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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.

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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

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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

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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

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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

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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.

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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

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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

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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

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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

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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

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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

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

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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)

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Answer : Long Division a. x+1

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b. x+2

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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

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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

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

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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

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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

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

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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

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

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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) = =

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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) =

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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.

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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

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

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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

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

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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

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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

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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

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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

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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)

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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

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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

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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)

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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

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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

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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

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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.

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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)

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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)

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

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

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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

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(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 =

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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 =

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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

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Answer no. 3 :

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If f(x) : (x-1) has remainder = 2 f(x) : (x-2) has remainder = 6 Find remainder of f(x) if f(x) : (x-1) (x-2) Solution : F(x) = P(x) H(x) + R Let R = ax + b F(1) = (x-1) H(x) + ax+b = 2 a + b = 2 F(2) = (x-2) H(x) + ax+b = 6 2a + b = 6

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Home Work 2. If F(x) : (x+1) has remainder = -3 F(x) : (x-1) has remainder = 5 Find remainder of F(x) if F(x) : (x2-1) Is divisible by (x+3). If it is divided by (x-2) it’s remainder is -4. Find a and b and their roots. If F(x) is divided by (x-2) then remainder is 5 F(x) divided by (x-5) then remainder is 11 Find remainder of F(x) if F(x) : (x2-7x+10) If F(x) : (x-1) then remainder is 6 F(x) : (x-3) then remainder is 18 F(x) : (x-4) then remainder is 27 Find remainder of F(x) : (x-1) (x-3) (x-4)