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Advanced Algebra

This chapter aims the following objectives:

1. To generalize Polynomial.2. To familiarize the operation used in polynomial3. To restate Synthetic Division4. To determine the zeros of Polynomial5. To derive Quadratic Formula

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To define polynomials.

To be aware of the terms of a polynomial.

To distinguish the degree of a polynomial.

Advanced Algebra

Hi Everybody! I want to introduce to you my friend, Polynomials. He is one variable of an algebraic expression of the form

anxn + an-1xn-1 + an-2xn-2 + … + a0,

where n is a non-negative integer, and an, an-1, an-2, …,a0 are constants, and an ≠ 0.

You want to know him better? Here is an example of a polynomial.

x2 + 4xy + 4y2

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LESSON 1: POLYNOMIALSObjective

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Advanced Algebra

After you have learned what polynomial is, you should also know his different terms and degrees.

I want you to meet Monomial. He is a polynomial consisting of only one term.

Another is Binomial. He is a polynomial consisting of two terms.

And the last but not the least is a Trinomial. He is a polynomial consisting of three terms.

Now you meet them, I want you to know more about them because it will help you a lot. Here are the examples.

Yes! This is it. You can now easily determine the terms of a polynomials. You’re now ready to take my challenges. But wait you’re in the half of our study we’re not yet done. You must know what the degree of polynomial is.

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Hi there. I ‘am monomial

5x; -2ab; 11a3; 28

And I ‘am binomial

3x2 + y; 8x2y2 – 4; ; a2 + 2

And we are the trinomials.

6x2 + 2xy + y2; m + n – p; x2 + 5x + 6

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Advanced Algebra

And now, this is the moment of truth you’re now ready to solve and answer all my challenges regarding our lesson which is polynomials. I’m sure that you’ll get excellent points if you really understand our lesson.

NAME: RATING:

A. Determine whether the following is a polynomial or not.

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5 is the degree of polynomial because on the term 6x2y x is in the highest

power, then x is in the 2nd degree and on the term 12x3y2, x is in the 3rd degree

so we add same variables exponent then we will get 5 as the degree of the

polynomial.

6x2y + 12x3y2 + 8

Remember that in determining the degree of a polynomial you must get the highest power of its terms after it has been simplified.

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Advanced Algebra

1. 3x2 – 5x + 6 ______2. A2b2 – 4ab + 8 ______3. x2 – y 2 ______4. a3 + b3 + c3 ______

5.______

6.______

7. x1/3 + 5 ______

8. q + c + c2 ______

9.______

10. 5x + ______

11.______

12. ______

B. Classify each polynomial according to the number of its terms.

1. 8a + 2 ______

2. -5xy + x + xy ______

3. + x ______

4. a2 + 4a + 16 ______

5. x + y3 ______

6. 5x2y – 12xy + y2 ______

7.xyz ______

8. m + 2n – 3p ______

9.______

10. 2a3 – 5a2 – 15 ______

C. Give the degree of each of the following polynomials.

1. 4x ______2. 4x2 + 4x – 8 ______3. 5xy + 8y2 + 13x2 ______

4. 6x2 + 4x + ______5. 3a4b2 + 4ab3 – 6b7 ______6. a3b4 + a2b2 – 6ab3 ______

7. 9 – 6xy + yz + xyz ______

8.v - ______

9. 8xyz2 - ______

10. m + 9m5n8 + 3m3n5p7 - 7np10 ______

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To explore the addition with subtraction operation.

To familiarize student the use of multiplication and division.

To perform the synthetic division method.

LESSON 2: FUNDAMENTAL

Advanced Algebra

After we discussed the polynomials terms and degree, I will now introduce to you the

different operation in solving polynomial expression. Meet addition and subtraction, and follow their

rules in solving polynomials.

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Objective

Remember that in adding polynomials, simply add the numerical coefficient of like terms. And in subtracting polynomials, just simply change the second polynomial

into its additive inverse and then proceed to addition.

Here is an example of adding polynomials.

(9x4 – x3 + 5x2 – 8x) + (4x4 – x3 – 7x + 7) =?

Solution:

(9x4 – x3 + 5x2 – 8x) + (4x4 – x3 – 7x + 7)

= 9x4 – x3 + 5x2 – 8x + 4x4 – x3 – 7x + 7

= 9x4+ 4x4– x3– x3+ 5x2 – 8x – 7x + 7

= 13x4 – 2x3 + 5x2 – 15x + 7

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Advanced Algebra

NAME: RATING:

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And here is the example of subtracting polynomials.

(15x2 – 4xy + 10y2) – (9x2 + 4xy + 5y2) =?

Solution:

(15x2 – 4xy + 10y2) – (9x2 + 4xy + 5y2)

= 15x2 – 4xy + 10y2 – 9x2 – 4xy – 5y2

= 15x2– 9x2 – 4xy – 4xy + 10y2 – 5y2

= 6x2 – 8xy + 5y

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Advanced Algebra

Perform the indicated operations.

1. (6xy – 2x +4) + (11 – 8xy + 6x)

2. (-8mn2 + 12n) – (m3 – 4m2 + 4n)

3. (3b2 – 2b + 9) + (b2 + 6b – 4)

4. (5x2 – 2y + 4z) – (x2 – 3y2 + z)

5. (a2 + 4a + 2) + (2a + 3)

6. (3x2 – y) – (2x2 + 5y)

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7. (-7x4y3 – 21x3y3 + 28x5y4) + (7x2y2 + 6xy + 2x2y)

8. (-x2 + 6x – 2) – (x2 – x + 3) – (x + 1) + (x + 2)

9. (3y2 + 3xy + 10) + (4y3 – 10xy – 15)

10. (-2m2 + mn + 5n2) – (-4m2 – 6mn + 3n2)

After we have discussed the two operations which are addition and subtraction, we now proceed to Multiplication and division, the other operation in polynomials. You will meet them later in the middle of our study.

We all know that mathematics have 4 major operations in solving an equation. Like in arithmetic calculation, polynomials have also those 4 operations. And we formerly discussed addition and subtraction. Now I will introduce to you multiplication.

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Advanced Algebra

Here is the example of multiplication for you to understand what multiplication meant to be.

(4x + 3) (2x + 5)

Solution:

(4x + 3) (2x + 5) = ( 4x)(2x) + (4x)(5) + (3)(2x) + (3)(5)

= (8x2 + 20x) + (6x + 15)

= 8x2 + (20x + 6x) + 15

= 8x2 + 26x + 15

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Remember that to get the product of polynomial; you must multiply each term of polynomial by each of the other terms of polynomial. Then combine like terms.

Rules in Dividing Polynomials

1) In dividing polynomials by monomial, divide each term of a polynomial by monomial.

2) In dividing polynomial by another polynomial:a. Arrange the term in descending power with respect to a variable.b. Get the common factor of the dividend and divisor.c. Cancelled out or divide its common factor.

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Advanced Algebra

To make it clear, here is the example that clarifies to your curiosity.

Solution:

=

=

= (a + 2)

And this is it. Now you answer again my challenges for me to know that your ability in learning my lesson is satisfy your knowledge.

NAME: RATING:

A. Perform the indicated operations.

1. (c2 – 16)(c +1)

2. (11p2 – 66p + 99)(p + 1)

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We factor the dividend with the common factor of the divisor.

Then I cancel it out

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Advanced Algebra

3. (a – b)2(a + b)2

4. (5x2y + 3xy2 – 7x2y3)(xy)

5. (a2 + 2a + 3)(a – 5) +

6. (x + 1)(x + 2)(x + 3)

7. (x2 + x + 2)(2x2 + 3) +

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8.

9.

10. (a2n – 3an + 5)(a + 2)(a2 + 4a + 2)

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Advanced Algebra

B. Find the Quotient.1. (a2 – 7a + 10) (a – 5)

2. (x3 – 4x2 – 2 + 5x) (x – 1)

3. (3a4 – 2a + 5) (a2 + 3)

4. (2x3 + 5x2 – x – 1) (x – 1)

5. (2x2 – 5x – 6) (2x – 1)

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To define Synthetic division.

To be aware of the procedure in using Synthetic division.

To perform synthetic division in solving polynomial function.

Advanced Algebra

I know that you have a difficulty in getting the common factor to divide the polynomials. I have here a friend that will help you to make your factoring in its easiest and shortest way.

I proudly introduce to you Synthetic division. He is a process of division for polynomials in one variable where the divisor is of the form x – c, and c is any real number.

So your difficulty in dividing polynomials will lessen through this method. I think that you will use this in the near future of your study in different branches of mathematics.

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LESSON 3: SYNTHETIC DIVISION

Objective

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Procedure in using synthetic division

1. Arrange the terms into descending power.2. Copy the numerical coefficient. (If the descending power of the terms is like this x3

+ 2x + 1 then the arrangement should be 1 0 2 1: because the degree of polynomials is in the 4th. )

3. Substitute the value of x in the divisor.4. Then, bring down 1st the numerical coefficient w/c is in his 1st term.5. Multiply to the value of x then subtract product to the 2nd term of numerical

coefficient and then multiply now the difference to the value of x and repeat the process of multiplying, subtracting until you meet the last term.

Advanced Algebra

Here is the example for you to apply this method.

(x3 + 9x2 + 17x – 19) (x + 4)

Solution:

-4 1 9 17 -19

-4 -20 12

1 5 -3 -7

= x2 + 5x – 3 –

So, we finally studied this method. And now, you can use it in your mathematics subject. And of course, after we end the lesson we might have a test regarding this lesson. Are you ready? But you should be ready because I rate you according to this scale:

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remainder

The final answer

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Advanced Algebra

NAME: RATING:

Get the quotient of the following by using the Synthetic Division Method.

1. (x4 + 4x3 + x2 + x + 17) (x – 2)

2. (x4 + 2x2 + x - 11) (x + 5)

3. (x2 + 4x + 21) (x + 7)

4. (2x4 + x3 – x – 12) (x – 2)

5. (x3 – 4x2 + 5x – 2) (x – 1)

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6. (x3 – 7x2 – 4x + 24) (x – 6)

7. (3x3 – 2x2 + 5x +1) (x + 5)

8. x5 – 2x4 + 3x3 – 2x2 + 1) (x – 2)

9. (x2 – x – 20) (x – 5)

10. (x3 – 4x2 – 2 + 5x) (x – 4)

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To determine the zeros of a polynomials.

To be aware in the different techniques in factoring a polynomials.

To perform the use of quadratic formula in solving polynomials.

Zero of a Polynomial Function is the value of t he variable x, which makes the polynomial function equal to zero. And it’s written in symbolic form:

f(x) = 0

Advanced Algebra

It’s a long discussion about the operations used in Polynomials. But now another mathematics word you will be meet. I introduced to you the Zeros of polynomial Function. And from the word itself you will be notice that he is pertaining to zero.

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LESSON 4: ZEROS OF A POLYNOMIAL FUNCTION

Objective

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x = -2; x2 + 4x + 4

f(x) = (-2)2 + 4(-2) + 4

= 4 + (-8) + 4

= 8 – 8

= 0

That’s the simplest explanation of it. And I think you already understand it or maybe I’ll give an example of that.

Now, take my challenges. You should get perfect score, because this lesson is very

easy anyone can perfect it. So grab it!

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Advanced Algebra

NAME: RATING:

Which numbers -3, -2, -1, 1, 2 and 3 are zeros of the following polynomial functions?

1. f(x) = x3 + 4x2 + x – 6

2. f(x) = 2x3 – 3x2 – 11x + 6

3. f(x) = x3 +3x 2 – x – 3

4. f(x) = x4 – 4x3 +6x2 – 4x + 1

5. f(x) = x3 – 2x2 – 5x + 6

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6. f(x) = x3 – x2 – 9x + 9

7. f(x) = x3 – 3x2 – 5x + 12

8. f(x) = 5x4 + 2x3 – 3x – 3

9. f(x) = 12x3 – 18x2 + 14x + 17

10. f(x) = x3 – 5x2 + 8x – 3

11. f(x) = x2 – 5x + 7

12. f(x) = 8x4 + 5x3 + 2x2 – 6x – 1

13. f(x) = 3x3 – 8x2 + 7x – 6

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To determine the different techniques used in factorization

To familiarize the use of such techniques

To evaluate equation using factorization techniques.

A. Common Factor of Highest Degree Ex: a2b + a2 = ab(a + b)

B. Difference of Squares

Ex: x2 – y2 = (x + y)(x – y)

C. Quadratic Trinomial

Ex: x2 + 7x + 12 = (x + 3)(x + 4)

Advanced Algebra

Under Zeros of polynomial function we have the factoring. He is the process of expressing a polynomial as a product of factors.

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Objective

D. Perfect Square TrinomialEx: a2 + 2ab + b2 = (a + b)2

E. Difference and Sum of CubesEx: x3 + y3 = (x + y)(X2 – xy + y2)

F. GroupingEx: am + bm – an – bn = (am + bm) – (an + bn) = m(a + b) – n(a + b)

= (m – n)(a + b)

LESSON 5: FACTORING TECHNIQUES

J. Perfect Square Trinomial

Ex: a2 + 2ab + b2 = (a + b)2

K. Difference and Sum of Cubes

Ex: x3 + y3 = (x + y)(X2 – xy + y2)

L. GroupingEx: am + bm – an – bn

= (am + bm) – (an + bn)

= m(a + b) – n(a + b)

G. Perfect Square Trinomial

Ex: a2 + 2ab + b2 = (a + b)2

H. Difference and Sum of Cubes

Ex: x3 + y3 = (x + y)(X2 – xy + y2)

I. GroupingEx: am + bm – an – bn

= (am + bm) – (an + bn)

= m(a + b) – n(a + b)

D. Perfect Square Trinomial

Ex: a2 + 2ab + b2 = (a + b)2

E. Difference and Sum of Cubes

Ex: x3 + y3 = (x + y)(X2 – xy + y2)

F. GroupingEx: am + bm – an – bn

= (am + bm) – (an + bn)

= m(a + b) – n(a + b)

TECHNIQUES

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Advanced Algebra

NAME: RATING:

Factor the given expressions.

1. a3 + 2a2 + 3a

2. 4x3 – 2x2 + x

3. 9p3q – 51p2q + 216pq2

4. x (x + 8) – 7(x + 8)

5. 3z2 – 2z + 6z – 4

6. 6x2 + x + 2 + 12x

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Advanced Algebra

7. a2 + 10ab + 25b2

8. x2 – 10xy + 25y2

9. 4c2 – 12cd + 9d2

10. 64xy2 – 9x3

11. (b2 – 2b + 1) – 100d2

12. m4 – 121n4

13. 4a2 – 28ac + 49c2

14. x2 – 3x + 2

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Advanced Algebra

15. 27a3 – 8b3

16. 4(x + y) – 64a4y2

17. x3 + 1

18. 16m4 – 25n6

19. 36y4 – 81x4

20. 24a3 + 27a2 + 64a + 12

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LESSON 6: QUADRATIC FORMULA

Objectives:

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To derive the quadratic formula

To analyze the quadratic formula

To use quadratic formula in solving polynomial function

Advanced Algebra

So you are now familiarized with the different techniques in factoring a polynomial function. Now we have to introduce the formula in getting a polynomial function zero.

The zero of a polynomial function is a real number that replace or substitute the value of x in f(x) and make f(x) = 0. This can be obtained through the last topics we discussed like synthetic and factoring techniques. But that method cannot get the exact value of x because of their remainder. And the appropriate method to use is get the exact value of x is the Quadratic formula. Its symbol form is

X =

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Advanced Algebra

Example:F(x) = x2 – 2x – 2Solution:

X = a = 1; b = -2; c = -2

=

=

=

=

= 1

This is the moment of truth, we done in this formula. So as you go beyond opening this workbook you encounter many challenges that help to develop your

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Always remember that Quadratic formula is use only in Quadratic form or in the 2nd

degree of terms. It always use in

ax2 + bx + c = 0

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Advanced Algebra

skills in Mathematics. And this is another challenge. Answer it carefully, because the one who careless in answering the more mistake he got.

NAME: RATING: A. Determine whether the given number is a zero of the given polynomial function.

1. f(x) = x3 + 4x – 5; 1

2. f(x) = 3x2 – 2x – 8; 3

3. f(x) = x4 + 3x2 – 2x – 2; 2

4. f(x) = 3x4 – 3x3 – 20x2 + 18; 3

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5. f(x) = x4 + x3 – 3x2 – 4x – 4; 2

6. f(x) = x4 – 5x3 + 8x2 + 15x – 2; 3

7. f(x) = 9x3 + 6x2 + 4x + 2; -

8. f(x) = x3 + 2x2 – 25x – 50; 5

B. Find the remaining zeros of the polynomial functions given one zero. Use Quadratic formula if the function is in quadratic form.1. f(x) = x3 + x2 – x – 1; x1 = 1

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2. f(x) = x3 – 3x2 – x + 3; x1 = 1

3. f(x) = x3 + 4x2 + x – 6; x1 = 1

4. f(x) = 3x3 – 4x2 – 13x – 6; x1 = -1

5. f(x) = 2x3 – 3x2 - - 3x + 2; x1 = -1

6. f(x) = x3 – 4x2 – 7x + 10; x1 = 5

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7. f(x) = x4 + x3 – 4x – 4; x1 = -1

8. f(x) = x4 + 5x3 + 5x2 – 5x – 6; x1 = -3

9. f(x) = x4 – 5x3 + 5x2 +5x – 6; x1 = 1

10. f(x) = x4 + 3x3 – x2 + 11x – 4; x1 -4

11. f(x) = x3 + 12x2 + 41x + 42; x1 = -2

12. f(x) = x3 + 2x2 – 2x – 4; x1 = -2

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Advanced Algebra

A. Encircle the letter of the correct answer.1. What is the graph of a linear function?

a. ellipse c. circle b. line d. parabola

2. What is the degree of the polynomial -6x4 + x5 – 3x7 + 2x – 1?a. 5 c. 7b. 4 d. 6

3. How many turning points does the graph of a polynomial function of degree n > 1 have?a. 2n c. nb. n + 1 d. n – 1

4. Given P(x) = 5x3 – 3x2 + x – 7, what is P(2)?a. -30 c. 30b. -23 d. 23

5. What is the remainder if the divisor x – c is a factor of a polynomial function P(x)?a. infinite c. positiveb. zero d. negative

6. How many real zeros do P(x) = x4 + 2x3 – 7x2 – 8x - 12 have?a. 3 c. 4b. 5 d. 7

7. In the polynomial 7x5 + 4x3 – 5x6 + x2 – 3x + 6, the coefficient of the term of highest degree is_____.a. 7 c. -5b. 1 d. -3

8. If P(x) = x3 + x2 – 5x + 3 is divided by D(x) = x – 1, the quotient Q(x) is_____.a. x2 + 2x – 3 c. 4x3 – 3x2 + 2x – 3b. x2 + 2x + 3 d. 4x3 – 3x2 – 2x – 3

9. What is the remainder in number 8?a. -6 c. 1b. 0 d. 6

10. Which of the following describes the graph of f(x) = 2x + 3?

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a. a line that rises to the right c. a parabola that opens upwardb. a line that falls to the right d. a parabola that opens downward

B. Solve each of the following:1. What is the remainder when 2x4 – 5x3 + 3x2 – 4x + 3 are divided by x – 2?

2. Determine if -2 is a zero of f(x) = x3 + 2x2 – x + 9.

3. If f(x) = x3 – 5x2 + 20x – 16 and f(2) = 0, what can be said about x – 2?

4. What are the zeros of f(x) = x3 – 2x2 – x + 6?

5. If one of the factors of x3 + 5x2 – 3x – 5 is x + 1, what are the other factors?

6. Find all the rational zeros of f(x) = 3x3 + 5x2 – 3x – 5.

7. If x – 1 is a factor of x3 – 2x2 – kx + 6, what is the value of k?

C. Challenge1. If P(x) = 3x4 – x3 + 5x – 7, find P(-2) and P(2).

2. Determine the value of m so that G(-3) = -1 for G(x) = 2x3 – 7x2 + 5x + m.

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3. Use synthetic division to find the quotient when P(x) = 2x4 – 3x3 – 12x– 32 is divided by x – 3.

4. What is the remainder if P(x) = 2x2 – 4x -5 is divided by x – 3?

5. Divide P(x) = x3 + 2x2 – 7x – 4 by x – c. Then write the given function in the form P(x) = (x – c) Q(x) + R where Q(x) is a quotient and R is the remainder.

6. For what values of p and q are x + 1 and x – 2 factors of x3 + px2 +2x + q?

7. Find the polynomial function whose zeros are 2, -2 and 3.

8. If -3 is a zero of the polynomial function P(x) = x3 + 3x2 – 2x – 6, find the other zeros.

9. Find all the zeros of the function P(x) = x3 + 4x2 +x – 6.

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10. Find a such that x – 3 will be a factor of x3 – 4x2 +ax – 9.

11. Find the rational zeros of P(x) = 2x4 – 3x3 + 2x2 – x – 3.

12. Which value(s) of k will the polynomial function P(a) = a2 – ka + k + 8 have exactly one real zero?

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Advanced Algebra

This chapter aims the following objectives:

1. To elaborate rational expression2. To familiarize the operation used in rational expression3. To synthesize variation4. To generalize Rational Function

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To define Rational Expression

To evaluate rational expression

To familiarize in simplifying Rational Expression

Advanced Algebra

And finally I will introduce to you RATIONAL EXPRESSION. It is a fraction containing polynomials on the numerator & denominator.

Analyze these examples:

All of the above examples are rational expressions.

Now who is that lowest term? Are you familiar with him?

Ok! He is a fraction having numerator & denominator with no common factor aside from one.

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Hi! I am a Rational Expression!!

This is my numerator having polynomial

This is my denominator

LESSON 1: RATIONAL EXPRESSION

Objective

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=

Let us try the following examples:

1. 2. 3.

Solution:

1. =

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I am the lowest term

We get the common factor w/c is.

Then we canceled out or divide the common factor.

And c is the common factor so I divide the expression by it.

Remember that b0 & c 0.

Remember that RATIONAL EXPRESSION should be in lowest term.

In reducing rational expression to lowest term, first, factor the numerator & the denominator, and then divide it by the common factor.

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=

=

Here is another one for you to familiarize with reducing rational expression to lowest term.

2. =

=

=

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Now, you get me, I am the lowest term! Remember!

Now, you get again the common factor w/c is x + y.

Then we canceled out or divide the

common factor.

You finally get again the lowest term!!

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Advanced Algebra

Some more? And you will do with your own after this.

3. =

=

=

Now you practice in these 3 examples, you’re ready to take challenge and test yourself how you understand the lesson.

NAME: RATING:

GET THE LOWEST TERM.

1.

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Common factor.

Cancel out or divide common factor.

Now, you get me, I am the lowest term! Remember!

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2.

3.

4.

5.

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6.

7.

8.

9.

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10.

11.

12.

13.

14.

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15.

16.

17.

18.

19.

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To define the rule of addition and subtraction

To determine the rule on multiplication

To be aware on the rule of division

To perform the different operations on rational expression.

Advanced Algebra

20.

The operations used in rational expression similar to ordinary arithmetic operations. Let me introduce to you the first operation.

Hi! We are addition and subtraction. And either we have the same procedure or rules in solving problem, you get the sum / difference in two rule which are the common and the no common denominator.

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LESSON 2: OPERATION ON

RATIONAL

Objective

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=

Now you know how to get our sum/difference because we have a common denominator then after you know us, we will introduce to you how to add and subtract rational expressions with different denominator; that isby finding the LCD.

=

=

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Here we are the sum and the difference with common denominator c.

Remember that c 0.

This is my LCD!

Now you get my sum/difference with

unlike denominator…

Remember that rational expressions with the same denominator can be added or subtracted by simply adding or subtracting their numerator and copy the

denominator.

To find the LCD factor completely the denominators and the product of each unique prime factor raised to the highest power. Then apply the procedures in getting the

sum/difference.

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Let’s try these examples and apply the procedure in our lesson.

1. +

-

2. +

-

3. -

+

Solution:

If you notice in number 1, we will use the common denominator.

1. +

-

=

=

=

=

You need one more? Here is another one.

2. +

-

=

=

=

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We will copy the common denominator w/c is 4a2.

Then we get the sum of the numerator

Then we simplify by getting the common factor and dividing it w/c is 2.

Then you get the sum.

Get again the common denominator

Then add / subtract the same term

You get now the sum / difference

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You will be challenged more because the next example is a rational expressions having unlike denominators.

3. –

+

=

=

=

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You will get the LCD of their denominators which is a3.

Then I’ll divide the LCD by their denominators and multiply to its

numerator.

Then you will combine similar terms.

Now you get their sum / difference

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NAME: RATING:

Start to find my sum / difference:

1. –

+

2. +

3. –

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4. –

5. +

6. +

7. –

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8. –

9. –

+

10. +

11. –

+ 1

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12. –

13. +

14. +

15. –

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16. –

17. –

18. +

19. +

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+

After we, introduced to you addition and subtraction, we now move on to the next level or next operation that will be used in solving rational expression. We all know that the next operation is the hardest among the two operations but now, we made t simpler for you to understand what multiplication and division are all about.

In solving rational expressions using multiplication and division, you must remember that there are rules and procedures to be followed.

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Steps in multiplying rational expression:

Multiply the expression from numerator by numerator and denominator by denominator.

Get the common factor then, cancel out or divide it to eliminate.

Simplify the remaining expression.

a x b = ab

c d cd

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=

=

x =

÷

÷

= x

Advanced Algebra

Here are the examples to make it clear for you.

1. 2. x2 + 6x + 9 x + 3 x2 + 3x – 10 x + 5

Solution:

1. 2a2d 9b2c . (2a2d) (9b2c) 3bc 16ad2 (3bc) (16ad2)

(2ad) (3bc) (ab)(3bc) (2ad) (8d)

ab8d

2. x2 + 3x – 10 x + 5 x2 + 3x – 10 x + 3 x2 + 6x + 9 x + 3 x2 + 6x + 9 x + 5

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Steps in dividing rational expression:

Get the reciprocal of the divisor then follow the steps in multiplication.

a ÷ b = a x d

c d c b

adcb

Multiply numerator by numerator then denominator by denominator

Find the common factor then divide or cancel it

Then you get the product of the expression

Get the reciprocal of the denominator

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=

=

=

Advanced Algebra

(x2 + 3x – 10) (x + 3)(x2 + 6x + 9) (x + 5)

(x + 5) (x – 2) (x + 3)(x + 3) (x + 3) (x + 5)

x – 2 x + 3

Now, you are fully charge with the examples I have given. So to know if you really understand what I’ am talking about you must answer my challenges so that I can measure your ability to learn.

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Then multiply numerator by numerator and denominator by

denominatorGet the common factor and

divide or cancel it

You have now the product of given expression

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x

x

x

÷

x

÷

xxxxxx

x

Advanced Algebra

Find the product or quotient of the following.

1. x3 + 6x2 + 5x – 12 x3 – 4 x – 1 x2 + 2x – 3

2. a + 3 a2 – 9

a2 + a – 12 a2 + 7a + 12

3. b2 – 25 2b – 10 (b + 5)2 4b + 20

4. 2x2 – 6x 9x + 81x2 + 18x + 81 x2 – 9

5. 2x2 + 5x + 2 2x2 + x – 14x2 – 1 x2 + x - 2

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÷

x

÷

÷

To define variation

To familiarize with the types of variation

To perform the different equation form in variation

To apply variation in Word problem.

x

Advanced Algebra

6. 6x3 – 6x2 3x2 – 15x + 12

x4 + 5x3 2x2 + 2x – 40

7. x2y x (y + 2)

3x (x – 1) y (x – 1)

8. 8x2y x + 1

x + 1 6xy2

9. 9m + 6n 12m + 8nm2n2 5m2

10. (x + 3) (2x – 1) (-x – 3) (2x + 1) x (x + 4) x

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LESSON 3: Variation

Objective

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Nowadays, topics in mathematics are really used, without knowing that math is involved. Particularly in the application of science that made a big role in our integral life. Using such proportionality in speed or motion, we all know that math is involved. This lesson will be used in science and in word problems application.

Variation is a relationship between variables that is defined using power function. It is in the form

f(x) = kxn

where k & n are real number.

DIRECT VARIATION Variable y is directly proportional to variable x, if the quotient of y divided by x is constant.

Written in the form y = kx

where k is called proportionality constant.

A variable y is said to be directly proportional to the nth power of variable x if

y=kxn

Where n > 0

INDIRECT (INVERSE) VARIATION

Variable y varies indirectly or inversely as the variable x if the product of x & y is constant written as

y=k/x or y=kx-1

Where k is not equal to 0

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Variable y is said to be inversely proportional to the nth power of he variable x if

y=k/xn or y=kx-n

Where n >0

JOINT VARIATIONVariable z varies jointly as the variable x & y if z varies directly as x when y is held

constant and varies directly as y when x is held constant written as

z=kxyWhere k is not equal to 0

Variable z said to be jointly proportional to the nth power of the variable x and the nth power of the variable y if

z=kyn xm

Where n & m > 0

Let’s proceed to our example.

Example

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Remember that two variables w/c increase or decrease in the same ratio is said to be direct variation. And if one of the variables increases while the other decreases in the same ratio is called indirect variation.

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Direct Variation The pressure of a given volume of gas varies directly as the temperature. Find the constant value if pressure of certain volume is 250 cm3 when its temperature is 170 C.

Solution:

P = kT formulate an equation using direct variation 250 cm3 = k 50C substitute the given value to the equation form K = 250 cm3/50 C divide pressure to temperature to get the constant value. K= 5 cm3/C final answer

Indirect Variation

If y varies inversely as x, and the constant of variation is k = , what is y when x = 10?

xy = k formulate an equation using indirect variation

10y = substitute the given value to the equation form

y = × divide k to y to get the constant value

= final answer

Joint VariationIf w varies jointly as x and y if w = 18 when x = 2 and y = 3, find the value of w if x= 4 and y = 5.

W = kxy formulate equation using joint variation18 = k (2) (3) substitute the first given value to look for the constant k = 18 / 6 divide product of x and y to w k = 3 this is the constant valuew = kxy formulate again equation to find ww = (3) (4) (5) substitute the second given value to look for ww = (12) (5) multiply k,x and yw = 60 you get the value of w.

After you studied variation, you must answer the following challenges for you to measure your capability in variation.

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NAME: RATING:

Evaluate the following.

1. Let y vary jointly as x and the cube root of z, and inversely as the square of w. What is the effect on y if x is increases by 20%, z is doubled and y is doubled?

2. The surface area of a sphere varies directly as the square of the radius. If the surface area is 36 in2 when the radius is 3 in, what is the surface area of a sphere with a radius of 5 in?

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3. When driver of a running vehicle applies the brakes abruptly, the resulting skid marks has a length (y) that varies directly as the square of the speed (x). A little boy crossing the street was struck by a car, leaving skid marks 15 cm long. If the police know that at a speed of 50 kph, the skid marks would be 12 meters long, how fast was the car running before the driver applied the break?

4. If y varies directly as x and if y is 15 when x = 5, find the value of y if x = 7.

5. If w varies jointly as x and y if w = 15 when x = 2 and y = 3, find the value of w if x= 3 and y = 4.

6. If an athlete could jump 23.5 ft when his takeoff velocity is 9.3 m/s, how far could he jump be if his takeoff velocity is only 9.0 m/s assuming that the jump is proportional to the square of the takeoff velocity.

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7. If z varies inversely as x and if z = 6 when x = 4, find the value of z when x = 3.

8. In an remote barangay with a population of only 300, the rate of growth of an epidemic (i.e., the rate of change of the number o f infected person), R is found to be jointly proportional to the number of people infected, x and the number of people who are not infected, (300 – x). Given that the epidemic is growing at the rate of 5 people per day when there are 75 people infected, how fast will the epidemic growing if the number of people infected in the barangay is doubled?

9. Suppose that the maximum number of bacteria that can be supported by particular environment is M and the rate of bacterial growth (y) is jointly proportional to the number of bacteria present (x) and the difference between M and the number of bacteria present. Write the defining equation for y as a function of x and give the domain.

10. The gravitational attraction (F) between two bodies varies jointly as their masses (m1) and (m2) and inversely as the square of the distance(d) between them. What is the effect on the gravitational attraction between two bodies if the masses are each halved and the distance between them is doubled?

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To define Rational Function

To distinguish the different types of Asymptotes

To perform operation used in finding Asymptotes

Advanced Algebra

A rational function f has the form

where g (x) and h (x) are polynomial functions and h(x) ≠ 0.

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Objective

LESSON 4: Rational

Function

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Asymptote is a straight line at singularity which graph of function tends to approach but never touches.

Vertical Asymptote

vertical lines which correspond to the zeroes of the denominator of a rational function

consist of vertical lines of the form x=a where "a" is any value of x resulting in division by zero

Horizontal Asymptote

horizontal line including x-axis to which graph of function comes closer and closer but never touches

consists of a horizontal line of the form y=b where "b" is the value of f(x) as x approaches positive or negative infinity

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Asymptote

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Proceed to our example:

1. Find the vertical asymptote of x2 + 2x – 8 = 0.

2. Find the horizontal asymptote of 5x3 -1 r(x) = ----------------

x2 + 3x + 2

Solution:

1. Use any kind of factorization to find the of x x2 + 2x – 8 = 0

(a) Horizontal Asymptote If m < n, there is no horizontal asymptote. If m = n, then y = an/bm is the horizontal asymptote. f m > n, then y = 0 is the horizontal asymptote.

(b) Vertical Asymptotes Every zero of the denominator q(x) determines a vertical

asymptote. If r1, r2, . . . , rk are zeros of q(x), then the lines x = r1, x = r2, . . . , x = rk are all vertical asymptotes of r(x).

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Synthetic Division (optional) 2 1 2 -8

2 81 4 0

So the value of x are 2 and -4 (x-2) (x+4) = 0And the value of vertical asymptote is the factor of the equation which is the value of x

2. Only the denominator will be use x2 + 3x + 2 = 0Use again any factorization to look for xSynthetic Division (optional) -2 1 3 2

-2 -21 1 0

So the value of x are -2 and -1 (x+1) (x+2) = 0And the value of horizontal asymptote is the value of x in which the denominator of a function becomes 0.

This is it. Now you are in the end of the lesson in my workbook. I admire for the patient you‘ve done in reading and analyzing the topics I presented to you. So hold you’re tight because I really admire if you answer all the challenges I prepared to you. Be careful for the last challenge because if you careless answer it the more mistake you’ve got.

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Find the vertical and horizontal asymptotes of the following if any:

1. f(x) = 1/(x + 2)

2. f(x) = (5x + 3) / (x – 2)

3. f(x) = (x + 2) / (x + 3)

4. f(x) = x / (x – 4)

5. f(x) = (3x2 – 15x + 12) / (2x2 + 2x – 40)

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6. f(x) = (10x – 1) / (2x2 + 1)

7. f(x) = (x2 + 1) / (1 – x2)

8. f(x) = (9x2 + 81) / (x2 + 18x + 81)

9. f(x) = (3x3 – 27x2 + 9x – 81) / (x2 + 4)

10. f(x) = (x2 + 10x + 25) / (x3 + 3x2 + 9x + 27)

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A. Simplify the following:1. (5x2 – 15x) / 10x2

2. (3t4 – 9t3) / 6t2

3. (2t2 +5t – 3) / (2t2 + 7t + 3)

4. (x3 + x2 – x – 1) / (x3 – x2 – x + 1)

5. (t4 + c4) / [(c + t)2 (c2 + t2)]

B. Find the sum/difference:1. [(a + b) / (a – b)] + [(a – b) / (a + b)] – [(b – a) / (a – b)] + (b – a) / (a + b)]

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2. (x – y)2 – (x + y)2

3. (4t2 – 4t + 1)-1 + (4t2 – 1)-1

C. Find the product/quotient:1. [(r2 + 4rs + 3s2) / (r2 + 5rs + 6s2)] × (r + 2s)-1 ÷ [(r + s) / (r2 + 4rs + 4s2)]

2. [(x2 + 3ax) / (3a – x)] ÷ [(x2 – 4ax + 3a2) / (a2 – x2)]

3. [(u2v) / (u+ v)] × [(u2 + 2uv + v2) / (uv2 – u2v)]

D. Solve the following:1. If r is inversely proportional to s and r = 5 when s = 4, find s when r= 7.

2. If y varies directly as x, and the constant of variation is 3, what is y when x = 7.

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3. If a varies jointly as c and d, and a = 20 when c = 2 and d = 4,find d when a = 25 and c = 8.

E. Find the vertical and horizontal asymptotes of the following if any:1. f(x) = 3x / (x +2)

2. f(x) = (x2 + 1) / x2

3. f(x) = 10x2 / (x2 + 1)

f(x) = (1 – x2) / x2

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