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Remainder and Factor Theorem

Remainder and Factor Theorem. Polynomials Combining polynomials Function notation Division of Polynomial Remainder Theorem Factor Theorem. Polynomials. An expression that can be written in the form a + bx + cx 2 + dx 3 +ex 4 + ….

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Remainder and Factor Theorem

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  1. Remainder and Factor Theorem Polynomials Combining polynomials Function notation Division of Polynomial Remainder Theorem Factor Theorem

  2. Polynomials • An expression that can be written in the form • a + bx + cx2 + dx3 +ex4 + …. • Things with Surds (e.g. x + 4x +1 ) and reciprocals (e.g. 1/x + x) are not polynomials • The degree is the highest index • e.g 4x5 + 13x3 + 27x is of degree 5

  3. Polynomials can be combined to give new polynomials They can be added: 2x2 - 5x - 3 + 2x4 + 4x2 + 2x Or multiplied: (2x4 + 4x2) (3x2 - 5x - 3) 3x2 -5x -3 Set-up a multiplication table 2x4 +4x2 = 2x4 + 6x2 -3x -3 6x6 -10x5 -6x4 +12x4 -20x3 -12x2 +12x4 6x6 -10x5 -20x3 -12x2 -6x4 Gather like terms = 6x6 - 10x5 + 6x4 - 20x3 -12x2

  4. f(x) x Function Notation • An polynomials function can be written as • f(x) = a + bx + cx2 + dx3 +ex4 + …. • f(x) means ‘function of x’ • instead of y = …. • e.g f(x) = 4x5 + 13x3 + 27x • f(3) means …. • “the value of the function when x=3” • e.g. for f(x) = 4x5 + 13x3 + 27x • f(3) = 4 x 35 + 13 x 33 + 27 x 3 • f(3) = 972 + 351 + 81 = 1404

  5. Combining Functions (1) • Suppose • f(x) = x3 + 2x +1 • g(x) = 3x2 - x - 2 • g(x) or p(x) or q(x) or ….. • Can all be used to define different functions • We can define a new function by any linear or multiplicative combination of these… • e.g. 2f(x) + 3g(x) • = 2(x3 + 2x +1) + 3(3x2 - x - 2) • e.g. 3 f(x) g(x) • = 3(x3 + 2x +1)(3x2 - x - 2)

  6. Combining Functions (2) GATHER LIKE TERMS x2 x3 x • We can define a new function by any linear or multiplicative combination of these… • e.g. 2f(x) + 3g(x) • = 2(x3 + 2x +1) + 3(3x2 - x - 2) • = 2x3 + 4x + 2 + 9x2 -3x -6 + 9 2 + - 4 • = 2x3 + 9x2 + x - 4

  7. x3 +2x +1 3x2 -x -2 Combining Functions (3) • e.g. 3 f(x) g(x) • = 3(x3 + 2x +1)(3x2 - x - 2) Do multiplication table; gather like terms and then multiply through by 3

  8. Finding one bracket given the other To get x2thexmust be multiplied by anotherx To get -20 the+4must be multiplied by -5 Fill in the empty bracket: x2 - x - 20 = (x + 4)( ) x2 - x - 20 = (x + 4)(x ) x2 - x - 20 = (x + 4)(x- 5) Expand it to check: (x + 4)(x- 5) = x2 + 4x - 5x -20 = x2 - x - 20

  9. We can do division now f(x) = (x2 - x - 20) (x + 4) x (x+4) x (x+4) It is exactly the same question as:= Fill in the empty bracket: x2 - x - 20 = (x + 4)( ) f(x) = (x2 - x - 20) (x + 4) (x + 4) x f(x) = (x2 - x - 20)

  10. Finding one bracket given the other - cubics Fill in the empty bracket: x3 + 3x2 - 12x + 4 = (x - 2)( ) To get x3thexmust be multiplied by x2 To get +4 the-2must be multiplied by -2 These 2 give us -2x2, but we need3x2 This x must be multiplied by 5x to give us another 5x2 x3 + 3x2 - 12x + 4 = (x - 2)(x2 ) x3 + 3x2 - 12x + 4 = (x - 2)(x2-2) x3 + 3x2 - 12x + 4 = (x - 2)(x2+5x-2)

  11. Finding one bracket given the other - cubics (2.1) Fill in the empty bracket: x3 + 3x2 - 12x + 4 = (x - 2)( ) ax2 +bx +c x -2 Using a multiplication table: x3 +3x2 -12x +4 x3 +4 Can put x3 and+4in. They can only come from 1 place. So a = 1, c = -2

  12. Finding one bracket given the other - cubics (2.2) Fill in the empty bracket: x3 + 3x2 - 12x + 4 = (x - 2)(x2 +bx -2 ) x2 +bx -2 x -2 +bx2 -2bx Complete more of the table by multiplying known values Complete the rest algebraically Using a multiplication table: +3x2 -12x -2x x3 -2x2 +4

  13. Finding one bracket given the other - cubics (2.3) Fill in the empty bracket: x3 + 3x2 - 12x + 4 = (x - 2)(x2 -2 ) x2 +bx -2 x -2 = bx2 - 2x2 +3x2 = b - 2 +3 -12x = -2bx - 2x -12 = -2b - 2 +5x Using a multiplication table: +3x2 -12x +bx2 -2x x3 -2x2 -2bx +4 Gather like terms Either way, b = 5

  14. Dealing with remainders Fill in the empty bracket: x3 - x2 + x + 15 = (x + 2)( )+ R x3 ax2 +bx +c -x2 x +2 +x +15 remainder Using a multiplication table: x3 Can put x3 This can only come from 1 place. So a = 1

  15. Dealing with remainders Fill in the empty bracket: x3 - x2 + x + 15 = (x + 2)( )+ R x2 +bx +c x +2 bx2 2x2 We can fill this bit in now The rest of the x2 term must come from here x2 remainder Using a multiplication table: x3 -x2 +x +15 x3 bx2 + 2x2 = -x2 b + 2 = -1 b = -3

  16. Dealing with remainders Fill in the empty bracket: x3 - x2 + x + 15 = (x + 2)( )+ R x2 -3x +c x +2 cx -6x We can fill this bit in now The rest of the x term must come from here x2 -3x remainder Using a multiplication table: x3 -x2 +x +15 x3 -3x2 2x2 cx - 6x = +x c - 6 = 1 c = 7

  17. Dealing with remainders Fill in the empty bracket: x3 - x2 + x + 15 = (x + 2)( )+R x2 -3x +7 x +2 +14 We can fill this bit in now - and we’ve got our 2nd function x2 – 3x +7 remainder Using a multiplication table: x3 -x2 +x 7x +15 x3 -3x2 -6x 2x2

  18. Dealing with remainders Fill in the empty bracket: x3 - x2 + x + 15 = (x + 2)( )+R x2 -3x +7 x +2 x2 – 3x +7 remainder Using a multiplication table: x3 -x2 +x 7x +15 x3 -3x2 +14 -6x 2x2 Remainder The numerical term (+15) comes from the +14 and the remainder R +15 = +14 + R So, R = 1

  19. Dealing with remainders Filled in the empty bracket: x3 - x2 + x + 15 = (x + 2)( )+1 x2 -3x +7 x +2 x2 – 3x +7 Using a multiplication table: x3 -x2 +x 7x +15 x3 -3x2 +14 -6x 2x2 Remainder The numerical term (+15) comes from the +14 and the remainder R +15 = +14 + R So, R = 1

  20. Division with Remainders Fill in the empty bracket: If: f(x) = x3 - x2 + x + 15 = (x + 2)( )+ R f(x) = x3 - x2 + x + 15 What is f(x) divided by x+2 …. and what is the remainder This is exactly the same as ……………

  21. We found: If: x3 - x2 + x + 15 = (x + 2)( )+1 f(x) = x3 - x2 + x + 15 x2 – 3x +7 What is f(x) divided by x+2? …. and what is the remainder? What is f(x) divided by x+2? (x2 – 3x +7) …. and what is the remainder? 1

  22. If: f(x) = x3 - x2 + x + 15 = (x + 2)( )+1 Our remainder -2 from (x+2)=0 The Remainder Theorem – example 1 x2 – 3x +7 What is f(x) divided by x+2? (x2 – 3x +7) …. and what is the remainder? 1 If we calculate f(-2) ….. f(-2) = (-2)3 – (-2)2 + -2 + 15 = -8 -4 - 2 +15 = 1

  23. If: p(x) = x3 + 2x2 - 9x + 10= (x - 2)( )+8 Our remainder 2 from (x-2)=0 The Remainder Theorem – example 2 x2 + 4x - 1 What is p(x) divided by x-2? (x2 + 4x - 1) …. and what is the remainder? 8 If we calculate p(2) ….. p(2) = (2)3 + 2(2)2 - 9(2) + 10 = 8 + 8 - 18 + 10 = 8

  24. The Remainder Theorem When p(x) is divided by (x-a) …. the remainder is p(a)

  25. Given: p(x) = 2x3 - 5x2 + x - 12 The Factor Theorem – example For bigger values of ‘x’ the x3 term will dominate and make p(x) larger What value of p(...)=0, hence will give no remainder? If we calculate p(0) = 2(0)3 – 5(0)2 + 0 - 12 = -12 p(1) = 2(1)3 – 5(1)2 + 1 - 12 = 2-5+1-12 = -14 p(2) = 2(2)3 – 5(2)2 + 2 - 12 = 16-20+2-12 = -16 p(3) = 2(3)3 – 5(3)2 + 3 - 12 = 54-45+3-12 = 0 By the Remainder Theorem :- the factor (x-3) gives no remainder

  26. Given: p(x) = 2x3 - 5x2 + x - 12 The Factor Theorem – example p(3) = 2(3)3 – 5(3)2 + 3 - 12 = 54-45+3-12 = 0 By the Remainder Theorem :- the factor (x-3) gives no remainder So (x-3) divides exactly into p(x) ……… (x-3) is a factor

  27. The Factor Theorem For a given polynomial p(x) If p(a) = 0 … then (x-a) is a factor of p(x)

  28. If: p(x) = x3 + bx2 + bx + 5 -2 from (x+2) Our remainder When is p(x) divided by x+2 the remainder is 5 Which theorem? The Remainder Theorem If we calculate p(-2) ….. p(-2) = (-2)3 + b(-2)2 + b(-2) + 5 = -8 + 4b - 2b + 5 = 2b - 3 By the Remainder theorem: 2b - 3 = 5 2b = 8 b = 4

  29. If: f(x) = x3 + 3x2 - 6x - 8 a) Find f(2) f(2) = (2)3 + 3(2)2 - 6(2) - 8 = 8 + 12 - 12 - 8 = 0 b) Use the Factor Theorem to write a factor of f(x) For a given polynomial p(x) If p(a) = 0 … then (x-a) is a factor of p(x) f(2) = 0 …. so (x-2) is a factor of x3 + 3x2 - 6x - 8

  30. If: f(x) = x3 + 3x2 - 6x - 8 b) (x-2) is a factor of x3 + 3x2 - 6x - 8 c) Express f(x) as a product of 3 linear factors .. means (x-a)(x-b)(x-c)=x3 + 3x2 - 6x - 8 We know (x-2)(x-b)(x-c)=x3 + 3x2 - 6x - 8 …. consider (x-2)(ax2+bx+c)=x3 + 3x2 - 6x - 8 a=? a=1 : so x x ax2 = x3 (x-2)(x2+bx+c)=x3 + 3x2 - 6x - 8 c=4 : so -2x4 = -8 c=? (x-2)(x2+bx+4)=x3 + 3x2 - 6x - 8

  31. If: f(x) = x3 + 3x2 - 6x - 8 b) (x-2) is a factor of x3 + 3x2 - 6x - 8 c) Express f(x) as a product of 3 linear factors (x-2)(x2+bx+4)=x3 + 3x2 - 6x - 8 Expand : need only check the x2 or x terms … + bx2 -2x2 + … = … + 3x2 + …. EASIER b - 2 = 3 b=5 Or … - 2bx + 4x + … = … - 6x + …. -2b + 4 = -6 b=5 HARD (x-2)(x2+5x+4)=x3 + 3x2 - 6x - 8

  32. If: f(x) = x3 + 3x2 - 6x - 8 b) (x-2) is a factor of x3 + 3x2 - 6x - 8 c) Express f(x) as a product of 3 linear factors (x-2)(x2+5x+4)=x3 + 3x2 - 6x - 8 (x2+5x+4) = (x+4)(x+1) So, (x-2)(x+4)(x+1) = x3 + 3x2 - 6x - 8 ……. a product of 3 linear factors

  33. (x-2)(x+4)(x+1) = x3 + 3x2 - 6x - 8 ……. Sketch x3 + 3x2 - 6x - 8 y = x3 + 3x2 - 6x - 8 y = (x-2)(x+4)(x+1) Where does it cross the x-axis (y=0) ? (x-2)(x+4)(x+1) = 0 Either (x-2) = 0 x=2 Or (x+4)= 0 x=-4 Or (x+1) = 0 x=-1 Where does it cross the y-axis (x=0) ? y = (0)3 + 3(0)2 - 6(0) - 8 = -8

  34. y x x x x -1 2 -4 x -8 Factor and Remainder Theorem Either (x-2) = 0 x=2 Where does it cross the x-axis (y=0) ? Or (x+4)= 0 x=-4 Or (x+1) = 0 x=-1 y = -8 Where does it cross the y-axis (x=0) ? Goes through these -sketch a nice curve

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