# polynomials

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### POLYNOMIALS :

POLYNOMIALS The General Form of Polynomials The general form of polynomials in x term is anxn + an-1xn-1 + an-2xn-2 + ….. + a1x + a0 With : an, an-1, an-2, …….. , a1 is called coefficient. a0 is called constant. Degree of polynomials can be looked of the highest exponent of x of that polynomials.

### Slide 2:

Example ! Find Coefficient of x4 degree of polynomials and constant of it 2X5 - 3X2 – 7X – 4 4X - 2X2 + 3X3 – 9X4 + ½ X6 5X5 - 2X7 + 10X8 – ½ X + X4 - 12 + 7X -2X2 + 6X3 - X4 + 8X5 –3X6

### Slide 3:

Answer : a. Coef. x4 = 0 constant = -4 Degree = 5 Coef. x4 = -9 constant = 0 Degree = 6 coef. x4 = constant = -12 degree = 8 coef. x4 = - constant = - Degree = 6

### Slide 4:

The Value of Polynomials If the polynomials F(x) = anxn + an-1xn-1 + an-2xn-2 + ............. + a1x + a0 So the value of polynomials F(x) for x=a can be Found by : Subtitution Scheme / Horner / Synthetic Division

### Slide 5:

Example : 1. Determine the value of polynomials below : F (x) = 2x3 – 3x2 + 4x + 2 For x = 5 F (x) = x5 – 3x2 – 6x + 2 For x = -2 F (x) = 2x3 -3x2 + 5x - 9 For x = 2 F (x) = x4 + 2x2 +5x -10 For x = -2 F (x) = 8x5 – x2 + 2x + 1 For x = ½ F (x) = x5 -2x2 + 2x + 1 For x = -3 F (x) = x3 -3x + 5 For x = 5

### Slide 6:

a. Substitution F(5) = 2(5)3 – 3(5)2 + 4(5) + 2 = 250 – 75 + 20 +2 = 197 Scheme x3 x2 x1 x0 x=5 2 -3 4 2 10 35 195 + 2 7 39 197 F(5) x Write down the value of coefficient from left to right (higher to smaller exponent of variable. For the highest exponent write down the result and continue with multiply with value of x that fulfilled.

### Slide 7:

b. F (x) = x5 – 3x2 – 6x + 2 For x = -2 x5 x4 x3 x2 x1 x0 -2 1 0 0 -3 -6 2 -2 4 -8 22 -32 1 -2 4 -11 76 -30 c. F (x) = 2x3 -3x2 + 5x - 9 For x = 2 x3 x2 x1 x0 2 2 -3 5 -9 4 2 14 2 1 7 5

### Slide 8:

2. Determine the value of p F(x) = x3+px2-3x+5 Has value 15 for x=2 F(x) = x3+(2p-1)x2+5 Has value 25 for x=2 F(x) = 1+2x+px3-7x4 has value 0 for x=1 F(x) = x5+4x4-6x2+2px+2 has value 6 for x=-2 F(x) = 4x7+5x3-2px2+7p-4 has value 12 for x=-1 F(x) = 7x9+px8+3x5+5x4+7 has value -3 for x=-1

### Slide 9:

Answer : F(x) = x3+px2-3x+5 F(2) = 23+4p-6+5 = 15 4p+7 = 15 4p = 8 p = 2 b. F(x) = x3+(2p-1)x2+5 F(2) = 25 23+(2p-1) 4+5 = 25 8+8p-4+5 = 25 8p = 16 p = 2

### Slide 10:

c. F(x) = 1+2x+px3-7x4 F(1) =0 1+2(1)+p(1)3-7(1)4 = 0 1+2+p-7 = 0 -4+p = 0 p = 4 d. F(x) = x5+4x4-6x2+2px+2 F(-2) = 6 (-2)5+4(-2)4-6(-2)2+2p(-2)+2 = 6 -32+64-24-4p+2 = 6 -4p+10 = 6 p = 1

### Slide 11:

F(x) = 4x7+5x3-2px2+7p-4 F(-1) = 12 4(-1)7 + 5(-1)3 – 2p(-1)2 + 7p – 4 = 12 5p – 13 = 12 5p = 25 p = 5 F(x) = 7x9+px8+3x5+5x4+7 F(-1) = -3 7(-1)9 + p(-1)8 + 3(-1)5 + 5(-1)4 + 7 = -3 P +2 = -3 p = -5

### Slide 12:

The Division of Polynomials If polynomials F(x) is divided by P(x) has Quotient is H(x) and Remainder is R so polynomials F(x) can be written in the form : Notes : P(x) = divisior/pembagi H(x) = Quotient/hasil bagi R = Remainder/sisa F(x) = P(x) H(x) + R

### Slide 13:

The methode of division of polynomials : Long division / porogapit. Scheme / horner / synthetic division.

### Slide 14:

Example : Determine quotient, remainder of polynomials below F(x) = x3 – x – 11 P(x) = (x+1) F(x) = 7x4 – 20x2 + 15x + 2 P(x) = (x+2) F(x) = 3x2 + 5x +9 P(x) = (x - 4) F(x) = 2x2 - x -1 P(x) = (x – 1) F(x) = x4 + 3x3 – 9x2 + 3x – 10 P(x) = (x - ½) F(x) = x4 – 81 P(x) = (x – 3)

### Slide 15:

Answer : Long Division a. x+1

b. x+2

### Slide 17:

Scheme a. x + 1= 0 1 0 -1 -11 x = -1 -1 1 0 1 -1 0 -11 = R x2 x1 x0 H(x) = (x + 1) (x2 – x) b. x = -2 7 0 -20 15 2 -14 28 -16 2 7 -14 8 -1 4 = R x3 x2 x1 x0 H(x) = 7x3 – 14x2 + 8x -1

### Slide 18:

Theorm Remainder If polynomials F(x) = P(x) H(x) + R and P(x) = (x-h), so we will get : F(x) = (x-h). H(x) + R THEORM : if the polynomials F(x) is divided by (x-h), the remainder is F(h) Prove : F(x) = (x-h) . H(x) + R F(h) = (h-h) . H(x) + R = 0 + R F(h) = R

### Slide 19:

Example : Find the remainder of polynomials : a. b. c. d. e.

### Slide 20:

Answer : a. x=3 3 7 -2 9 48 + 3 16 46 = R b. x=2 2 -4 1 2 -1 4 0 2 8 + 2 0 1 4 7 = R

### Slide 21:

c. x=-1 1 5 -4 1 -1 -4 8 + 1 4 -8 9 = R d. 3 -4 7 -3 1 -1 2 + 3 -3 6 -1 = R

### Slide 22:

e. 4 -3 7 -2 -1 -1 1 -2 1 + 4 -4 8 -4 2 = R

### Slide 23:

Division of Polynomials by (ax - b) , a ? 1 Suppose : F (x) = (x - ) . H(x) + R ? f ( ) = S = (x - ) . H (x) + f ( ) = (ax – b ) . + f( ) Conclusion : If polynomials F(x) = (ax-b) , a ? 1 So Quotient = Remainder = f

### Slide 24:

Example : Find H(x) and R a. b. c. d.

### Slide 25:

a. 9 6 4 2 3 3 9 9 7 = R H(x) = = b. 4 6 -2 2 4 4 8 2 = R H(x) = =

### Slide 26:

c. 2 1 0 10 -3 -3 2 -2 3 = R H(x) = = d. -1 1 0 0 16 -16 -1 -1 0 0 0 16 -1 0 0 0 16 0 = R H(x) =

### Slide 27:

Division of Polynomials by (ax2 + bx + c) If divisior of Polynomials ax2 + bx + c can’t be factorized then division of polynomials can be done by long division. If division polynomials ax2 + bx + c can be factorized then division of polynomials can be done by long division or horner.

### Slide 28:

Notes : If division of polynomials by horner so : The remainder of it : R = R2 .P1(x) + R2 R = Remainder of Polynomials R1 = 1st remainder of polynomials R2 = 2nd remainder of polynomials P1(x) = 1st division Quotient = if a ? 1 R = R2 .P1(x) + R2

### Slide 29:

Example : Determine the quotient and remainder of polynomials below : a. b. c. d. e. f.

### Slide 30:

Answer : P(x) = x=2 1 0 -3 1 -2 2 4 2 6 + 1 2 1 3 4 = R1 x=-1 -1 -1 0 1 1 0 3 = R2 H(x) = x2 + x R = R2.P1(x)+R1 = 3(x-2)+4 = 3x-2

### Slide 31:

b. P(x) = (x2+x-1) x2 + x-1

### Slide 32:

c. P(x) = (x2-1) = (x+1) (x-1) 1 0 -4 0 3 x=1 -1 1 3 -3 + 1 -1 -3 3 0 = R1 x=-1 -1 2 1 1 -2 -1 4 = R2 H(x) = x2 – 2x – 1 R = 4 . (x-1) + 0 = 4x - 4

### Slide 33:

d. P(x) = (x2—2x-3) = (x-3) (x+1) 1 0 -3 -14 -10 3 9 18 12 1 3 6 4 2 = R1 -1 -2 -4 1 2 4 0 = R2 H(x) = x2 + 2x + 4 R = 0 . (x-3) + 2 = 2

### Slide 34:

e. P(x) = x2-2x+1 = (x-1) (x-1) 1 -1 -5 -2 -1 1 1 0 -5 -7 1 0 -5 -7 -8 = R1 1 1 1 -4 1 1 -4 -11 = R2 H(x) = x2 + x -4 R = -11 (x-1) – 8 = -11x + 11 -8 = -11x + 3

### Slide 35:

f. P(x) = (2x2-x-1) = (2x+1) (x-1) 2 1 -8 -4 -1 0 4 2 0 -8 0 = R1 1 2 2 2 2 -6 = R2 H(x) = P(x) = -6 + 0 = -6x - 3

### Slide 36:

Factor Theorm Theorm : If F(x) a polynomials then F(h) = 0 If and only if (x-h) is factor of F(x) If F(h) =0 then (x-h) is factor of F(x) Prove: F(x) = P(x) H(x) + R = (x-h) H(x) + R F(h) = 0 = (x-h) H(x) + 0 = (x-h) H(x) Proved (x-h) is factor F(x)

### Slide 37:

If (x-h) is factor of F(x) then F(h) = 0 Prove : F(x) = P(x) H(x) = (x-h) H(x) F(h) = (h-h) H(x) F(h) = 0

### Slide 38:

Example : Prove (x+4) is factor polynomials (x-2) is factor polynomials (x+3) is factor polynomials Determine the value a then Has factor (x+1) Is divisible by (x-1) Find a then polynomials Divisible by (x-2) Divisible by (x-1) Find a and b if polynomials is divisible by

### Slide 39:

a. 2 9 5 3 -4 -4 -8 -4 -4 4 + 2 1 1 -1 0 (Proved) b. 1 5 -2 -24 2 2 14 24 + 1 7 12 0 (Proved) c. 1 -3 -10 24 -3 -3 18 -24 1 -6 8 0 (Proved)

### Slide 40:

2. a. 1 4 -a 4 1 -1 -1 -3 3+a -7-a 1 3 -3-a 7+a -6-a = 0 -a = 6 a= -6 b. 1 4 -a 4 1 1 1 5 5-a 9-a 1 5 5-a 9-a 10-a = 0 a = 10

### Slide 41:

3 a. 2 5 3 a 10 26 5 13 26+a = 0 a = -26 b. 1 5 3 a 5 8 5 8 8+a = 0 a = -8

### Slide 42:

4. P(x) = = (x+3) (x-1) F(-3) = 0 (-3)4 + 2(-3)3 - 7(-3)2 + a(-3) + b = 0 81 – 54 -63 – 3a + b = 0 -36 - 3a + b = 0 F(1) = 0 (1)4 + 2(1)3 – 7(1)2 + a(1) + b = 0 1 + 2 - 7 + a + b = 0 -4 + a + b = 0 a + b = 4 b = 4 +8 b = 12

### Slide 43:

Determined factor of Polynomials Algorithm : If the sum coefficient xodd = the sum of coefficient xeven so x = 1 is the factor of polynomials. If the sum of coefficient xodd = sum of coefficient xeven so x=-1 is the factor of polynomials. If the 1st and 2nd step not fulfill look at factor of constant (by chance). Until find R of it is 0.

### Slide 44:

Example : Determine the factor of polynomials x3 – 11x2 + 30x – 8 = 0 2x3 + 5x2 + x – 2 = 0 10x3 – 19x2 + 9 = 0 Answer : Constant = -8 Factor -8 = 1 2 4 8 4 1 -11 30 -8 4 -28 8 1 -7 2 0 Factor = (x-4) (x2-7x+2)

### Slide 45:

b. 2x3 + 5x2 + x – 2 = 0 coefficient xodd = xeven -1 2 5 1 -2 -2 -3 2 2 3 -2 0 -2 -4 2 2 -1 0 (x+1) (x+2) (2x-1) c. 10x3 – 19x2 + 9 = 0 sum coefficient = 0 1 10 -19 0 9 10 -9 -9 10 -9 -9 0 Factor = (x-1) (10x2 – 9x – 9) = (x-1) (2x-3) (5x+3)

### Slide 46:

Home Work d. x3 + 2x2 – 5x – 6 e. x3 – 8x2 + 19x -12 f. x4 – 6x3 + 12x2 – 10x + 3 g. 2x3 + 7x2 + 2x -3

### Slide 47:

Determined the roots of Polynomials Algorithm to determined the roots of polynomials same with algorithm factor of polynomials.

### Slide 48:

Example : Find the solution set x4 + 3x3 – 5x2 – 3x + 4 = 0 x3 + 2x2 – 5x – 6 = 0 x3 – 9x + 20 – 12 = 0 Answer : x4 + 3x3 – 5x2 – 3x + 4 = 0 1 1 3 -5 -3 4 1 4 -1 -4 1 1 4 -1 -4 0 1 5 4 -1 1 5 4 0 -1 -4 1 4 0

### Slide 49:

(x-1) (x-1) (x+1) (x+4) Hp = b. x3 + 2x2 – 5x – 6 = 0 -1 1 2 -5 -6 -1 -1 6 1 1 -6 0 -3 -3 6 1 -2 0 Hp =

### Slide 50:

c. x3 – 9x + 20 – 12 = 0 1 1 -9 20 -12 1 -8 12 1 -8 12 0 2 2 -12 1 -6 0 Hp =

### Slide 51:

Home Work d. x4 – 6x3 + 12x2 – 10x + 3 = 0 e. 6x3 + 25x2 +2x – 8 -4 is the roots of polynomials 5x3 + 7x2 – 58x – 24 = 0 The roots of polynomials x4 – 8x3 + ax2 – bx + c = 0 Form aritmathic squence with the difference is 2. Find a, b, c