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Division with polynomials

Division with polynomials. from the remainder theorem to the Factor Theorem. The Remainder Theorem. When p(x) is divided by (x-a) …. the remainder is p(a). Given:. p(x) = 2x 3 - 5x 2 + x - 12. The Factor Theorem – example.

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Division with polynomials

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  1. Division with polynomials from the remainder theorem to the Factor Theorem

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

  3. 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

  4. 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

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

  6. Page 95; Ex.G; Q3 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

  7. Page 97; Q2 – AQA 2002 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

  8. Page 97; Q2 – AQA 2002 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

  9. Page 97; Q2 – AQA 2002 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

  10. Page 97; Q2 – AQA 2002 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

  11. Page 97; Q2 – AQA 2002 - extra bit (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

  12. y x x x x -1 2 -4 x -8 Page 97; Q2 – AQA 2002 - extra bit 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

  13. Have a go • Page 93 • Exercise F • Q1, 2, 5, 8

  14. Good News Everybody You have a test on Monday (6th November), covering the following topics ……. • Simultaneous equations • linear and quadratic • mainly using the substitution method • Inequalities • linear and quadratic • Polynomials and functions • division • remainder theorem • factor theorem

  15. Some targeted revision questions(do these as an absolute minimum) • Simultaneous equations • Page 59-60; Exercise C; Q3, Q6 • Inequalities • Page 70; Exercise C; Q2, Q9 • Polynomials • Page 93; Exercise F; Q4 • Page 95; Exercise G; Q1 + Q2

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