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Remainder and Factor Theorem Polynomials Polynomials Polynomials Combining polynomials Combining polynomials Combining polynomials Combining polynomials.

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Presentation on theme: "Remainder and Factor Theorem Polynomials Polynomials Polynomials Combining polynomials Combining polynomials Combining polynomials Combining polynomials."— Presentation transcript:

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

2 Polynomials An expression that can be written in the form a + bx + cx 2 + dx 3 +ex 4 + …. 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 4x 5 + 13x 3 + 27x is of degree 5

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

4 Function Notation An polynomials function can be written as f(x) = a + bx + cx 2 + dx 3 +ex 4 + …. f(x) means ‘function of x’ instead of y = …. e.g f(x) = 4x 5 + 13x 3 + 27x f(3) means …. – “the value of the function when x=3” – e.g. for f(x) = 4x 5 + 13x 3 + 27x – f(3) = 4 x 3 5 + 13 x 3 3 + 27 x 3 – f(3) = 972 + 351 + 81 = 1404 x f(x)

5 Combining Functions (1) Suppose – f(x) = x 3 + 2x +1 – g(x) = 3x 2 - 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(x 3 + 2x +1) + 3(3x 2 - x - 2) e.g. 3 f(x) g(x) = 3(x 3 + 2x +1)(3x 2 - x - 2)

6 Combining Functions (2) We can define a new function by any linear or multiplicative combination of these… e.g. 2f(x) + 3g(x) = 2(x 3 + 2x +1) + 3(3x 2 - x - 2) = 2x 3 + 4x + 2 + 9x 2 -3x -6 GATHER LIKE TERMS x3x3 x2x2 x = 2x 3 + 9x 2 + x - 4 2 + 9 + - 4

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

8 Finding one bracket given the other Fill in the empty bracket: x 2 - x - 20 = (x + 4)( ) To get x 2 the x must be multiplied by another x x 2 - x - 20 = (x + 4)(x ) To get -20 the +4 must be multiplied by -5 x 2 - x - 20 = (x + 4)(x - 5) Expand it to check: (x + 4)(x - 5) = x 2 + 4x - 5x -20 = x 2 - x - 20

9 We can do division now f(x) = (x 2 - x - 20)  (x + 4) f(x) = (x 2 - x - 20) (x + 4) x (x+4) (x + 4) x f(x) = (x 2 - x - 20) It is exactly the same question as:= Fill in the empty bracket: x 2 - x - 20 = (x + 4)( )

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

11 Finding one bracket given the other - cubics (2.1) Fill in the empty bracket: x 3 + 3x 2 - 12x + 4 = (x - 2)( ) Using a multiplication table: +4 -12x +3x 2 x3x3 x -2 ax 2 +bx +c x3x3 +4 Can put x 3 and +4 in. 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: x 3 + 3x 2 - 12x + 4 = (x - 2)(x 2 +bx -2 ) Using a multiplication table: -12x +3x 2 x -2 x 2 +bx -2 x3x3 +4 -2x -2x 2 Complete more of the table by multiplying known values Complete the rest algebraically +bx 2 -2bx

13 Finding one bracket given the other - cubics (2.3) Fill in the empty bracket: x 3 + 3x 2 - 12x + 4 = (x - 2)(x 2 -2 ) Using a multiplication table: -12x +3x 2 x -2 x 2 +bx -2 x3x3 +4 -2x -2x 2 +bx 2 -2bx +3x 2 = bx 2 - 2x 2 -12x = -2bx - 2x +3 = b - 2 -12 = -2b - 2 Either way, b = 5 +5x Gather like terms

14 Dealing with remainders Fill in the empty bracket: x 3 - x 2 + x + 15 = (x + 2)( ) + R Using a multiplication table: x +2 ax 2 +bx +c x3x3 remainder +x -x 2 +15 x3x3 Can put x 3 This can only come from 1 place. So a = 1

15 Dealing with remainders Fill in the empty bracket: x 3 - x 2 + x + 15 = (x + 2)( ) + R Using a multiplication table: +x -x 2 x +2 x 2 +bx +c x3x3 remainder +15 x3x3 x2 x2 2x 2 We can fill this bit in now The rest of the x 2 term must come from here bx 2 bx 2 + 2x 2 = -x 2 b + 2 = -1 b = -3

16 Dealing with remainders Fill in the empty bracket: x 3 - x 2 + x + 15 = (x + 2)( ) + R Using a multiplication table: +x -x 2 x +2 x 2 -3x +c x3x3 remainder +15 x3x3 x 2 -3x 2x 2 -3x 2 -6x We can fill this bit in now The rest of the x term must come from here cx cx - 6x = +x c - 6 = 1 c = 7

17 Dealing with remainders Fill in the empty bracket: x 3 - x 2 + x + 15 = (x + 2)( ) +R Using a multiplication table: +x -x 2 x +2 x 2 -3x +7 x3x3 remainder +15 x3x3 x 2 – 3x +7 2x 2 -3x 2 -6x 7x +14 We can fill this bit in now - and we’ve got our 2 nd function

18 Dealing with remainders Fill in the empty bracket: x 3 - x 2 + x + 15 = (x + 2)( ) +R Using a multiplication table: +x -x 2 x +2 x 2 -3x +7 x3x3 remainder +15 x3x3 x 2 – 3x +7 2x 2 -3x 2 -6x 7x +14 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: x 3 - x 2 + x + 15 = (x + 2)( ) +1 Using a multiplication table: +x -x 2 x +2 x 2 -3x +7 x3x3 +15 x3x3 x 2 – 3x +7 2x 2 -3x 2 -6x 7x +14 Remainder The numerical term (+15) comes from the +14 and the remainder R +15 = +14 + R So, R = 1

20 Fill in the empty bracket: f(x) = x 3 - x 2 + x + 15 = (x + 2)( ) + R If: f(x) = x 3 - x 2 + x + 15 Division with Remainders What is f(x) divided by x+2 …. and what is the remainder This is exactly the same as ……………

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

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

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

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

25 Given: p(x) = 2x 3 - 5x 2 + x - 12 What value of p(...)=0, hence will give no remainder? The Factor Theorem – example 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 For bigger values of ‘x’ the x 3 term will dominate and make p(x) larger

26 Given: p(x) = 2x 3 - 5x 2 + 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 For a given polynomial p(x) If p(a) = 0 … then (x-a) is a factor of p(x) The Factor Theorem

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

29 If: f(x) = x 3 + 3x 2 - 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 x 3 + 3x 2 - 6x - 8

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

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

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

33 (x-2)(x+4)(x+1) = x 3 + 3x 2 - 6x - 8 ……. Sketch x 3 + 3x 2 - 6x - 8 y = x 3 + 3x 2 - 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 Factor and Remainder Theorem Where does it cross the x-axis (y=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 = -8 x y Goes through these -sketch a nice curve x x -4 x 2 x -8

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