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MATHS POWERPOINT PRESENTATION Made by – NANDEESH LAXETTI AND VEDANT PARIKH
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Factorising polynomials This PowerPoint presentation demonstrates two methods of factorising a polynomial when you know one factor (perhaps by using the factor theorem). Click here to see factorising by inspection Click here to see factorising using a table
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Factorising by inspection If you divide 2 x ³ - 5 x ² - 4 x – 3 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax ² + bx + c. 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)( ax ² + bx + c )
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Factorising by inspection Imagine multiplying out the brackets. The only way of getting a term in x ³ is by multiplying x by ax ², giving ax ³. So a must be 2. 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)( ax ² + bx + c )
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Factorising by inspection Imagine multiplying out the brackets. The only way of getting a term in x ³ is by multiplying x by ax ², giving ax ³. So a must be 2. 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)( 2 x ² + bx + c )
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Factorising by inspection Now think about the constant term. You can only get a constant term by multiplying –3 by c , giving –3 c . So c must be -1. 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)(2 x ² + bx + c )
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Factorising by inspection Now think about the constant term. You can only get a constant term by multiplying –3 by c , giving –3 c . So c must be -1. 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)(2 x ² + bx - 1 )
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Factorising by inspection Now think about the x ² term. When you multiply out the brackets, you get two x ² terms. -3 multiplied by 2 x ² gives –6 x ² x multiplied by bx gives bx ² So –6 x ² + bx ² = -5 x ² therefore b must be 1. 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)(2 x ² + bx - 1)
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Factorising by inspection Now think about the x ² term. When you multiply out the brackets, you get two x ² terms. -3 multiplied by 2 x ² gives –6 x ² x multiplied by bx gives bx ² So –6 x ² + bx ² = -5 x ² therefore b must be 1. 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)(2 x ² + 1 x - 1)
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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 x gives –3 x x multiplied by –1 gives - x -3 x – x = -4 x as it should be! 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)(2 x ² + x - 1)
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Factorising by inspection 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)(2 x ² + x - 1) Now factorise the quadratic in the usual way. = ( x – 3)(2 x – 1)( x + 1)
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Factorising polynomials Click here to see this example of factorising by inspection again Click here to see factorising using a table Click here to end the presentation
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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: 2 x ³ -6 x ² -8 x 3 x ² -9 x -12 So (2 x + 3)( x ² - 3 x – 4) = 2 x ³ - 3 x ² - 17 x - 12 The method you are going to see now is basically the reverse of this process. 2 x 3 x ² -3 x - 4
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Factorising using a table If you divide 2 x ³ - 5 x ² - 4 x + 3 (cubic) by x – 3 (linear), then the result must be quadratic. Write the quadratic as ax ² + bx + c. x -3 ax ² bx c
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Factorising using a table The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 The only x ³ term appears here, so this must be 2 x ³. 2 x ³ x -3 ax ² bx c
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Factorising using a table This means that a must be 2. The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 x -3 ax ² bx c 2 x ³
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Factorising using a table This means that a must be 2. The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 x -3 2 x ² bx c 2 x ³
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Factorising using a table The constant term, 3, must appear here 3 The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 x -3 2 x ² bx c 2 x ³
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Factorising using a table so c must be –1. 3 The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 x -3 2 x ² bx c 2 x ³
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Factorising using a table so c must be –1. 3 The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 x -3 2 x ² bx -1 2 x ³
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Factorising using a table Two more spaces in the table can now be filled in -6 x ² - x The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 3 x -3 2 x ² bx -1 2 x ³
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Factorising using a table This space must contain an x ² term and to make a total of –5 x ², this must be x ² x ² -6 x ² - x The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 3 x -3 2 x ² bx -1 2 x ³
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Factorising using a table This shows that b must be 1. x ² The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 -6 x ² - x 3 x -3 2 x ² bx -1 2 x ³
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Factorising using a table This shows that b must be 1. x ² The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 -6 x ² - x 3 x -3 2 x ² 1 x -1 2 x ³
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Factorising using a table Now the last space in the table can be filled in The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 x ² -3 x 3 x -3 2 x ² x -1 2 x ³ -6 x ² - x
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Factorising using a table and you can see that the term in x is –4 x , as it should be. So 2 x ³ - 5 x ² - 4 x + 3 = ( x – 3)(2 x ² + x – 1) x ² The result of multiplying out using this table has to be 2 x ³ - 5 x ² - 4 x + 3 -6 x ² - x -3 x 3 x -3 2 x ² x -1 2 x ³
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Factorising by inspection 2 x ³ – 5 x ² – 4 x + 3 = ( x – 3)(2 x ² + x - 1) Now factorise the quadratic in the usual way. = ( x – 3)(2 x – 1)( x + 1)
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Factorising polynomials Click here to see this example of factorising using a table again Click here to see factorising by inspection Click here to end the presentation