POLYNOMIALS

Maths 4th ESO José Jaime Noguera

1

Algebraic expressions

• Book, page 26

• YOUR TURN: exercises 1, 2, 3.

• Exercise: Find the numerical value of the algebraic expression 𝑥𝑦2 − 8𝑥 + 𝑦, knowing that x=-2 and y=3.

2

Definitions

• A monomial is an algebraic expression as 𝑎𝑥𝑛, where a is a real number, and n is a non negative integer.

• A polynomial is an algebraic expression that consists of a sum of several monomials.

3

Standard form of a polynomial

• Example: 𝑃 𝑥 = 3𝑥2 − 5 + 5𝑥3 − 8𝑥5 a) Write in standard form: 𝑃 𝑥 = −8𝑥5 + 5𝑥3 + 3𝑥2 − 5 b) Degree: 5 c) Constant term: -5 d) Leading coefficient: -8

4

Exercise

• Find the standard form, the leading term, the leading coefficient, the linear term, the quadratic term, the constant term and the degree of:

a) 𝑃 𝑥 = 5𝑥3 − 7𝑥2 + 5𝑥4 − 8 + 9𝑥

b) 𝑃 𝑥 = 2𝑥7 − 3𝑥5 + 7𝑥 − 4

c) 𝑃 𝑥 = 4𝑥3 − 7𝑥5 + 𝑥4 − 5𝑥

5

Reviewing operations with monomials

• ADDITION/SUBTRACTIONS. You can sum or subtract monomials WITH THE SAME LITERAL PART (called similar monomials or like terms in a polynomial): 𝑎𝑥𝑛 + 𝑏𝑥𝑛 = 𝑎 + 𝑏 𝑥𝑛

– 3𝑥5 − 5𝑥5 = −2𝑥5

– 2𝑥3 − 5𝑥2 you can’t do anything

• PRODUCT: 𝑎𝑥𝑛 · 𝑏𝑥𝑚 = 𝑎 · 𝑏 𝑥𝑛+𝑚 – 3𝑥2 · 5𝑥7 = 15𝑥9

• DIVISION: 𝑎𝑥𝑛: 𝑏𝑥𝑚 = 𝑎: 𝑏 𝑥𝑛−𝑚

– 8𝑥5: 5𝑥7 =8

5𝑥−2

• POWER: 𝑎𝑥𝑛 𝑚 = 𝑎𝑚 · 𝑥𝑛·𝑚

– 3𝑥2 3 = 33 · 𝑥6 = 27𝑥6

6

Exercises

a) 5x2+3x2 =

b) 3x5·9x2 =

c) 8x-3x =

d) 21x6:3x3 =

e) -3x3·5x7 =

f) 30x5: (-5x3) =

g) 5x3-3x2-7x3+3x2+5x =

h) 5x2-3x+7x2+6x-2x2 =

7

Extracting common factors

1. Factorize all the coefficients

2. Find the greatest common factor of the terms (considering both coefficients and variables).

3. Divide each term of the expression in parentheses by the greatest common factor, and write the variable factor outside the parentheses.

Example: 21𝑥3𝑦2 − 15𝑥2𝑦4 + 3𝑥2𝑦2

1. 3 · 7𝑥3 𝑦2 − 3 · 5𝑥2𝑦4 + 3𝑥2𝑦2

2. 𝟑 · 7𝒙𝒙𝑥𝒚𝒚 − 𝟑 · 5𝒙𝒙𝒚𝒚𝑦𝑦 + 𝟑𝒙𝒙𝒚𝒚

3. 3𝑥2𝑦2(7𝑥 − 5𝑦2 + 1)

8

Exercise

• Extract common factors:

a) 7𝑥2 + 49𝑥 − 14𝑥3

b) 28𝑎3𝑏2𝑐 − 12𝑎2𝑏𝑐3 − 8 𝑎4𝑏2

c) 5𝑥𝑦3 − 35𝑥2𝑦3 + 15𝑥𝑧

d) (x-2)(x+7)-(x-2)(2x-6)

e) 3(x+2)-(2x+4)

9

Binomial theorem

Where the binomial coefficients can be calculated using the Pascal’s triangle (or Tartaglia’s triangle):

10

Click on the image

11

What happens with 𝑎 − 𝑏 𝑛?

It’s the same but you have to change the signs:

𝑎 − 𝑏 5 = +1𝑎5𝑏0 − 5𝑎4𝑏1 + 10𝑎3𝑏2 − 10𝑎2𝑏3 + 5𝑎1𝑏4 − 1𝑎0𝑏5

Exercise: Find the following powers: 1. 𝑥 − 2 2 2. 𝑥 + 5 3 3. 3𝑥 − 1 2 4. 𝑥 + 2𝑦 3 5. 3𝑥 − 5 4 6. 2𝑥 + 1 5

12

Now, we are going to follow the book:

• Page 27. 3.1 Important equalities

• Exercises 5, and

• Exercise: Write as a power:

a) 4𝑥2 + 20𝑥 + 25

b) 25𝑥2 − 1

c) 4𝑥2 − 12𝑥𝑦 + 9𝑦2

d) 9𝑥4 − 49

13

Operations with polynomials

• Page 27, 28, 29

3.2 Operations with polynomials: +, -, ·, : , Ruffini’s rule or synthetic division.

• Exercises:

– Pag 27: 6, 7

– Pag 28: 8, 9

– Pag 29: 10

14

Remainder theorem:

P(x) x-a

Remainder quotient

Remainder = P(a)

Example: 𝑃 𝑥 = 2𝑥3 − 𝑥2 − 7𝑥 + 5

If we calculate P(x) : x-2 we obtain • Quocient: 2𝑥2 + 3𝑥 − 1

• Remainder: 3

If we calculate 𝑃 2 = 2 · 23 − 22 − 72 + 5 = 3

15

Roots of a polynomial

• We say that a is a root of P(x) if P(a)=0

The roots of a polynomial fulfills:

• If P(x):(x-a) is exact, then a is a root of P(x)

• If a is a root of P(x), then P(x):(x-a) is exact

• The number of roots of a polynomial is less than or equal to the degree of P(x)

• If a is a root of P(x), then a has to be a divisor of the constant term of P(x)

16

Examples

• Is 3 a root of 𝑃 𝑥 = 3𝑥2 − 5𝑥 + 1? No, because 𝑃 3 = 3 · 32 −5 · 3 + 1 = 13 ≠ 0

• Is 3 a root of 𝑃 𝑥 = 3𝑥2 − 8𝑥 − 3? Yes, because 𝑃 3 = 3 · 32 −8 · 3 − 3 = 0

• What is the remainder of (3𝑥2 − 8𝑥 − 3):(x-3)?

The remainder is 0 because 3 is a root of 3𝑥2 − 8𝑥 − 3

17

EXERCISES

• Pag 29: 11, 12, 13

EXERCISE: True of false?

a) The division 3𝑥3 − 2𝑥2 + 1: 𝑥 − 1 is exact.

b) The polynomial P x = 3𝑥3 − 2𝑥2 + 1 has 4 roots.

c) 3 is a possible root of 3𝑥4 − 2𝑥2 + 12

d) If P(-2)=0, then the remainder of P(x):(x+2) is 0.

18

Factorization of a polynomial

• The goal is to write the polynomial as a product of polynomials with the lowest possible degree.

• Examples:

a) 𝑥2 − 6𝑥 + 9 = 𝑥 − 3 · 𝑥 − 3

b) 2𝑥4 − 6𝑥3 − 26𝑥2 + 30𝑥 =

=2𝑥(𝑥 − 1)(𝑥 + 3)(𝑥 − 5)

19

How can we do that?

• There are several ways but you can follow the steps: STEP 1: Extract common factors. Now we are going to work with the polynomial inside the parentheses. (If x is a common factor, then 0 is a root) STEP 2: The degree of the polynomial inside the parentheses is 1. Great!, you have finished. GO TO STEP 5. STEP 3: The degree of the polynomial inside the parentheses is 2. Solve the equation polynomial=0 . You will obtain 2 solutions=two roots of the polynomial. Save them. You have finished. GO TO STEP 5. If the equation has no solution, you have finished, GO TO STEP 5. (In this step you can also use the important equalities)

20

STEP 4: The degree of the polynomial inside the parentheses is greater than or equal to 3. You have to use the Ruffini’s rule:

– The possible roots are the divisors of the constant term. – When you obtain REMAINDER=0 you have found a root. Save

it. – Dividend = Divisor·Quotient (+remainder=0). Working with

the quotient GO TO STEP 2. – If you don’t find any root, you have finished. GO TO STEP 5.

STEP 5: The factorized polynomial is: a(x-root_1)(x-root_2)….(x-root_n)(maybe a polynomial that you

cannot factorize) VERY IMPORTANT. YOU HAVE TO CHECK WHETER THE LEADING COEFFICIENT IS THE SAME THAT THE ORIGINAL ONE. If THIS IS NOT THE CASE, YOU JUST HAVE TO MULTIPLY THE FACTORIZED EXPRESSION BY THE ORIGINAL LEADING COEFFICIENT.

21

EXAMPLES

• 3𝑥2 − 15𝑥

STEP 1: 3x·(x-5)

STEP 2: The degree of (x-5) is 1, so you have finished.

STEP 5 The leading term of 3𝑥2 − 15𝑥 is the same as the leading term of 3x·(x-5), so this is the correct result.

22

4𝑥3 − 12𝑥2 + 9𝑥

• STEP 1: 𝑥 4𝑥2 − 12𝑥 + 9 (0 is a root) • STEP 2: 4𝑥2 − 12𝑥 + 9 don’t have degree 1 • STEP 3: We have to solve 4𝑥2 − 12𝑥 + 9 = 0

𝑥 =12 ± −12 2 − 4 · 4 · 9

2 · 4=

𝑥1 =3

2

𝑥2 =3

2

• STEP 5

4𝑥3 − 12𝑥2 + 9𝑥 = 𝟒 · x · x −3

2· 𝑥 −

3

2= 4x 𝑥 −

3

2

2

In step 3 is also possible 𝟒𝒙𝟐 − 𝟏𝟐𝒙 + 𝟗 = 𝟐𝐱 − 𝟑 𝟐 In this case 𝟒𝒙𝟑 − 𝟏𝟐𝒙𝟐 + 𝟗𝒙 = 𝐱 · 𝟐𝐱 − 𝟑 𝟐 (pay attention with the leading coefficient).

23

𝑥3 − 3𝑥2 − 𝑥 + 3

• STEP 1: There is no common factor

• STEP 2: The degree isn’t 1

• STEP 3: The degree isn’t 2

• STEP 4: The degree is 3. Ruffini.The divisors of the constant term are: +1,-1,+3,-3

Let’s begin with 1:

QUOTIENT=𝑥2 − 2𝑥 − 3 REMAINDER=0 Then, 1 is a root of 𝑥3 − 3𝑥2 − 𝑥 + 3. We go to step 2 using the quotient.

24

• We are working with 𝑥2 − 2𝑥 − 3

• STEP 2: The degree isn’t 1

• STEP 3: The degree is 2. We have to solve

𝑥2 − 2𝑥 − 3=0

𝑥 =2 ± −2 2 − 4 · 1 · (−3)

2 · 1=

𝑥1 = −1𝑥2 = 3

The roots are -1 and 3.

• STEP 5: We have the roots: 1, -1 and 3, so:

𝑥3 − 3𝑥2 − 𝑥 + 3=(x-1)(x+1)(x-3)

25

2𝑥4 + 14𝑥3 + 16𝑥2 − 32𝑥

• STEP 1: 2𝑥 𝑥3 + 7𝑥2 + 8𝑥 − 16 It means that 0 is also a root

• STEP 2: NO

• STEP 3: NO

• STEP 4: Ruffini. Possible roots:

+1,-1,2,-2,4,-4,8,-8,16,-16

QUOTIENT: 𝑥2 + 8𝑥 + 16 REMAINDER: 0

1 is a root of 𝑥3 + 7𝑥2 + 8𝑥 − 16

26

• Working with 𝑥2 + 8𝑥 + 16 • Step 2: No • Step 3: We have to solve 𝑥2 + 8𝑥 + 16=0

𝑥 =8 ± 8 2 − 4 · 1 · (16)

2 · 1=

𝑥1 = 4𝑥2 = 4

• Step 5: The roots are 0, 1, 4 (double): 𝟐 𝑥 − 0 𝑥 − 1 (𝑥 − 4)(𝑥 − 4)=

𝟐𝑥(𝑥 − 1)(𝑥 − 4)2 Sometimes it’s clearer to write all the steps:

𝟐𝑥4 + 14𝑥3 + 16𝑥2 − 32𝑥 𝑥 𝑥3 + 7𝑥2 + 8𝑥 − 16

x(x-1)(𝑥2 + 8𝑥 + 16 ) 𝟐𝑥(𝑥 − 1)(𝑥 − 4)2

27

Exercises

• Pag 30: 14

Exercise: Factorize:

28

Algebraic fractions

• Book. Page 30 and 31: 3.6 Algebraic fractions.

• Exercises:

– PAG 30: 15

– PAG 31: 16

29