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Factorising polynomials

Factorising polynomials

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Factorising polynomials

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  1. Factorising polynomials This PowerPoint presentation demonstrates three methods of factorising a polynomial when you know one linear factor. Click here to see factorising by inspection Click here to see factorising using a table Click here to see polynomial division

  2. Factorising by inspection If you divide x³ - x² - 4x – 6 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax² + bx + c. x³ – x² – 4x - 6 = (x – 3)(ax² + bx + c)

  3. Factorising by inspection Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³. x³ – x² – 4x - 6 = (x – 3)(ax² + bx + c) So a must be 1.

  4. Factorising by inspection Imagine multiplying out the brackets. The only way of getting a term in x³ is by multiplying x by ax², giving ax³. x³ – x² – 4x - 6 = (x – 3)(1x² + bx + c) So a must be 1.

  5. Factorising by inspection Now think about the constant term. You can only get a constant term by multiplying –3 by c, giving –3c. x³ – x² – 4x - 6 = (x – 3)(x² + bx + c) So c must be 2.

  6. Factorising by inspection Now think about the constant term. You can only get a constant term by multiplying –3 by c, giving –3c. x³ – x² – 4x - 6 = (x – 3)(x² + bx+ 2) So c must be 2.

  7. Factorising by inspection Now think about the x² term. When you multiply out the brackets, you get two x² terms. -3 multiplied by x² gives –3x² x³ – x² – 4x - 6 = (x – 3)(x² + bx + 2) x multiplied by bx gives bx² So –3x² + bx² = -1x² therefore b must be 2.

  8. Factorising by inspection Now think about the x² term. When you multiply out the brackets, you get two x² terms. -3 multiplied by x² gives –3x² x³ – x² – 4x - 6 = (x – 3)(x² +2x + 2) x multiplied by bx gives bx² So –3x² + bx² = -1x² therefore b must be 2.

  9. Factorising by inspection You can check by looking at the x term. When you multiply out the brackets, you get two terms in x. -3 multiplied by 2x gives -6x x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2) x multiplied by 2 gives 2x -6x + 2x = -4x as it should be!

  10. Factorising by inspection Now you can solve the equation by applying the quadratic formula to x²+ 2x + 2 = 0. x³ – x² – 4x - 6 = (x – 3)(x² + 2x + 2) The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.

  11. Factorising polynomials Click here to see this example of factorising by inspection again Click here to see factorising using a table Click here to see polynomial division Click here to end the presentation

  12. x² -3x - 4 2x 3 Factorising using a table If you find factorising by inspection difficult, you may find this method easier. Some people like to multiply out brackets using a table, like this: 2x³ -6x² -8x 3x² -9x -12 So (2x + 3)(x² - 3x – 4) = 2x³ - 3x² - 17x - 12 The method you are going to see now is basically the reverse of this process.

  13. ax² bxc x -3 Factorising using a table If you divide x³ - x² - 4x - 6 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax² + bx + c.

  14. Factorising using a table ax² bxc x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ The only x³ term appears here, so this must be x³.

  15. Factorising using a table ax² bxc x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ This means that a must be 1.

  16. Factorising using a table 1x² bxc x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ This means that a must be 1.

  17. Factorising using a table x² bxc x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ -6 The constant term, -6, must appear here

  18. Factorising using a table x² bxc x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ -6 so c must be 2

  19. Factorising using a table x² bx2 x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ -6 so c must be 2

  20. Factorising using a table x² bx 2 x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ 2x -3x² -6 Two more spaces in the table can now be filled in

  21. Factorising using a table x² bx 2 x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ 2x² 2x -3x² -6 This space must contain an x² term and to make a total of –x², this must be 2x²

  22. Factorising using a table x² bx 2 x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ 2x² 2x -3x² -6 This shows that b must be 2

  23. Factorising using a table x² 2x 2 x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ 2x² 2x -3x² -6 This shows that b must be 2

  24. Factorising using a table x² 2x 2 x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ 2x² 2x -3x² -6x -6 Now the last space in the table can be filled in

  25. Factorising using a table x² 2x 2 x -3 The result of multiplying out using this table has to be x³ - x² - 4x - 6 x³ 2x² 2x -3x² -6x -6 and you can see that the term in x is -4x, as it should be. So x³ - x² - 4x - 6 = (x – 3)(x² + 2x + 2)

  26. Factorising by inspection Now you can solve the equation by applying the quadratic formula to x²- 2x + 2 = 0. x³ – x² – 4x - 6 = (x – 3)(x² - 2x + 2) The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.

  27. Factorising polynomials Click here to see this example of factorising using a table again Click here to see factorising by inspection Click here to see polynomial division Click here to end the presentation

  28. Algebraic long division Divide x³ - x² - 4x - 6 by x - 3 x - 3 is the divisor x³ - x² - 4x - 6 is the dividend The quotient will be here.

  29. Algebraic long division First divide the first term of the dividend, x³, by x (the first term of the divisor). x² This gives x². This will be the first term of the quotient.

  30. Algebraic long division x² Now multiply x² by x - 3 and subtract

  31. Algebraic long division x² - 4x Bring down the next term, -4x

  32. Algebraic long division Now divide 2x², the first term of 2x² - 4x, by x, the first term of the divisor x² + 2x - 4x which gives 2x

  33. Algebraic long division x² + 2x - 4x Multiply 2x by x - 3 2x² - 6x 2x and subtract

  34. Algebraic long division x² + 2x - 6 - 4x Bring down the next term, -6 2x² - 6x 2x

  35. Algebraic long division x² + 2x + 2 Divide 2x, the first term of 2x - 6, by x, the first term of the divisor - 4x 2x² - 6x 2x - 6 which gives 2

  36. Algebraic long division x² + 2x + 2 - 4x 2x² - 6x Multiply x - 3 by 2 2x - 6 Subtracting gives 0 as there is no remainder. 2x - 6 0

  37. Factorising by inspection So x³ – x² – 4x - 6 = (x – 3)(x² - 2x + 2) Now you can solve the equation by applying the quadratic formula to x²- 2x + 2 = 0. The solutions of the equation are x = 3, x = -1 + i, x = -1 – i.

  38. Factorising polynomials Click here to see this example of polynomial division again Click here to see factorising by inspection Click here to see factorising using a table Click here to end the presentation