1 / 34

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 + ….

karly-mcintyre
Download Presentation

Remainder and Factor Theorem

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related