)2 (3y) + 3(2x)(3y)2 - (3y)3 = 8x3 - 36x2 y + 54xy2 - 27x3 8. (x + y + z)2 = x2 + y2 + z2 + 2(xy + xz + yz) Ex. (2x - 3y + 4z)2 = (2x)2 + (-3y)2 + (4z)2 + 2[(2x)(-3y) + (2x)(4z) –(3y)(4z) = (2x)2 + (-3y)2 + (4z)2 + 2(-6xy + 8xz – 12yz) = 4x2 + 9y2 + 16z2 + 2(-6xy + 8xz – 12yz) 9. (ax+by)(cx+dy)=acx2 +(ad+bc)xy+bdy2 Ex. (2x + 3y)( 4x - 5y), a = 2, b = 3, c = 4, d = -5 = (2)(4)x2 +[(2)(-5)+(3)(4)]xy + (3)(-5)y2 = 8x2 + (-10 + 12)xy – 15y2 = 8x2 + 2xy – 15y2
- 15. CHAPTER IV: Algebra Divisionof Polynomials To divide polynomials, make sure that the degree of the numerator is higher than the degree of the numerator and arranged in ascending order. For the numerator/dividend, the missing term will be fill in by 0 and the literal coefficient. In dividing polynomials, we can use factoring, long division or the synthetic division. Steps in long division of polynomials: 1. Divide the first term of the numerator by the first term of the denominator and put that in the answer. 2. Multiply the denominator by that answer, put that below the numerator. 3. Subtract to create a new polynomial. 4. Repeat steps 1-3, using the new polynomial if the degree of the polynomial is still higher than the denominator, repeat the process until the degree of the new polynomial is less than the degree of the numerator. This would now be the remainder.
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- 17. CHAPTER IV: Algebra Stepsrequired for Synthetic Division of a Polynomial: 1. To set up the problem, first, set the denominator equal to zero to find the number to put in the division box. Next, make sure the numerator is written in descending order and if any terms are missing you must use a zero to fill in the missing term, finally list only the coefficient in the division problem. 2. Once the problem is set up correctly, bring the leading coefficient (first number) straight down. 3. Multiply the number in the division box with the number you brought down and put the result in the next column. 4. Subtract the two numbers together and write the result in the bottom of the row. 5. Repeat steps 3 and 4 until you reach the end of the problem. 6. Write the final answer. The final answer is made up of the numbers in the bottom row with the last number being the remainder and the remainder must be written as a fraction. The variables or x’s start off one power less than the original denominator and go down one with each term.
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- 21. CHAPTER IV: Algebra ComplexFractions/Compound fractions Complex fraction is a single fraction with more than one fractional line. There are two methods used to solve complex fraction. Method 1 1. Create a single fraction in the numerator and denominator. 2. Apply the division rule of fractions by multiplying the numerator by the reciprocal or inverse of the denominator. 3. Simplify, if necessary. Method 2 1. Least Common Denominator (LCD) of all the denominators in the complex fractions. 2. Multiply this LCD to the numerator and denominator of the complex fraction. 3. Simplify, if necessary.
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