Polynomials

POLYNOMIALS

Polynomial is an algebraic expression that is a sum of terms contains only variables with whole number exponents and integer coefficients. It also contains four fundamental operations applied on polynomials.

Definitions of Basic Terms in Polynomials

1. A constant is a symbol that assumes one specific value.

2. A variable is a symbol that assumes many values.

3. An algebraic expression, or simply an expression, is a collection of constants and variables involving atleast one of the basic operations in mathematics.

4. A term is an expression preceeded by plus or minus sign.

Example:

One Term: r2 ; (a + b)4

Two Terms: ab- 3; 2-r2

5. A monomial is a term involving only the product of a real number and variables with non-negative integral exponents.

Example:

Monomial: 6, 3b, 15xyz2, x2y4

Not Monomial: x+2 ; x/y ; ½

6. A polynomial is a sum of finite numbers of monomials. The general polynomial in one variable of degree n is of the form

Anxn + … + a1x+a0

A binomial is a polynomial consisting of the sum of two monomials. Example: x2 – x.

A trinomial is a polynomial consisting the sum of 3 monomials. Example: x3 – x2 + 7.

7. The degree of polynomial in x is the greatest exponent occurring in the variable x.

For example x4- x 7and 2-x3 – x4 have the degrees 7 and 4, respectively. A constant has 0 degree. The degree of (2x)0 is 0, while x20 is 1. The expressions √x and 1/x have no degree since they are not polynomials.

8. If a monomial is expressed as a product of two or more symbols, each of the symbols is called the coefficient of the rest of the product. In 2xy, 2 is called a numerical coefficient, and xy is called the literal coefficient.

9. Two monomials (or two terms) are similar if they have the same literal coefficient. Example: 2x and -3x are similar terms; -3x and 4xy are dissimilar terms.

ADDITION OF EXPRESSIONS AND POLYNOMIALS

Rule 1: To add two or more monomials with the same literal coefficient, add only their numerical coefficients and affix the literal coefficient. For Example:

A. -8x +15x = (-8+15) x = 7x

B. -8y – ( -15y) = (-8 +15)y = 7y

Rule 2: To add two or more polynomials, add similar or like terms together.

Example:

A. 3x2 – 4x -4y, 7x2 – 2y – 2 and -4x2 + x – y – 7

B. 8x2 – 7x – 2y, 5x2 – 6x – 15y, and -4x2 + 11x – 9y

Solution: We write the polynomials in horizontal form and perform the addition.

3x2 – 4x - 4y 7x2 – 2y – 2 -4x2 + x – y – 7 ________________ 6x2 -3x – 7y – 9Example2. a. Subtract 4x – y – 3 from 2x – y -4.b. Subtract 4x + 3y +5 from the sum of -3x – y +5 and x +8y -3 c. Subtract the sum of 2x – 9y – 8 and 6x +4y from 2x -5y – 7d. Subtract the sum of 12x – 14y and 6y -9 from the sum of 7x-2y +3 and 4x -5y – 8.Solutions:a. 2x – y- 4 2x- y – 4 b. -3x– y + 5- (4x – y – 3) -4x + y + 3 + x +8y -3 -2x -1 -2x +7y +2 -2x +7y +2

-(4x +3y +5) -4x -3y -5 (+) -6x +4y -3

SYMBOLS OF GROUPING EXPRESSIONS

Addition of algebraic expressions frequently involves the symbols of grouping such as parentheses (), Brackets [], and braces {}.

Rule 3. To move a grouping symbol preceded by a

i) minus sign, change the sign of each terms;

ii.) plus sign, no further change is done;

iii.) factor, use the distributive law.

Example: Perform the indicated operations

a. –(2x –y +10) + (4x – 3y) – 2(3x – 4y +6)

b. 4x – 2y – 5 - 2(8x - 7y) – (x – 4y – 1)

Solution:

(i) (ii) (iii)

– (2x –y +10) + (4x – 3y) – 2(3x – 4y +6)

= -2x + y - 10 + 4x – 3y - 6x + 8y – 12

= -2x +4x -6x +y -3y +8y – 10 -12 = -4x+ 6y – 22

Rule 4. When one symbol of grouping is within another symbol of grouping the innermost symbol must be removed first.

Example: Simplify the following

a. – {-2 –y – [3x – (4x + y – 3) – y] – 7}

b. – {-3x +4y – (7x – 8)} – {3x – [-4y - (x – 5)]}

c. – {9x – 6 – [3x – (4x – 5y – 7)] – 8y}

Solutions:

a. – {-2 –y – [3x – (4x + y – 3) – y] – 7} b. – {-3x +4y – (7x – 8)} – {3x – [-4y - (x – 5)]}

= - {- 2x – y – [3x – 4x – y + 3 – y] – 7} = – {-3x +4y – (7x – 8)} – {3x – [-4y - (x –-5)]}

= - {-2x – y – [-x – 2y +3] -7} = - {-4x +4y + 8} – {3x +4y + x -5}

= - {-2x-y +x +2y – 3 -7} = 4x – 4y – 8 – {4x + 4y – 5}

= - {-x +y -10} = 4x – 4y – 8 – 4x – 4y + 5

= x-y +10 = -8y - 3

Value of Polynomial

The value of Polynomial or expression in x is obtained by substituting a certain given value for x.

Example, the value of 2x3 – 4x2 –x – 5 at x is = -1

2(-1)3 – 4(-1)2 – (-1) – 5 = -10

The value of Polynomial or Expression at two limits denoted by f (x) |x=bx =a is defined as the difference

of the values of f (x) at x + b and that of f (x) at x = a. The number b is called the upper limit of the polynomial or expression, while a is the lower limit. For Example:

[-(2)2 -4(2) – 1] – [-(-1)2 – 4(-1) – 1] = -15

Also, the value of x-1/ 2x + 5 |10 is 0 - (-1/5) = 1/5; x-a /2x + 5 | 2-1 is 1/9 – (2/3) = 7/9

POWERS WITH THE POSITIVE INTEGRAL EXPONENTS

A compact notation for the product of n factors each of which is a is a given in the following definition.

The power an is defined as follows:

an = a • a • a • . . .• a, n is a positive integer |_____ n factors____|

We call the exponent n the exponent of a as the base.

The power is one strong concept in mathematics to put a number or expression in a more compact form. Some notations employed in this concept which convey different notations such as the following

-32 = -3 • 3 = 9 ; (-3)2 = (-3)(-3) = 9; -(3)2 = -(3 • 3) = -9

(-2)3 = (-2) (-2) (-2) = -8 ; -54 = -5 • 5 • 5 • 5 = -625 ; -23 = -2 • 2• 2 = 8

LAWS OF EXPONENTS

1. The product of powers.

am • an = am + n

Illustrations: 1. x2 . x3 = x5 2. x4 • x5 = x9 3. (2x3)(-3x4) = -6x7

2. The quotient of powers.

a m an = am-n

Illustrations: 1. (x2)3 = x 6 2. (4x3)2 = 16x6 3. (3x4c) 5 = 243x2() c5

3. The power of product.

(ab)m = ambm

Illustrations: 1. (2x)3 = 23 x3 = 8x3 2. -2(x2)3 = -2x6

4. The power of power.

(am )n = amn

5. The power of quotient.

(a / b)m = am / bm

Illustration: (-2x3/3)4 = (-2)4(x3)4/ 34 = 16x12/81

THE PRODUCTS OF POLYNOMIAL

Rule1. To multiply two monomials, use commutative, associative and the laws of exponents in multiplication. Example:

(3x2y4z2w9) (-4xyz4wv)

= 3 (-4) (x2x) (y4y) (z2z4) (w9w) v

= -12 x10 y6 w10 v

Rule 2. To multiply two polynomials, use the distributive law and apply rule 1.

Example:

(2x – 3y) (4x + 5y)

= 2x (4x +5y) – 3y (4x + 5y)

= 8x2 – 2xy – 15y2

DIVISION OF POLYNOMIALS

Rule 1. To divide monomials, use the laws of exponents in division.

a. -6x 4 y 6 zw = -6 . x 4 . y 6 . z . w = x 3 w 12xy8 zu 12 x y8 z u 2y2 u

Rule 2. To divide polynomial by a monomial, we use : a+b = a + b c c + c

The preceding theorem is applied to a finite number of terms in a given polynomial.

Example: (4x5+18x2+16x) ÷ (2x) = (4x5÷ 2x) + (18x2÷2x) + (16x÷ 2x) = (2x4) + (9x) + (8) = 2x4+ 9x +8

Rule 3. The last rule is to divide a polynomial by another polynomial with at least two terms. This type of divisions is applied only when the degree of the polynomial in the numerator is greater or equal to the degree of the polynomial in the denominator.

1. Arrange the terms of the dividend in descending powers of the variable.

2. Divide the first term in the dividend by the first term of the divisior, giving the first term of the quotient.

3. Multiply each term of the divisor by the first term of the quotient and subtract the product of the dividend.

4. Use the remainder obtained in step 3 as a new dividend, and repeat steps 2 and 3.

5. Continue the process until the remainder is reached whose degree should be less than the degree of the divisor.

The result of division is expressed as follow:

a. for exact division (remainder 0) b. for remainder ≠ 0

dividend = quotient dividend = quotient(Q) + remainder (R )divisor divisor divisor (D)

Example : x 2 + 2x – 3 = x +3 x 2 + 2x – 3 = x+ 4 + _5_ x-1 x-2 x -2

x2 + 2x – 3 = (x+4)(x-2) +5

SYNTHETIC DIVISION

Another method of division which has a very short and simple procedure is called Synthetic Division. Unlike the usual division which involves the four fundamental operations, this method requires only addition and multiplication applied to the coefficients. This method is applied when the divisor is of the form x + a.

Steps to in Synthetic Divison

1. Arrange the terms of the dividend in descending powers of variable.

2. Write the numerical coefficients of each term of the dividend in a row indicating the coefficients of powers. Replace the missing power with the zero coefficient.

3. Replace the divisor x – r by r: for divisor x + r, replace it with – r (constant divisor).

4. Multiply the coefficient of the largest power of x, written on the third row, by the constant divisor. Place the product beneath the coefficient. Multiply the sum by the constant divisor and place it beneath the coefficient of the next largest power. Continue this procedure until there is a product added to the constant of the last term.

5. The last number on the third tow is called the remainder, the rest of the numbers, starting from the left to right, are the coefficients of the terms in the quotient, which is one degree less than that of the dividend.

Example:

f(x) = 4x3 - 3x2 + x - 4

We will picture it evaluated at the input value x = 2. Arrange the input value, the coefficients, and a line like this:

2) 4 -3 1 -4

---------------

Now drop down the 4:

2) 4 -3 1 -4

--------------- 4

Multiply the input 2 times the 4. Place this product, 8, under the -3:

2) 4 -3 1 -4

8 --------------- 4

Add the -3 and the 8. Place this sum, 5, under the line:

2) 4 -3 1 -4 8 --------------- 4 5

Multiply the input 2 times the 5. Place this product, 10, under the 1:

2) 4 -3 1 -4 8 10 --------------- 4 5

Add the 1 and the 10. Place this sum, 11, under the line:

2) 4 -3 1 -4 8 10 --------------- 4 5 11

Multiply the input 2 times the 11. Place this product, 22, under the -4:

2) 4 -3 1 -4 8 10 22 --------------- 4 5 11

Add the -4 and the 22. Place this sum, 18, under the line:

2) 4 -3 1 -4 8 10 22 --------------- 4 5 11 18

Thus, the answer is 4x2 + 5x + 11.

FACTORING SPECIAL PRODUCTS

A. Factoring using the GCF

1. Find the largest number common to every coefficient or number.

2. Find the GCF of each variable. It will always be the variable raised to the smallest exponent.

3. Find the terms that the GCF would be multiplied by to equal the original polynomial.

It looks like the distributive property when in factored form ….. GCF(terms).

Examples:

3x2 - 6x

2x2 - 4x + 8

5x2y3 + 10x3y

B. Factoring the Difference of Two Squares

1. The factors will always be (a + b)(a – b).

2. "a" is the square root of the first term

3. "b" is the square root of the second term.

Examples:

9x2 - 49

121x2 - 100

25x2 - 64y2

C. Factoring a Perfect Square Trinomial

1. Characteristics

"ax2" term is a perfect square.

"c" term is a perfect square.

"c" term is positive.

Factors into two identical binomials: (a ± b)2.

2. Steps to Factor

Write it as (a ± b)2 since it factors into two identical binomials.

"a" is the square root of the "ax2" term.

"b" is the square root of the "c" terml.

The operation in the binomial factor is the same as the operation in front of the "x" term.

Examples:

9x2 - 30x + 25

4x2 + 28xy + 49y2

2x2 + 16x + 32

16x3 + 80x2 + 100x

RATIONAL EXPRESSIONS

A rational expression is one that can be written in the form

where P and Q are polynomials and Q does not equal 0.

An example of a rational expression is:

Domain of a Rational Expression

With rational functions, we need to watch out for values that cause our denominator to be 0. If our denominator is 0, then we have an undefined value.

So, when looking for the domain of a given rational function, we use a back door approach. We find the values that we cannot use, which would be values that make the denominator 0.

Example 1: Find all numbers that must be excluded from the domain of .

Our restriction is that the denominator of a fraction can never be equal to 0.

So to find what values we need to exclude, think of what value(s) of x, if any, would cause the denominator to be 0.

*Factor the den.

This give us a better look at it.

Since 1 would make the first factor in the denominator 0, then 1 would have to be excluded.

Since - 4 would make the second factor in the denominator 0, then - 4 would also have to be excluded.

Fundamental Principle of Rational Expressions

For any rational expression , and any polynomial R, where , , then

In other words, if you multiply the EXACT SAME thing to the numerator and denominator, then you have an equivalent rational expression.

This will come in handy when we simplify rational expressions, which is coming up next.

Simplifying (or reducing) a Rational Expression

Step 1: Factor the numerator and the denominator.

Step 2: Divide out all common factors that the numerator and the denominator have.

Example 2: Simplify and find all numbers that must be excluded from the domain of the simplified

rational expression: .

Step 1: Factor the numerator and the denominator

AND


*Factor the trinomials in the num. and den.

*Divide out the common factor of (x + 3)

*Rational expression simplified

To find the value(s) needed to be excluded from the domain, we need to ask ourselves, what value(s) of x would cause our denominator to be 0?

Looking at the denominator x - 9, I would say it would have to be x = 9. Don’t you agree?

9 would be our excluded value.

Example 3: Simplify and find all numbers that must be excluded from the domain of the

simplified rational expression:

Step 1: Factor the numerator and the denominator

AND


*Factor the diff. of squares in the num. and *Factor the trinomial in the den.

*Factor out a -1 from (5 - x)

*Divide out the common factor of (x - 5)

*Rational expression simplified

Note that 5 - x and x - 5 only differ by signs, in other words they are opposites of each other. In that case, you can factor a -1 out of one of those factors and rewrite it with opposite signs, as shown in line 3 above.

To find the value(s) needed to be excluded from the domain, we need to ask ourselves, what value(s) of x would cause our denominator to be 0?

Looking at the denominator x - 5, I would say it would have to be x = 5. Don’t you agree?

5 would be our excluded value.

Multiplying Rational Expressions

Q and S do not equal 0.

Step 1: Factor both the numerator and the denominator.

Step 2: Write as one fraction.Write it as a product of the factors of the numerators over the product of the factors of the denominators. DO NOT multiply anything out at this point.Step 3: Simplify the rational expression.

Step 4: Multiply any remaining factors in the numerator and/or denominator.

Example 1: Multiply .

Step 1: Factor both the numerator and the denominator

AND

Step 2: Write as one fraction.

*Factor the num. and den.

In the numerator we factored a difference of squares.

In the denominator we factored a GCF and a trinomial.

Step 3: Simplify the rational expression.

AND


*Simplify by div. out the common factors of (y + 3), (y - 3) and y

*Excluded values of the original den.

Note that even though all of the factors in the numerator were divided out there is still a 1 in there. It is easy to think there there is "nothing" left and make the numerator disappear. But when you divide a factor by itself there is actually a 1 there. Just like 2/2 = 1 or 5/5 = 1.

Also note that the values that would be excluded from the domain are 0, 3, -6, and -3. Those are the values that makes the original denominator equal to 0.

Example 2: Multiply .

Step 1: Factor both the numerator and the denominator

AND

Step 2: Write as one fraction.


In the numerator we factored a difference of cubes and a GCF.

In the denominator we factored a trinomial.

Step 3: Simplify the rational expression.

AND


*Simplify by div. out the common factors of (x - 3), 2, and (x + 2)


Note that the values that would be excluded from the domain are 0, 3, and -2. Those are

the values that makes the original denominator equal to 0.

Dividing Rational Expressions

where Q, S, and R do not equal 0.

Step 1: Write as multiplication of the reciprocal.

Step 2: Multiply the rational expressions as shown above.

Example 3: Divide .

Step 1: Write as multiplication of the reciprocal

AND


*Rewrite as mult. of reciprocal


*Simplify by div. out the common factors of 3x and (x + 6)

*Multiply the den. out

*Excluded values of the original den. of product

In the numerator of the product we factored a GCF.

In the denominator we factored a trinomial.

Note that the values that would be excluded from the domain are -6 and 0. Those are the values that makes the original denominator of the product equal to 0.

Example 4: Divide .

Step 1: Write as multiplication of the reciprocal

AND


*Rewrite as mult. of reciprocal


*Simplifyby div. out the common factors of y, (y + 4), and (y - 4)

*Multiply the num. and den. out

*Excluded values of the original den. of quotient & product

In the numerator of the product we factored a GCF and a trinomial.

In the denominator we factored a GCF and a difference of squares.Note that the values that would be excluded from the domain are 0, 2, - 4, 4, and -3. Those are the values that make the original denominator of the quotient and the product equal to 0.

Adding or Subtracting Rational Expressions with Common Denominators

Step 1: Combine the numerators together.

Step 2: Put the sum or difference found in step 1 over the common denominator.

Step 3: Reduce to lowest terms as shown in Tutorial 8: Simplifying Rational Expressions.

Why do we have to have a common denominator when we add or subtract rational expressions?????

Good question. The denominator indicates what type of fraction that you have and the numerator is counting up how many of that type you have. You can only directly combine fractions that are of the same type (have the same denominator). For example if 2 was my denominator, I would be counting up how many halves I had. If 3 was my denominator, I would be counting up how many thirds I had. But I would not be able to add a fraction with a denominator of 2 directly with a fraction that had a denominator of 3 because they are not the same type of fraction. I would have to find a common denominator first, which we will cover after the next two examples.

Example 1: Add .

Since the two denominators are the same, we can go right into adding these two rational expressions.

Step 1: Combine the numerators together

AND


*Common denominator of 5x - 2

*Combine the numerators *Write over common denominator


Step 3: Reduce to lowest terms.

Note that neither the numerator nor the denominator will factor. The rational expression is as simplified as it gets.

Also note that the value that would be excluded from the domain is 2/5. This is the value that makes the original denominator equal to 0.

Example 2: Subtract

Since the two denominators are the same, we can go right into subtracting these two rational expressions.

Step 1: Combine the numerators together

AND


*Common denominator of y - 1



http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut8_simrat.htm

*Factor the num.

*Simplify by div. out the common factor of (y - 1)


Note that the value that would be excluded from the domain is 1. This is the value that makes the original denominator equal to 0. Least Common Denominator (LCD)

Step 1: Factor all the denominators

Step 2: The LCD is the list of all the DIFFERENT factors in the denominators raised to the highest power that there is of each factor.

Adding and Subtracting Rational Expressions Without a Common Denominator

Step 1: Find the LCD as shown above if needed.

Step 2: Write equivalent fractions using the LCD if needed.If we multiply the numerator and denominator by the exact same expression it is the same as multiplying it by the number 1. If that is the case, we will have equivalent expressions when we do this.

Now the question is WHAT do we multiply top and bottom by to get what we want? We need to have the LCD, so you look to see what factor(s) are missing from the original denominator that is in the LCD. If there are any missing factors then that is what you need to multiply the numerator AND denominator by.

Step 3: Combine the rational expressions as shown above.

Step 4: Reduce to lowest terms as shown in Tutorial 8: Simplifying Rational Expressions.

Example 3: Add .


The first denominator has the following two factors:

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut10_addrat.htm#lcd


http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut10_addrat.htm#add

*Factor the GCF

The second denominator has the following factor:

Putting all the different factors together and using the highest exponent, we get the following LCD:

Step 2: Write equivalent fractions using the LCD if needed.

Since the first rational expression already has the LCD, we do not need to change this fraction.

*Rewriting denominator in factored form

Rewriting the second expression with the LCD:

*Missing the factor of (y - 4) in the den.

*Mult. top and bottom by (y - 4)


http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut10_addrat.htm#equiv



*Simplify by div. out the common factor of y


Note that the values that would be excluded from the domain are 0 and 4. These are the values that make the original denominator equal to 0.

Example 4: Add .

Step 1: Find the LCD as shown above if needed.The first denominator has the following factor:

The second denominator has the following two factors:

*Factor the difference of squares


Step 2: Write equivalent fractions using the LCD if needed.Rewriting the first expression with the LCD:

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut10_addrat.htm#equiv


*Missing the factor of (x + 1) in the den. *Mult. top and bottom by (x + 1)

Since the second rational expression already has the LCD, we do not need to change this fraction

*Rewriting denominator in factored form





This rational expression cannot be simplified down any farther.

Also note that the values that would be excluded from the domain are -1 and 1. These are the values that make the original denominator equal to 0.

Example 5: Subtract .


The first denominator has the following two factors:



http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut10_addrat.htm#add

*Factor the trinomial

The second denominator has the following two factors:

*Factor the trinomial


Step 2: Write equivalent fractions using the LCD if needed.

Rewriting the first expression with the LCD:

*Missing the factor of (x - 8) in the den. *Mult. top and bottom by (x - 8)

Rewriting the second expression with the LCD:

*Missing the factor of (x + 5) in the den. *Mult. top and bottom by (x + 5)



*Distribute the minus sign through the ( )


*Factor the num.

*No common factors to divide out


Note that the values that would be excluded from the domain are -5, -1 and 8. These are the values that make the original denominator equal to 0.

Full Name: Pioray, Claire E.

Sex: Female

Age : 17 y/o

Birthday: January 27, 1993

Year: Second Year

Course: Bachelor of Science in Computer Science (BSCS)

School: Cainta Catholic College

Address: Blk 11 Lot 15 Ciudad Del Sol I Subd., San Juan, Taytay, Rizal

E-mail Add: [email protected]

Facebook Account: [email protected]

Contact Number: (0935)1063283

Affiliation(s): Youth For Christ- East A Campus Based

Bio:Hello!

Claire Pioray is my full name.

My friends often call me BUMBAY.

I’m a Family Oriented person (Yes, I love spending my time with my parents and siblings).

And a person who is very positive in every aspects of life.

There are many things I love doing such as eating plenty of foods, hanging out with my friends but

What I loved most is serving my papa God.

Bring back all the glory in Him.

Worshipping Him through singing.

And trying to be a good example to others.

Full Name: Celajes, Ron Daniel

Sex: Male

Age : 18y/o

mailto:[email protected]


Birthday: March 9, 1992

Year: Second Year



Address: Karangalan Gate 5, Pasig City, Metro Manila



Contact Number: ( (0916)4115084


Bio:

I dream to live my dream life, stay in my dream house, drive my dream car, and love my dream family..I♥Coco Jiang

I♥Jeon Ji HyunI♥coffee

I♥TVI♥Science

I♥ComputerI♥AnimeI♥ComSci

I♥medicineI♥foodand

I♥her<thenextchapter>

25

Full Name: Ceriola, Gemarly C.

Sex: Female

Age : 17y/o

Birthday: February 23, 1993





Year: Second Year

Address: DM II Subd., San Roque, Cainta, Rizal



Contact Number: (0909)1422418


Bio :

Gemarly Ceriola is the complete name.

I AM HUMAN ..

I cry.I laugh.I fear.I love ' yeah I fell inlove for several times ;) But now i found the greatest love from my Lord ' Yes I'm so inlove with my papa God :)

----------------I like everything that makes SENSE!

I love worshipping my God ;)----------------

no hatred.no insecurities.no selfishess.I LOVE BEING ME ;)



Student’s

Profile

Cainta Catholic College College Department

A. Bonifacio Ave. Cainta, Rizal

College Algebra

Prepared by:Ron Daniel Celajes

Gemarly Ceriola Claire Pioray

Submitted To:Mr. RV. CruzInstructor- College Math

References:

•College Algebra by: Felix M. Panopio and Sergio E. Ymas Jr.

•Intermediate Algebra Work Book by: Ferdinand Pascual

•http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/index.htm

Polynomials

Documents

Transcript of Polynomials